Solution Manual for Introduction to Nonlinear Finite Element Analysis Nam-Ho Kim 1st edition

SOLUTION MANUAL

1.6.

Exercises

P1.1 Using Cartesian bases, show that (u Ä v) ⋅ (w Ä x) = (v ⋅ w)u Ä x where u, v, w, and x are rank 1 tensor.

Solution: Using the Cartesian basis, (u Ä v)(w Ä x) = (ui ei Ä v j e j ) ⋅ (wk ek Ä xl el ) .

Since the dot product occurs between adjacent bases, we have

(ui ei Ä v j e j ) ⋅ (wk ek Ä xl el ) = uiv j wk xl (e j ⋅ ek )(ei Ä el ) = uiv j wk xl djk (ei Ä el ) = uiv j w j xl (ei Ä el ) = v j w j (ui ei Ä xl el ) = (v ⋅ w)(u Ä x) In the above equation, we used the following properties: e j ⋅ ek = d jk , wk d jk = w j , and v jw j = v ⋅ w .

P1.2 Any rank 2 tensor T can be decomposed by T = S + W, where S is the symmetric part of T and W is the skew part of T. Let A be a symmetric rank 2 tensor. Show A : W = 0 and A : T = A : S .

Solution: Since A is symmetric and W is skew, we have A : W = AijWij = -AijW ji = -AjiW ji

Since in the above equation, the repeated indices i and j are dummy, the above equation can be rewritten as AijWij = -AijWij = 0

In addition, from the relation T = S + W,

A : T = A : (S + W) = A : S + A : W = A : S

P1.3 For a symmetric rank-two tensor E , using the index notation, show that I : E = E , where I = 21 [dik djl + dil djk ] is a symmetric unit tensor of rank-4.

Solution: Using index notation, the contraction operator can be written as

(I : E)ij = 12 [dik djl + dil djk ]Ekl Since the Kronecker-delta symbol replaces indices, the above equation can be written as

(I : E)ij = 12 [Eij + E ji ] = Eij = (E)ij The symmetric property of E is used.

P1.4 The deviator of a symmetric rank-2 tensor is defined as Adev = A - Am 1 where Am = 13 (A11 + A22 + A33 ) . Find the rank-4 deviatoric identity tensor Idev that satisfies Adev = Idev : A .

Solution: From Problem P1.3, it can be shown that I : A = A . In addition, Am can be written in the tensor notation as Am = 13 1 : A . Therefore, Adev = A - Am 1 and it can be written as

Adev = éêë I - 13 1 Ä 1ùúû : A = Idev : A The last equality defined the rank-4 deviatoric identity tensor Idev .

P1.5 The norm of a rank-2 tensor is defined as A = A : A . Calculate the following derivative ¶ A / ¶A . What is the rank of the derivative?

Solution: From the definition

¶ A ¶A

=

¶ é A 1 (A : A)1/2 ùú = (A : A)-1/2(2A : I) = ê û 2 ¶A ë A

The result is a rank-2 tensor. Note that the property that ¶A / ¶A = I is used.

P1.6 A unit rank-2 tensor in the direction of rank-2 tensor A can be defined as N = A / A . Show that ¶N / ¶A = [I - N Ä N] / A .

Solution: Using chain-rule of differentiation, the unit normal tensor can be differentiated as æ ö æ ¶ A ö÷ ¶N ¶ çç A ÷÷ 1 çç ¶A = ÷= A -AÄ ÷÷÷ ç 2 ç çç ¶A ÷÷ø ¶A ¶A ççè A ÷ø÷÷ ¶ A A è

It is straightforward to show that ¶A / ¶A = I . From Problem 1.5, we have

¶ A ¶A Therefore, we have

=

1 ¶ é A (A : A)1/2 ùú = (A : A)-1/2 (2A) = ê û 2 ¶A ë A

¶N 1 = ( I - N Ä N) ¶A A

P1.7 Through direct calculation of a rank-2 tensor, show that the following identity erst det[A] = eijk Air Ajs Akt is true Solution: In the index notation, (r, s, t) are real indices, while (i, j, k) are dummy indices.

Since (r, s, t) only appears in the permutation symbol, it is enough to show the cases of even and odd permutation. Consider the following case of even permutation: (r, s, t) = (1, 2, 3). In such a case, non-zero components of the right-hand side can be written as eijk Ai 1Aj 2Ak 3 = e123A11A22A33 + e132A11A32A23 +e231A21A32A13 + e213A21A12A33 +e312A31A12A23 + e321A31A22A13

In the above equation, we have e123 = e231 = e312 = 1 and e132 = e213 = e 321 = -1 .

Therefore, the above equation becomes

eijk Ai 1Aj 2Ak 3 = A11(A22A33 - A32A23 ) + A21(A32A13 - A12A33 ) + A31(A12A23 - A22A13 ) which is the definition of det[A ] . By following a similar approach, it can be shown that the odd permutation of (r, s, t) will yield - det[A] .

P1.8

For a vector r = x1e1 + x 2e2 + x 3 e3 and its norm r = r , prove  ⋅ (rr) = 4r .

Solution: From the product rule,

 ⋅ (rr) = r ⋅ r + r  ⋅ r Now consider

(r )i =

¶x j x ¶ ¶ 1 1 ¶x j 1 xj + xj (x j x j )1/2 = (x j x j ) = ( ) = dij x j = i 1/2 ¶x i ¶x i r r 2r ¶x i 2(x k x k ) ¶x i

Therefore,  ⋅ (rr) = r ⋅ r + r  ⋅ r =

This completes the proof.

¶x xi r2 xi + r i = + 3r = 4r r ¶x i r

P1.9 A velocity gradient is decomposed into symmetric and skew parts, v = d + w , where

¶v j ö÷ 1 æ ¶v ÷÷, dij = ççç i + ¶x i ø÷ 2 èç ¶x j

¶v j ö÷ 1 æ ¶v ÷÷ wij = ççç i ¶x i ø÷ 2 èç ¶x j

Show that (a) For a symmetric stress tensor, s : v = s : d . ¶v 1 (b) wij = eijkemnk m ¶x n 2

Solution: (a) From Prob. 1.2, Since stress tensor is symmetric, s : w = 0 . Therefore, it is obvious that s : v = s : w + s : d = s : d . (b) The direct substitution method can be used to show the identity. We will show the case when i = 1, j = 2 . The other cases can also be shown in the same way. Knowing that the permutation symbol becomes zero when indices are repeated, in this case the only nonzero situation happens when k = 3 . For the second permutation symbol, the only non-zero situations are m = 1, n = 2 and m = 2, n = 1 , where the former is even permutation and the latter is odd permutation. Therefore, w12 =

¶v ¶v ö 1 1 æ ¶v e123emn 3 m = çç 1 - 2 ÷÷÷ 2 2 çè ¶x 2 ¶x1 ÷ø ¶x n

Other cases can also be shown in the same way.

P1.10 A symmetric rank four tensor is defined by D = l1 Ä 1 + 2mI where 1 = [dij ] is a unit tensor of rank-two and I = 21 [dik djl + dil djk ] is a symmetric unit tensor of rank-four.

When E is an arbitrary symmetric rank-two tensor, calculate S = D : E in terms of E .

Solution: Using index notation, the contraction can be written as Sij = Dijkl Ekl = éêldij dkl + m(dik djl + dil djk )ùú Ekl ë û

Since the Kronecker-delta symbol replaces indices, the above equation can be simplified as

Sij = Dijkl Ekl = lEkk dij + m(Eij + E ji ) = lEkk dij + 2mEij In the tensor notation, the above relation can be written as

S = D : E = ltr (E)1 + 2mE

P1.11 Using integration by parts, calculate I = ò x cos(x )dx .

Solution: Let u = x and v ¢ = cos(x ) . Then

ò x cos(x )dx = ò uv ¢ dx = uv - ò u ¢v dx = x sin(x ) - ò sin(x )dx = x sin(x ) + cos(x ) + C

P1.12 Using integration by parts, calculate I = ò e x cos(x )dx .

Solution: Let u = cos(x ) and v ¢ = e x . Then

ò e x cos(x )dx = ex cos(x ) + ò ex sin(x )dx Now, to evaluate the second terms on the right-hand side using additional integration by parts with u = sin(x ) and v ¢ = e x , as

ò e x sin(x )dx = e x sin(x ) - ò ex cos(x )dx Therefore, putting these together, we have

ò e x cos(x )dx = e x cos(x ) + ex sin(x ) - ò ex cos(x )dx After rearranging, the original integral can be obtained as 1

ò e x cos(x )dx = 2 (ex cos(x ) + ex sin(x )) + C

P1.13 Calculate the surface integral of the vector function F = xe1 + ye2 over the portion of the surface of the unit sphere, S : x 2 + y 2 + z 2 = 1 , above the xy plane; i.e., z ³ 0.

òS F ⋅ n dS Solution: If we close the surface of integration by adding the portion of the xy plane which spans the hemisphere, we notice that the surface integral of F over the added surface is zero, since

F ⋅ n = F ⋅ (-e3 ) = 0

over this area. Thus, the divergence theorem states that we may calculate the required surface integral of F by evaluating

òS F ⋅ n dS = òòV  ⋅ F dV where V is the volume interior of the hemisphere. Since  ⋅ F = 2 , the result is merely twice the volume of the unit hemisphere, or 4p/3.

P1.14 Evaluate the surface integral of a vector, F = xe1 + ye2 + ze3 , over the closed surface of the cube bounded by the planes, x = 1, y = 1, z = 1 , using the divergence theorem.

òS F ⋅ n dS Solution: Using the divergence theorem and  ⋅ F = 3 ,

òS F ⋅ n dS = òòV  ⋅ F dV = òòV 3 dV = 24

P1.15 Consider a unit-depth (in z-axis) infinitesimal element as shown in the figure.

Using force equilibrium, derive the governing differential equation in two-dimension (equilibrium in x- and y-directions). Assume that a uniform body force, f B = [ f1B , f2B ] , is applied to the infinitesimal element.  22 y  dy 2

 21 y  dy 2

 12 x  dx

y

2

 11 x  dx

x

2

 12 x  dx

 11 x  dx 2

2

 21 y  dy 2

 22 y  dy 2

Solution: Equilibrium in the x–direction yields the following equation:

æ ö÷ æ ö÷ æ ö÷ ÷÷ödy + æçç t çç s çç s çç t dy dx ÷ ÷ ÷dx + f1Bdxdy = 0 dx dx dy dy èç 11 x + 2 ÷ø èç 11 x - 2 ÷ø èç 21 y + 2 ø÷ èç 21 y - 2 ø÷ If the first-order Taylor series expansion is used to represent stresses on the surfaces of the rectangle in terms of stresses at the center, the first two terms in the above equation can be approximated by

æ ö÷ æ ö÷ çç s s dx ÷dy - ç dx ÷dy ç 11 11 x+ ÷ x- ÷ çè èç 2 ø 2 ø æ æ s ¶ ¶s dx ö ¶s dx ö = çç s11 x + 11 ÷÷÷dy - çç s11 x - 11 ÷÷÷dy = 11 dxdy çè çè ¶x 2 ø÷ ¶x 2 ÷ø ¶x Similarly, the next two terms can be approximated by

æ ö÷ æ ö÷ çç t t dy ÷dx - ç dy ÷dx ç 21 21 ÷ y+ ø y- ÷ çè èç 2 2 ø æ æ t ¶ ¶t dy ö ¶t dy ö = çç t21 y + 21 ÷÷÷dx - çç t21 y - 21 ÷÷÷dx = 21 dxdy çè çè ¶y 2 ÷ø ¶y 2 ÷ø ¶y By substituting these two equations into the original equation, we obtain an equilibrium equation in the x–direction as

¶s11 ¶t21 + + f1B = 0 ¶x ¶y Similarly, equilibrium in the y-direction yields the following equation:

¶t12 ¶s22 + + f2B = 0 ¶x ¶y

P1.16 In the above unit-depth (in z-axis) infinitesimal element, show that the stress tensor is symmetric using moment equilibrium.

Solution: Moment equilibrium with respect to the center of the element becomes

æ ö÷ dxdy æ ö÷ dxdy æ ö÷ dxdy æ ö÷ dxdy çç t çç t çç t çç t + =0 dx ÷ dx ÷ dy ÷ dy ÷ 12 12 21 21 x+ ø x- ø y+ ø y- ø ÷ 2 ÷ 2 ÷ 2 ÷ 2 çè èç èç èç 2 2 2 2 If the first-order Taylor series expansion is used to represent stresses on the surfaces of the rectangle in terms of stresses at the center,

t12dxdy - t21dxdy = 0 Thus, the stress tensor is symmetric. The same relation can be shown for 3-D stress tensor.

P1.17 The principal stresses at a point in a body are given by s1 = 4, s2 = 2, s3 = 1 , and the principal directions of the first two principal stresses are given by n(1) = 1 (0,1, -1) and n(2) = 1 (0,1,1) . Determine the state of stress at the point; i.e., 6 2 2 components of stress tensor.

Solution: Since the three principal directions are mutually orthogonal, the third principal direction can be calculated by using the cross-product of the two principal directions, as n(3) = n(1) ´ n(2) = (1, 0, 0)

Since these three principal directions are mutually orthogonal, they can be considered as a basis of coordinate system. In this new coordinate system, the stress tensor will only have diagonal components, which is the same as the three principal stresses. Then, the transformation between the two coordinate systems for a rank-2 tensor can be written as [ s]123 = [Q]T[ s]xyz [Q]

where [Q] = [ n(1) n(2) n(3) ] is the orthogonal transformation matrix between the two coordinate systems. Using the property that the inverse of an orthogonal matrix is the same as the transpose, the reverse relationship can be obtained as [ s]xyz = [Q][ s]123[Q]T

Or,

é 0 ê ê [ s]xyz = ê 1 ê 2 ê- 1 ë 2

0 1 2 1 2

1 ùú éê 4 0 0 ùú éê 0 ú 0 ú êê 0 2 0 úú êê 0 úê úê 0 ú ëê 0 0 1 ûú êë 1 û

1 2 1 2

0

é1 0 - 1 ùú 0 ùú ê 2 1 ú = ê0 3 -1 úú ê 2 ú ú ê 0 -1 3 ú 0 ú ëê ûú û

The last matrix defines all 6 components of stress tensor.

P1.18 Find the principal stresses and the corresponding principal stress directions for the following cases of plane stress: s22 = 0 MPa, s12 = 80 MPa (a) s11 = 40 MPa, (b) s11 = 140 MPa, s22 = 20 MPa, s12 = −60 MPa (c) s11 = −120 MPa, s22 = 50 MPa, s12 = 100 MPa Solution: (a) The stress matrix becomes é sxx ê êt êë xy

é 40 80 ù txy ù ú=ê ú ê 80 0 ú MPa syy úú ê úû û ë

To find the principal stresses, the standard eigen value problem can be written as é s - sI ù { n } = 0 ë û

The above problem will have non-trivial solution when the determinant of the coefficient matrix becomes zero:

sxx - s txy 40 - s 80 = =0 txy syy - s 80 0-s

The equation of the determinant becomes:

( ( 40 - s ) ⋅ -s ) - ( 80 ⋅ 80 ) = s 2 - 40s - 6400 = 0 The above quadratic equation yields two principal stresses, as

s1 = 102.46 MPa and s2 = -62.46 MPa .

To determine the orientation of the first principal stresses, substitute 1 in the original eigen value problem to obtain é 40 - 102.46 ù ïì nx üï ìï 0 üï 80 ê ú ïí ïý = ïí ïý ê 80 0 - 102.46 úú ïï ny ïï ïï 0 ïï êë ûî þ î þ

Since the determinant is zero, two equations are not independent 62.46 ⋅ nx = 80 ⋅ ny

and

80 ⋅ nx = -102.46 ⋅ ny

.

Thus, we can only get the relation between nx and ny. Then using the condition |n| = 1 we obtain ìï nx üï(1) ì ü ïí ïý = ïïí 0.788 ïïý ïï ny ïï ïï 0.615 ïï î þ î þ

To determine the orientation of the second principal stress, substitute 2 in the original eigen value problem to obtain é 40 + 62.46 ù ìï nx üï ìï 0 üï 80 ê ú ïí ïý = ïí ïý ê ú ïn ï ï 0 ï 80 0 62.46 + ëê ûú îï y þï îï þï 102.46 ⋅ nx = -80 ⋅ ny

and

80 ⋅ nx = -62.46 ⋅ ny

.

Using similar procedures as above, the eigen vector of 2 can be obtained as ì nx ï ü(2) ì 0.615 ï ü ï ï ï ï í ï ý =ï í ý ï ï ï ï n 0.788 ï ï ï î yï þ î þ

Note that if n is a principal direction, −n is also a principal direction (b) Repeat the procedure in (a) to obtain s1 = 164.85 MPa

and

s2 = -4.85 MPa

.

ì nx ï ü(1) ü ì nx ï ü(2) ï ì 0.383 ï ü ï ïìï -0.924 ï ï ï ï ï ï ï í ý =í ý í ï ý =ï í ý ï ï ï ï ï ï ï ï n 0.383 n 0.924 y y ï þ ï ï and î ï þ ï ï ï î îï þ î þ

(c) Repeat the procedure in (a) to obtain s1 = 96.24 MPa

and

s2 = -166.24 MPa

.

(2) ì nx ï ü(1) ì 0.420 ïü ï ï ïìï nx ïüï ïì -0.908 ïüï ï ï ï ï í ý =í ý í ý = ïí ý ï ï ïn ï ïï 0.420 ïï n ï 0.908 ïï ï ï î yï þ î þ and ïî y ïþ î þ

Note that for the case of plane stress 3=0 is also a principal stress and the corresponding principal stress direction is given by n(3) =(0,0,1)

P1.19 Determine the principal stresses and their associated directions, when the stress matrix at a point is given by é1 1 1 ù ê ú [ s ] = êê 1 1 2 úú MPa ê1 2 1 ú úû ëê

Solution: Use Eq. (1.50) with the coefficients of I1=3, I2= −3, and I3 = −1, l 3 - 3l 2 - 3l + 1 = 0

By solving the above cubic equation, s1 = 3.73 MPa,

s2 = 0.268 MPa,

s3 = -1.00 MPa

(a) Principal direction corresponding to 1: (1 - 3.7321)nx1 + ny1 + nz1 = 0 nx1 + (1 - 3.7321)ny1 + 2nz1 = 0 nx1 + 2ny1 + (1 - 3.7321)nz1 = 0

Solving the above equations with |n1| = 1 yields n1 = {0.4597, 0.6280, 0.6280}

(b) Principal direction corresponding to 2: (1 - 0.2679)nx1 + ny1 + nz1 = 0 nx2 + (1 - 0.2679)ny2 + 2nz2 = 0 nx2 + 2ny2 + (1 - 0.2679)nz2 = 0

Solving the above equations with |n2| = 1 yields n2 = {0.8881, 0.3251, 0.3251}

(c) Principal direction corresponding to 3:

(1 + 1)nx3 + ny3 + nz3 = 0 nx3 + (1 + 1)ny3 + 2nz3 = 0 nx3 + 2ny3 + (1 + 1)nz3 = 0

Solving the above equations with |n2| = 1 yields n 3 = {0, 0.7071, 0.7071}

P1.20 Let x′y′z′ coordinate system be defined using the three principal directions obtained from Problem P1.19. Determine the transformed stress matrix [σ]x′y′z′ in the new coordinates system.

Solution: The three principal directions in Problem 6 can be used for the coordinate transformation matrix: é (1) n (2) n (3) ù é 0.460 -0.888 0 ùú x x ú ê nx ê ê ú (1) (2) (3) é N ù = ê ny 0.707 úú ny ny ú = êê 0.628 0.325 ë û ê (1) ú ê ú ê nz nz(2) nz(3) ú êë 0.628 0.325 -0.707 úû ë û

To determine the stress components in the new coordinates we use Eq. (1.30): é 3.732 0 0 ùú ê ê ésù é ù é ùé ù .268 0 úú ë û x ¢y ¢z ¢ = ë N û ë s û ë N û = ê 0 ê 0 0 -1 úú ëê û T

Note that the transformed stress matrix is a diagonal matrix with the original principal stresses on the diagonal.

P1.21 The stress-strain relationship for three-dimensional isotropic solid is given as sij = éêK dij dkl + 2m(dik djl - 13 dij dkl )ùú ekl where K is the bulk modulus and m is the shear ë û modulus. In practice, stress and strain are written in the vector forms such that {s} = {s11, s22 , s33 , s12, s23 , s12 }T and {e} = {e11, e22, e33, g12, g23, g12 }T . Then, the stress-strain can be written as {s} = [D]{e} . Write the expression of 6x6 elasticity matrix [ D] in terms of K and m .

Solution: Based on the arrangement of stress and strain vector, the indices are arranged as

éD ê 1111 êD ê 2211 êD [D] = êê 1 3311 ê 2 D1211 ê1 ê 2 D2311 ê1 ê 2 D1311 ë

D1122 D2222 D3322 D1222

1 2 1 2 1 2

D2322 D1322

D1133 D2233 D3333 D1233

1 2 1 2 1 2

D2333 D1333

1 2 1 2 1 2 1 2 1 2 1 2

D1112 D2212 D3312 D1212 D2312 D1312

1 2 1 2 1 2 1 2 1 2 1 2

D1123 D2223

1 2 1 2 1 2 1 2 1 2 1 2

D3323 D1223 D2323 D1323

D1113 ùú D2213 úú D3313 úú D1213 úú ú D2313 ú ú D1313 ú û

Note that the components corresponding to shear strains are divided by two because gij = 2eij . Due to Kronecker-delta symbol, many components are zero. Non-zero components are D1111 = D2222 = D3333 = K + 43 m D1122 = D1133 = D2211 = D2233 = D3311 = D3322 = K - 23 m D1212 = D2323 = D1313 = 2m

Therefore, the elasticity matrix can be written as éK + 4 m K - 2 m K - 2 m 0 0 0 ù 3 3 3 ê ú êK - 2 m K + 4 m K - 2 m 0 0 0 ú 3 3 3 ê ú êK - 2 m K - 2 m K + 4 m 0 0 0 ú ú 3 3 3 [D] = êê 0 0 m 0 0 úú ê 0 ê ú 0 0 0 m 0ú ê 0 ê ú 0 0 0 0 mú ê 0 ë û

P1.22 For steel, the following material data are applicable: Young’s modulus E = 207 GPa and shear modulus G = 80 GPa. For the strain matrix at a point shown below, determine the symmetric 3×3 stress matrix. é 0.003 0 -0.006 ùú ê [e] = êê 0 -0.001 0.003 úú ê -0.006 0.003 0.0015 ú êë úû

Solution: From Eq. (1.81) the elasticity matrix becomes é1 - n n n ê ê n 1-n n ê ê n 1-n n E ê [ D] = ê 0 0 (1 + n )(1 - 2n ) ê 0 ê 0 0 ê 0 ê ê 0 0 0 ë

0 0

0 0

0

0

1 -n 2

0

0

1 -n 2

0

0

ù ú ú ú 0 úú ú 0 ú ú 0 ú ú 1 -nú 2 û 0 0

From the relation G = E / 2(1 + n ) , we calculate n = (E / 2G ) - 1 = 0.294 . ì sxx ï ü ï ïìï 0.003 ïüï ïìï 0.879 ïüï ï ï ï ï ïï ï ï ï ï s ïï -0.001 ïï ïï 0.239 ïï ï ï yy ï ï ïï ï ï ï ï szz ï ï ï ïï 0.0015 ïïï ïïï 0.639 ïïï ï ï í ý = [D] í ý=í ý GPa ï ï ï ï ï ïï t 0.006 0.480 yz ï ï ïï ïï ïï ï ï ï ïï -0.012 ïï ïï -0.960 ïïï txz ï ï ï ï ï ïï ïï ïï ïï ï ï ï ï ï ï ï ï t 0 0 ï xy þ ï î îï þï îï þï

In the matrix notation é 0.879 0 -0.960 ùú ê ésù = ê 0 0.239 0.480 úú GPa ê ë û ê -0.960 0.480 0.639 ú ëê ûú

P1.23 A strain rosette consisting of three strain gages was used to measure the strains at a point in a thin-walled plate. The measured strains in the three gages are: A = 0.001, B = −0.0006, and C = 0.0007. Not that Gage C is at 45o with respect to the x-axis. (a) Determine the complete state of strains and stresses (all six components) at that point. Assume E = 70 GPa, and = 0.3. (b) What are the principal strains and their directions? (c) What are the principal stresses and their directions? (d) Show that the principal strains and stresses satisfy the stress-strain relations.

B

y

C A

x Solution: (a) From figure it is obvious xx = A = 0.001 and yy = B = −0.0006. Shear strain can be found using the strain version of the stress transformation relation in Eq. (1.38). The 2-D version becomes enn = exx nx2 + eyy ny2 + gxy nx ny

where nx = cos(45o) and ny = sin(45o). Thus, eC = enn (45 ) = exx cos2 45 + eyy sin2 45 + gxy sin 45 cos 45 = 0.0007

By solving the above equation, we obtain xy = 0.001. Since the strain rosette only measures plane stress state, zz is unknown. But, there is no shear strain in the zdirection, xz = yz = 0. In order to calculate the unknown stress zz , we use the constitutive relation for plane stress. Since the plate is in a state of plane stress, zz = xz

= yz = 0. Other stresses can be obtained from stress-strain relations for plane stress conditions as shown below: ü ìï 63.1 üï ïìï sx ïüï E éê 1 n ùú ìï ïí ex ï ïý = ïí ïý MPa í ý= ê ú ïï sy ïï 1 - n 2 ê n 1 ú ï ï ï ï 23.1 e î þ ë û îï y þï îï þï txy = G gxy = 13.5 MPa

For plane stress condition the through-the-thickness strain is obtained, as ezz =

-n ( sxx + syy ) = -0.000171 E

(b) For a state of plane stress, zz = −0.000171 is a principal stress and the z-axis (0,0,1) is the corresponding principal strain direction. The other two principal strains can be found from the eigen value problem in 2D strain state: ée - l exy ù ïìï nx ïüï ïïì 0 ïïü úí ý = í ý [e - lI]{n} = êê xx eyy - l úú ïï ny ïï ïï 0 ïï êë exy ûî þ î þ

Two principal strains are calculated from the condition that the determinant of the 2 coefficient matrix is zero: (exx - l)(eyy - l) - exy = 0 . The solution of the quadratic equation becomes 1 = 0.0011 and 2 = −0.0007. Thus, the three principal strains are 1 = 0.0011, 2 = −0.000171, and 3 = −0.0007. Two principal directions can be obtained from the original eigen value problem. Adding z-axis, the three principal directions are ìï -0.961 ü ï ïï ï ï 1 ï n = í -0.276 ï ý, ïï ï ïï 0 ï ï ï î þ

ì ï ï 0ü ï ï ï ï 2 ï n = í0ï ý, ï ï ï ï 1 ï ï ï ï î þ

ì ï ï 0.276 ü ï ï ï ï 3 ï n = í -0.961ï ý ï ï ï ï 0 ï ï ï ï î þ

(c) Principal stresses For plane stress condition, z = 0 is a principal stress and the z-axis (0,0,1) is the corresponding principal direction. The other principal stresses and the directions can be found by solving the following eigen value problem: és - l txy ù ïìï nx ïüï ïïì 0 ïïü ê xx úí ý = í ý é ù ë s - lI û {n} = ê t syy - l úú ïï ny ïï ïï 0 ïï xy êë ûî þ î þ

Two principal stresses are calculated from the condition that the determinant of the 2 coefficient matrix is zero: (sxx - l)(syy - l) - txy = 0 . The solution of the quadratic equation becomes 1 = 70.8 and 2 = −30.8. Thus, the three principal stresses are 1 = 70.8 MPa, 2 = 0.0 MPa, and 3 = −30.8 MPa. Two principal directions can be obtained from the original eigen value problem. Adding z-axis, the three principal directions are ìï -0.961 ü ï ïï ï ï 1 ï n = í -0.276 ï ý, ïï ï ïï 0 ï ï ï î þ

ì ï ï 0ü ï ï ï ï 2 ï n = í0ï ý, ï ï ï 1ï ï ï ï î ï þ

ì ï ï 0.276 ü ï ï ï ï 3 ï n = í -0.961ï ý ï ï ï 0 ï ï ï ï ï î þ

For isotropic materials, principal stress directions and principal strain directions are the same. (d) Principal Stress-strain relations From Eq. (1.55), the stress-strain relation can be written as ìe ï ü é ù ïì s ïü ïì 0.0011 ïü ï 1ï ï ï ê 1 -n -n ú ï ï ï ï 1ï ï ï ï ï 1 íï e2 ýï = êê -n 1 -n úú íï s2 ýï = íï -0.0002 ýï ï ï ï ï E ê -n -n 1 ú ï ïs ï ï ï ï -0.0007 ï ï ï ï êë úû ï ï e3 ï ï ï 3ï ï ï ï ï î þ î þ î þ

Also, all shear strains and stresses are zero because they are in the principal directions.

Thus, the stress-strain relation satisfies in the principal stresses and strains.

P1.24 A rectangular plastic specimen of size 100×100×10 mm3 is placed in a rectangular metal cavity. The dimensions of the cavity are 101×101×9 mm3. The plastic is compressed by a rigid punch until it is completely inside the cavity. Due to Poisson effect, the plastic also expands in the x and y directions and fills the cavity. Calculate all stress and strain components and the force exerted by the punch. Assume there is no friction between all contacting surfaces. The metal cavity is rigid. Elastic constants of the plastic are E = 10 GPa, = 0.3.

Rigid punch

Rigid punch Plastic Rigid die

Plastic F

Rigid die

Solution: The strains in the specimen are calculated as the ratio of change in length to original length. ezz =

( 9 - 10 ) 10

= -0.1, exx = eyy =

( 101 - 100 ) 100

= +0.01

We have assumed that the plastic expands laterally and fill the cavity completely. If it does not, then we will get positive values for xx and/or yy, which will indicate that our assumption is wrong. Then we can assume xx and/or yy = 0, and redo the problem and obtain corresponding strains xx and/or yy which will be less than that calculated above.

Since there is no friction between contacting surfaces, all shear stresses and hence all shear strains will be identically equal to zero.

The normal stresses can be obtained from three-dimensional stress strain relations:

ìs ï ü é (1 - n ) ï xï ï ê ï ï E ï ê n í sy ï ý= ê ï ï + n n 1 1 2 ( )( ) ê n ï ï s ï ï ï zþ ï î ëê

n (1 - n ) n

n ùú ìï ex üï ï ï ï ï ï ú n ú í ey ï ý ï ï ú ï ( 1 - n ) ûú îïï ez þïïï

Substituting for the strains and elastic constants E and  we obtain the stresses as {sxx

syy

sxx } = {-385 -385 -1, 231} MPa

Since xx and yy are negative (compressive), our initial assumption about the strains is correct. The punch force is obtained from z and the area of cross section: F = Asz = 0.1 ´ 0.1 ´ 1, 231 = 12.31 MN

P1.25 Repeat Problem P1.24 with elastic constants of the plastic as E = 10 GPa and  = 0.485.

Solution: The strains in the plastic specimen are calculated as the ratio of change in length to original length. ez =

( 9 - 10 )

10

= -0.1, ex = ey =

( 101 - 100 )

100

= +0.01

We have assumed that the plastic expands laterally and fill the cavity completely. If it does not, then we will get positive values for xx and/or yy, which will indicate that our assumption is wrong. Then we can assume xx and/or yy = 0, and reiterate the problem and obtain corresponding strains xx and/or yy which will be less than that calculated above.

Since there is no friction between contacting surfaces, all shear stresses and hence all shear strains will be identically equal to zero.

The normal stresses can be obtained from three-dimensional stress strain relations: ìs ï ü é (1 - n ) ï xï ï ê ï ï E ï ê n í sy ï ý= ê ï ï + n n 1 1 2 ( )( ) ê n ï ï s ï ï ï zþ ï î ëê

n (1 - n ) n

n ùú ìï ex üï ï ï ï ï ï ú n ú í ey ï ý ï ï ú ï ( 1 - n ) ûú îïï ez þïïï

Substituting for the strains and elastic constants E and  we obtain the stresses as {sxx

syy

sxx } = {-8, 642 -8, 642 -9, 383} MPa

Since xx and yy are negative (compressive), our initial assumption about the strains is correct. The punch force is obtained from zz and the area of cross section: F = Asz = 0.1 ´ 0.1 ´ 9, 383 = 93.83 MN

Note: Punch force for this problem is almost 8 times that for Problem 24. The increase is due to Poisson’s ratio. As the material compressibility decreases, Poisson’s ratio increases. For example, as n  0.5 the material becomes incompressible, i.e., its volume cannot be changed, and the stresses become unbounded. Note the term ( 1 - 2n ) in the denominator of the above constitutive relation.

P1.26 The strain energy and work done by applied load are given in the following equations. When the solution is expressed by u(x) = c1x + c2x2, calculate the solution using the principle of minimum potential energy.

U =

1 1 (u ¢)2 dx , 2 ò0

1

W = ò u dx + u(1) 0

Solution: From the given form of displacement, the virtual displacement can be expressed as u (x ) = c1x + c2x 2 . The variation of the potential energy can be written as 1

1

0 1

0

dP = ò u ¢u ¢ dx - ò u dx - u(1) 1

= ò (c1 + 2c2x )(c1 + 2c2x )dx - ò (c1x + c2x 2 )dx - c1 - c2 =0

0

0

The above variational equation must satisfy for all u (x ) Î  . Since the virtual displacement is expressed by u (x ) = c1x + c2x 2 , it is possible that the above variational equation must satisfy for arbitrary coefficients c1 and c2 . Since c1 and c2 are independent, those terms that contain them must vanish independently; that is, c1(c1 + c2 - 23 ) = 0

c2 (c1 + 43 c2 - 43 ) = 0

By solving the above two equations, we obtain c1 = 2 and c 2 = - 21 . Thus, the solution becomes u(x ) = 2x -

x2 2

P1.27 The governing differential equation for the bar component in the figure is given as -(EA(x )u,1 ),1 = f (x ),

x Î (0, l )

u(0) = 0 u,1(l ) = 0,

where the subscribed comma denotes differentiation with respect to the spatial coordinate, i.e., u,1 = du/dx. Derive the weak form using the principle of virtual work.

x, u E, A(x) f(x) l

x=0

x=l

Solution: The principle of virtual work can be obtained by multiplying the governing differential equation with an arbitrary function u (called virtual displacement) and then integrating over the domain as l

l

l

ò0 u,1u,1 dx - ò0 f u dx = [EAu,1u ] 0 , where integration by parts is used once. The above equation is called the variational identity. Among arbitrary u , let us choose those that satisfy the homogeneous essential boundary condition, that is, u (0) = 0 . Thus, the space of kinematically admissible displacements is defined as  = { u Î H 1(0, l ) | u(0) = 0 },

where H1 is the Sobolev space of the first order. Note that  contains the homogeneous essential boundary condition but not the natural boundary condition. Since the derivative of the solution vanishes at x = l, the following variational equation can be obtained l

l

0

0

ò EAu,1u,1 dx = ò f u dx, for all u in . Note that the above variational problem is well defined for the integrable cross-sectional area A(x) as well as for the continuous displacement function u(x) whose first derivative is in L2(W). Therefore, smoothness requirements for this variational problem are much less than for the classical differential equation.

For the homogeneous boundary condition, the solution space is the same as .

Therefore, the structural energy bilinear and load linear forms are defined as l

a(u, u ) = ò EAu,1u,1 dx 0

and l

(u ) = ò fu dx .

0

Then, the variational equation of the bar component can be represented using the energy bilinear and load linear forms as a(u, u ) = (u ),

"u Î .

Note that a(•,•) is symmetrical with respect to its arguments.

P1.28 Derive the weak form of two-dimensional, steady-state heat transfer problem.

Solution: The governing differential equation of a steady-state heat transfer problem in two-dimension is ¶ æç ¶T ÷ö ¶ æç ¶T ö÷ ÷÷ + ÷ +Q = 0 çç k1 çk ¶x1 è ¶x1 ÷ø ¶x 2 çè 2 ¶x 2 ÷ø÷

 (k1T,1 ),1 + (k2T,2 ),2 + Q = 0

For given boundary conditions, the space of kinematically admissible temperatures becomes Z = {T Î [H 1(W)]2 T (x ) = 0 on ST }

Multiplying the governing equation with a virtual temperature and integrating over the domain, we obtain the following equation:

òòWT [(k1T,1),1 + (k2T,2 ),2 + Q ]dW = 0

"T Î Z

After integrating by parts, we have

òòW [(k1T,1T ),1 + (k2T,2T ),2 ]dW - òòW (k1T,1T,1 + k2T,2T,2 )dW + òòWTQ dW = 0  ò (k1T,1n1T + k2T,2n2T )dG - òò (k1T,1T,1 + k2T,2T,2 )dW + òò TQ dW = 0 S W W  ò (-f1n1 - f2n2 )T dG - òò (k1T,1T,1 + k2T,2T,2 )dW + òò TQ dW = 0 S W W  ò qnT dG - òò (k1T,1T,1 + k2T,2T,2 )dW + òò TQ dW = 0 S W W q

In the above equation, we used the Green-Gauss theorem and the following properties:

f1 = -k1T,1 f2 = -k2T,2 Thus, the weak form becomes

òòW (k1T,1T,1 + k2T,2T,2 )dW = òS qnT dG + òòWTQ dW, q

P1.29 Derive the weak form of simply-supported beam problem.

Solution: The governing differential equation of a beam becomes EIv (4) = f (x )

"T Î Z

where v(4) is the forth-order derivative of deflection, and f (x ) is the distributed load.

Boundary conditions are given as v(0) = v(L ) = 0,

EIv ¢¢(0) = EIv ¢¢(L) = 0

By including homogeneous essential boundary conditions, the space of kinematically admissible displacements becomes Z = { v Î H 2 (0, L) v (0) = v (L) = 0 }

The weak form can be obtained by multiplying the governing equation by a virtual deflection and integrating over the domain, as L

ò0 [EIv(4) - f ]v dx = 0 After performing integration-by-parts twice for the first term, we have L

ò0

L

L

L

EIv ¢¢v ¢¢ dx - ò fv dx + [EIv ¢¢¢v ] 0 - [EIv ¢¢v ¢ ] 0 = 0 0

Since the virtual displacements are zero at the both boundaries and the bending moments are zero at the both boundaries, the boundary terms in the above equation vanish. Thus, the weak form becomes L

ò0

L

EIv ¢¢v ¢¢ dx = ò fv dx , 0

"v Î Z

P1.30 When the potential energy of P1.29 is given, derive the variational equation using the principle of minimum potential energy.

Læ1 ö P = ò çç EI (v,11 )2 - fv ÷÷÷ dx 0 ç è2 ø

Solution: It is clear that the potential energy is well defined as long as u,11 Î L2(0,l) and it does not require u to be C4(0,l), as in the original differential equation. Equating the first variation of  to zero, in which the variation u (x ) has the second-order derivative, u,11 Î L2 (0, l ) , and assuming that u satisfies the essential boundary conditions, we obtain

dP =

ù d l é1 2 ê ú dx t t EI ( u u ) f ( u u ) + + ,11 ,11 úû d t ò0 êë 2 t =0 l

= ò éë EIu,11u,11 - fu ùû dx = 0.

0

In order to make a consistent notation, the following energy bilinear and load linear forms are defined:

l

a(u, u ) = ò EIu,11u,11 dx 0

and l

(u ) = ò fu dx .

0

Then, the variational equation can be written as a(u, u ) = (u ),

"u Î .

(0.1)

which is identical to the weak form obtained using the principle of virtual work.

P1.31 Derive the principle of virtual work for the simply-supported Kirchhoff plate element from the governing equation: [D(u,11 + n u,22 )],11 + [D (u,22 + n u,11 )],22 + 2(1 - n )[Du,12 ],12 = f

Solution: For a simply supported plate, the space of kinematically admissible displacements is  = {u Î [H 2 (W)]2 u = 0 on G }

The principle of virtual work can be obtained by multiplying the governing differential equation with a virtual displacements u and integrate over the domain as

òòW {[D(u,11 + nu,22 )],11 + [D(u,22 + nu,11)],22 + 2(1 - n )[Du,12 ],12 } u dW = òòW fu dW Applying the integration by parts once and using Green-Gauss theorem, we have

òG {[D(u,11 + nu,22 )],1un1 + [D(u,22 + nu,11)],2 un2 + 2(1 - n )[Du,12 ],1un2 } dG -òò {[D(u,11 + nu,22 )],1u,1 + [D(u,22 + nu,11 )],2 u,2 + 2(1 - n )[Du,12 ],1u,2 } dW = òò fu dW W W In the above equation, the first boundary integral term becomes

òG {[D(u,11 + nu,22 )],1un1 + [D(u,22 + nu,11)],2 un2 + 2(1 - n )[Du,12 ],1un2 } dG = -òG uNu dG where Nu is the transverse shear force on the boundary. Applying the integration by parts again and using Green-Gauss theorem, we have -ò uNu dG - ò { D(u,11 + nu,22 )u,1n1 + D(u,22 + nu,11 )u,2n2 + 2(1 - n )Du,12u,2n1 } dG G

G

+òò { D(u,11 + nu,22 )u,11 + D(u,22 + nu,11 )u,22 + 2(1 - n )Du,12u,12 } dW = òò fu dW W

where the second boundary integral becomes

W

¶u Mu dG G ¶n

-ò { D(u,11 + nu,22 )u,1n1 + D(u,22 + nu,11 )u,2n2 + 2(1 - n )Du,12u,2n1 } dG = -ò G

In addition, the domain integrals in the above equation can be converted to ¶u

òòW k(u )T Cb k(u)dW = òòW uf dW + òG uNu dG + òG ¶n Mu dG For the simply supported plate, u = Mu = 0 on the boundary. Thus, the variational equation becomes

òòW k(u )T Cb k(u)dW = òòW uf dW P1.32 Consider a bar element as shown in the figure. The cross sectional areas are A1 and A2 at Nodes 1 and 2, respectively, and vary linearly. In addition, the gravitational acceleration is applied along the axial direction of the bar, such that the distributed load per unit length is f (x ) = rgA(x ) , where r is the density and g is gravitational acceleration. Construct the discrete variational equation for the element. x, u1

u2

E, A(x)

1



2

f(x)





L (a) Finite Element

(b) Reference Element

Solution: The discrete variational equation of a bar element is L é L ù dT ê ò EA(x )BT B dx ú d = dT ò NT f (x )dx 0 ëê 0 ûú

In the problem statement, both cross-sectional area A(x ) and distributed load f (x ) are not constant. Since f (x ) is also a function of A(x ) , it is necessary to integrate the area over the length of the element. Note that the area varies linearly between A1 at Node 1 and A2 at Node 2. Or, it can be considered that the cross-sectional area is interpolated using shape function, as A(x ) = N 1 (x )A1 + N 2 (x )A2

Then, the integral of the cross-sectional area will be L

1

L

ò0 A(x )dx = ò-1 (N 1(x)A1 + N 2 (x)A2 ) 2 dx =

(A1 + A2 )L 2

Also, for the distributed load term, we need to calculate the following integrals:

L

L

1

2

L

1

ò0 N1(x)A(x )dx = ò-1 N1(x)(N1(x)A1 + N 2(x)A2 ) 2 dx = (3 A1 + 3 A2 ) 2

1 L 1 2 L N ( x ) A ( x )d x = 2 ò0 ò-1 N 2 (x)(N1(x)A1 + N 2(x)A2 ) 2 dx = (3 A1 + 3 A2 ) 2 L

T

T

Then, the above discrete variational equation becomes d kd = d f , where the element stiffness matrix and nodal force vector are defined as

k=

E (A1 + A2 ) éê 1 -1 ùú ê -1 1 ú 2L êë úû

f =

ì 2A1 + A2 ï ü rgL ï ï ï í ý 6 ï ï A1 + 2A2 ï ï î þ

P1.33 For the Euler beam element shown in the figure, derive interpolation functions N I (x ) , stiffness matrix k , nodal force vector f . Assume uniformly distributed load f (x ) = f . Note that the reference element is defined in the domain x = [-1, 1] . u1

f(x)

u2

1

2

1

E, , I, A



2





l x2

x1 (a) Finite Element

(b) Reference Element

Solution: The mapping relation between the physical and reference elements is

x = 21 (1 - x)x1 + 21 (1 + x)x 2 Thus, the Jacobian becomes J =

dx L = 21 (x 2 - x 1 ) = dx 2

Since the Euler beam element has four DOFs, the transverse deflection can be assumed as u(x ) = a 0 + a1x + a 2 x 2 + a 3 x 3

Accordingly, the rotational angle can be assumed as q(x ) =

du(x ) du(x ) dx 2 = = (a1 + 2a2 x + 3a 3 x 2 ) dx dx dx L

Now, we want to express the four coefficients in terms of four DOFs, by

u1 = u(-1) = a 0 - a1 + a2 - a 3 u2 = u(1) = a 0 + a1 + a2 + a 3 du(-1) 2 = (a1 - 2a2 + 3a 3 ) L dx du(1) 2 = (a1 + 2a2 + 3a 3 ) q2 = L dx q1 =

By solving the above equation for four coefficients, we have u(x ) = 14 (2 - 3x + x 3 )u1 + L8 (1 - x - x 2 + x 3 )q1

+ 14 (2 + 3x - x 3 )u2 + L8 (-1 - x + x 2 + x 3 )q2

Thus,

N 1(x ) = 14 (2 - 3x + x 3 )

N 2 (x ) = L8 (1 - x - x 2 + x 3 )

N 3 (x ) = 14 (2 + 3x - x 3 )

N 4 (x ) = L8 (-1 - x + x 2 + x 3 ) The displacement-strain relation for the Euler beam element becomes d2u dx

2

=

2 d2u æç 2 ö÷ 4 3 ç ÷÷ = éëê 2 x 2ç L dx è L ø

L (-2 + 6x ) 8

- 23 x

L (2 + 6x ) ù d = Bd 8 ûú

Then, the element stiffness matrix becomes

é 12 6L -12 6L ù ê ú ê 6L 4L2 -6L 2L2 ú 1 EI ê ú k = ò EIBT BJ dx = ê ú 3 -1 L ê -12 -6L 12 -6L ú ê ú êë 6L 2L2 -6L 4L2 úû For the distributed load, the nodal force vector becomes

ì fL / 2 ü ï ï ï ï ï ï 2 ï ï 1 fL / 12 ï ï ï f = ò NT fJ dx = ï í ý -1 ï ï fL / 2 ï ï ï ï 2 ï ï fL / 12 ï ï ï ï î þ Note that the element stiffness matrix and the force vector are the same with the case when the reference element domain x = [0, 1] is used.

P1.34 Below is the governing differential equation of one-dimensional bar under uniformly distributed load. Using one bar element, calculate displacement at x = L and

x = 21 L . Compare these displacements with that of exact calculation. (Note: exact solution can be calculated by integrating the differential equation twice).

-EAu,11 = f ,

x Î (0, L)

u(0) = 0 u,1(L) = 0

Solution: From the textbook, the finite element equation becomes ìu ï ü ì fL / 2ï ü EA éê 1 -1ùú ï ï 1ï = ï ï ï í ý í ý ï ï ï / 2 u fL L êê-1 1 úú ï 2 ï ï ï ï ë ûî þ î þ

Since Node 1 is fixed, the first row and the first column can be deleted, yielding the following displacement at Node 2

u2 = u(L) =

fL2 2EA

Since u1 = 0 , the displacement at x = 21 L can be calculated by

u( 12 ) = N 1( 12 )u1 + N 2 ( 21 )u2 =

fL2 4EA

The exact displacement can be obtained by integrating the differential equation by twice and applying two boundary conditions as u(x ) =

f x (L - 21 x ) EA

Therefore, the displacements at the end and at the center are

fL2 u(L) = 2EA u( L2 ) =

3 fL2 8EA

Note that the displacement at the tip is exact, but that at the center is different. This happens because the finite element method uses linear interpolation, while the exact displacement is a quadratic function.

P1.35 An Euler beam element shown in the figure is under uniformly distributed couple C . Calculate equivalent nodal forces. Using a simply-supported beam under uniform couple, show that the reaction forces are equal and opposite directions with the equivalent nodal forces.

C 1

2 L

x1

x2

Solution: The loading form becomes L

(u ) = ò qC dx 0

By using the mapping relation and interpolation, the load form becomes T

æ dN ö÷ ÷÷ CJ dx = d T f (u ) = d ò ççç 0 è dx ø ÷ 1

T

where f is the equivalent nodal force. The derivatives of interpolation function becomes ïìï -6x + 6x 2 ïüï ïï ï T 2 ï æ dN ö÷ + L (1 4 x 3 x ) ï ïï 1 ï çç ÷ ý çè dx ÷÷ø = L íï 6x - 6x 2 ïï ïï 2 ïï L(-2x + 3x ) ïïï ïî ïþ

By integrating the derivatives of interpolation function, the equivalent nodal forces can be obtained as

{

f T = -C

0 C

}

0

The following figure illustrates the equivalent nodal forces.

C 1

2 C

C

If simply-supported boundary conditions are given at Nodes 1 and 2, the reaction forces can be calculated using the equilibrium of moments at Nodes 1 and 2, respectively:

åM åM

Node 1

= CL + R2L = 0



R2 = -C

Node 2

= CL - R1L = 0



R1 = C

Therefore, the reactions are equal and opposite directions with the equivalent nodal forces.

P1.36 Integrate the following function using one–point and two–point numerical

integration (Gauss quadrature). Explain how to integrate it. The exact integral is equal to 2. Compare the accuracy of the numerical integration with the exact one. p

I = ò sin(x ) dx 0

Solution: NG 1 2

Integration Points (si) 0.0 .57735

Weights (wi) 2.0 1.0

Exact for polynomial of degree 1 3

Since the numerical integration must be between the bounds [-1,1], a change of variable is needed.

x = as + b p = a +b p b= 2

0 = -a + b p a= 2

p (s + 1) 2 p dx = ds 2

x=

æp ö p sin ççç (s + 1)÷÷÷ds 2 è2 ø÷ -1

p

1

I = ò sin(x )dx = ò 0

f (s ) =

æp ö p sin ççç (s + 1)÷÷÷ 2 è2 ø÷

One Point Integration: s = 0, w = 2 I » 2 f (0) = 2

æpö p sin ççç ÷÷÷ = p = 3.1415 2 è 2 ø÷

Error = π-2 = 1.1415 Two Point Integration: s =  .57735, w = 1 p p I » 1f (-.57735) + 1f (.57735) = 1 (.6162) + 1 (.6162) = 1.9358 2 2

Error = 2 – 1.9358 = .0642

2.6.

Exercise

P2.1 Find the roots of the following nonlinear vector equations using the NewtonRaphson method: ìïu + u üï ï ì3ï ü 2ï ï ï ïºf.

P(u) º í 21 = ý í ý ï ïï9ï u1 + u22 ï ï ï þ ï ï î ï î þ

Use the initial estimate u0 = {1, 5}T and convergence tolerance = 10-5. Discuss the convergence rate.

Solution: The Jacobian matrix and the residual term are defined first as KT =

1 ùú ¶P éê 1 =ê ú.

2 2 u u ¶u 2 úû ëê 1

ì ï 3 - u1 - u2 ü ï ï R = f - P(u) = ï í 2 2ý . ï ï 9 u u 1 2ï ï ï ï î þ

Iteration 0; ïì1ïü u 0 = ïí ïý ïï5ïï î þ

é1 1 ù ú KT0 = êê ú 2 10 êë úû

ü ïì -3 ï ï R0 = ï í ý ï ïî-17ï ïþ

conv = 3.25

Iteration 1; ïì-1.625ïüï Du 0 = ïí ý ïîï-1.375ïþï

ü ïì-0.625ï ï u1 = ïí ý ï ïîï 3.625 ï þ

é 1 1 úù KT0 = êê ú êë-1.25 7.25úû

ìï 5.266 üï ï R0 = ï í ý ï-4.531ïï ï î þ

ìï-0.092üï ï u1 = ïí ý ï 3.092 ïï îï þ

é 1 1 úù KT0 = êê ú êë-0.184 6.184úû

ì ï 0.0 üïï R0 = ï í ý ïï-0.568þïï î

ìï-0.003ü ï ï u1 = ïí ý ï ï 3.0003 ïî ï þ

é 1 1 úù KT0 = êê ú êë-0.0053 6.005úû

ïì 0.0 ïïü R0 = ï í ý ïîï-0.016ïþï

Iteration 2; ìï 0.533 üï ï Du 0 = ïí ý ï -0.533ïï îï þ

Iteration 3; ìï 0.089 üï ï Du 0 = ïí ý ï 0.089 ïî ïþï

Iteration 4; ïì 0.003 ïüï Du 0 = ïí ý ïîï-0.003ïþï

ïì-0.0ïüï u1 = ïí ý ïîï 3.0 ïþï

é 1 1 úù KT0 = êê ú êë0.0 6.0úû

ü ïì 0.0 ï ï R 0 = ïí ý ïîï-0.00001ïþï

3‐2

Below is the list of MATLAB programs that solve the nonlinear equations.

% % P2.1 Newton-Raphson method % tol = 1.0e-5; iter = 0; c = 0; u = [1; 5]; uold = u; f = [3; 9]; P = [u(1)+u(2); u(1)^2+u(2)^2]; R = f - P; conv= (R(1)^2+R(2)^2)/(1+f(1)^2+f(2)^2); fprintf('\n iter u1 u2 conv c'); fprintf('\n %3d %8.5f %8.5f %12.3e %7.5f',iter,u(1),u(2),conv,c); while conv > tol && iter < 20 Kt = [1 1; 2*u(1) 2*u(2)]; delu = Kt\R; u = uold + delu; P = [u(1)+u(2); u(1)^2+u(2)^2]; R = f - P; conv= (R(1)^2+R(2)^2)/(1+f(1)^2+f(2)^2); c = abs(3-u(2))/abs(3-uold(2))^2; iter = iter + 1; uold = u; fprintf('\n %3d %8.5f %8.5f %12.3e %7.5f',iter,u(1),u(2),conv,c); end

The following table shows the convergence history of the Newton-Raphson method. Note that the residual reduction is approximately quadratic, and the parameter c converges to a constant. Thus, the convergence rate is two.

Iteration 0 1 2 3 4 5

u1 1.0000 -0.6250 -0.0919 -0.0027 -0.0000 -0.0000

u2 5.0000 3.6250 3.0919 3.0027 3.0000 3.0000

conv 3.275E+00 2.256E-01 3.550E-03 2.790E-06 2.171E-12 1.324E-24

c 0.000 0.156 0.235 0.314 0.333 0.333

P2.2 Using the modified Newton-Raphson method, solve the nonlinear equations in P2.1. Compare the convergence rate with the Newton-Raphson method.

Solution: In the modified Newton-Raphson method, a constant Jacobian matrix is used for all iterations. Below is the list of MATLAB programs for solving the problem. % % P2.2 Modified Newton-Raphson method % tol = 1.0e-5; iter = 0; c = 0; u = [1; 5]; uold = u; f = [3; 9]; P = [u(1)+u(2); u(1)^2+u(2)^2];

3‐3 R = f - P; conv= (R(1)^2+R(2)^2)/(1+f(1)^2+f(2)^2); Kt = [1 1; 2*u(1) 2*u(2)]; fprintf('\n iter u1 u2 conv c'); fprintf('\n %3d %8.5f %8.5f %12.3e %7.5f',iter,u(1),u(2),conv,c); while conv > tol && iter < 20 delu = Kt\R; u = uold + delu; P = [u(1)+u(2); u(1)^2+u(2)^2]; R = f - P; conv= (R(1)^2+R(2)^2)/(1+f(1)^2+f(2)^2); c = abs(3-u(2))/abs(3-uold(2))^2; iter = iter + 1; uold = u; fprintf('\n %3d %8.5f %8.5f %12.3e %7.5f',iter,u(1),u(2),conv,c); end fprintf('\n');

The following table shows the convergence history of the modified Newton-Raphson method. Note that the rate of residual reduction is slower than that of the standard Newton-Raphson method, and the parameter c does not converges to a constant.

Iteration 0 1 2 3 4 5

u1 1.0000 -0.6250 -0.0586 -0.0138 -0.0034 -0.0009

u2 5.0000 3.6250 3.0586 3.0138 3.0034 3.0009

conv 3.275E+00 2.256E-01 1.412E-03 7.592E-05 4.584E-06 2.840E-07

c 0.000 0.156 0.150 4.017 17.879 73.280

P2.3 Using the Broyden method, solve the nonlinear equations in P2.1. Compare the convergence rate with the Newton-Raphson method.

Solution: The Broyden method uses the exact Jacobian at the first iteration, and progressively update it at every iteration using the increments of solution and increments of residuals.

Below is the list of MATLAB programs that solves the nonlinear equations. % % P2.3 Broyden method % tol = 1.0e-7; iter = 0; c = 0; u = [1; 5]; uold = u; f = [3; 9]; P = [u(1)+u(2); u(1)^2+u(2)^2]; R = P - f; Rold = R; conv= (R(1)^2+R(2)^2)/(1+f(1)^2+f(2)^2); Kt = [1 1; 2*u(1) 2*u(2)]; fprintf('\n iter u1 u2 conv c'); fprintf('\n %3d %8.5f %8.5f %12.3e %7.5f',iter,u(1),u(2),conv,c); while conv > tol && iter < 20 delu = -Kt\R; u = uold + delu; P = [u(1)+u(2); u(1)^2+u(2)^2];

3‐4 R = P - f; conv= (R(1)^2+R(2)^2)/(1+f(1)^2+f(2)^2); c = abs(3-u(2))/abs(3-uold(2))^2; delR = R - Rold; Kt = Kt + (delR-Kt*delu)*delu'/norm(delu)^2; uold = u; Rold = R; iter = iter + 1; fprintf('\n %3d %8.5f %8.5f %12.3e %7.5f',iter,u(1),u(2),conv,c); end fprintf('\n');

The following table shows the convergence history of the Broyden method. Note that the rate of residual reduction is slower than that of the standard Newton-Raphson method, and the parameter c does not converges to a constant. However, the method performs a little better than the modified Newton-Raphson method.

Iteration 0 1 2 3 4

u1 1.0000 -0.6250 -0.0758 -0.0128 -0.0003

u2 5.0000 3.6250 3.0758 3.0128 3.0003

conv 3.275E+00 2.256E-01 2.387E-03 6.531E-05 3.897E-08

c 0.000 0.156 0.194 2.229 1.917

P2.4 Using the incremental force method, solve the equations in P2.1. Use five equalinterval load steps.

Solution: Below is the list of MATLAB programs for solving the problem. % % P2.4 Incremental force method % tol = 1.0e-5; u = [1; 5]; fprintf('\n inc u1 u2 f1 f2'); for i=1:5 f = i*0.2*[3; 9]; P = [u(1)+u(2); u(1)^2+u(2)^2]; R = f - P; conv= (R(1)^2+R(2)^2)/(1+f(1)^2+f(2)^2); iter = 0; while norm(R) > tol && iter < 20 Kt = [1 1; 2*u(1) 2*u(2)]; delu = Kt\R; u = u + delu; P = [u(1)+u(2); u(1)^2+u(2)^2]; R = f - P; conv= (R(1)^2+R(2)^2)/(1+f(1)^2+f(2)^2); iter = iter + 1; end fprintf('\n %3d %8.5f %8.5f %8.5f %8.5f',i,u(1),u(2),f(1),f(2)); end fprintf('\n');

3‐5

The following table shows the convergence history of the incremental force method. Note that the force terms increase regularly as they are inputs, while the displacement terms are irregular.

Increment 1 2 3 4 5

u1 -0.6000 -0.6000 -0.4748 -0.2697 -0.0003

u2 1.2000 1.8000 2.2748 2.6697 3.0000

f1 0.6000 1.2000 1.8000 2.4000 3.0000

f2 1.8000 3.6000 5.4000 7.2000 9.0000

P2.5 Consider a uniform bar with a constant Young’s modulus E = 100MPa, crosssectional area A = 2´10-4m2, and a unit length L = 1m. The applied force F = 10kN is large enough such that the relation between displacement and strain is nonlinear: 2

du 1 æç du ö÷ e(u ) = + ç ÷ dx 2 çè dx ÷÷ø

Using a single two-node bar element, calculate the displacement at the tip and strain of the element. Assume stress-strain relation is linear and constant cross-sectional area and length. Use an increment force method with ten equal force increments.

Hint: The virtual strain can be obtained through variation of the strain as

e(u ) =

1

du du du + dx dx dx

2

F = 10kN

x L = 1m

Solution: For a bar element, the displacement is interpolated by

u(x ) = N 1(x )d1 + N 2 (x )d2 =

x d L 2

Note that the boundary condition, d1 = 0, is applied. From the displacement-strain relation, the strain is interpolated by e=

1 (d22 + 2Ld2 ) 2 2L

In the weak form of the bar requires the virtual strain e(u ) , which can be written in terms of the virtual nodal displacement.

3‐6

e(u ) =

1 (d2 + L)d2 L2

The relation between stress and strain is linear: s = Ee =

E (d22 + 2Ld2 ) 2 2L

The weak form of the nonlinear equation becomes

é L 1 ù E d2 ê ò 2 (d2 + L) 2 (d22 + 2Ld2 )Adx = F ú ê 0 L ú 2L ë û After integration, the nonlinear finite element equation can be obtained as EA 3 (d2 + 3Ld22 + 2L2d2 ) = F 3 2L

Below is the list of MATLAB programs that solves for the above nonlinear equation. % % P2.5 Nonlinear strain bar % fprintf('\n inc F u strain'); tol = 1.e-5; u = 0; for i=1:10 strain = 0.5*u^2 + u; f = 0.1*i; iter = 0; P = u^3+3*u^2+2*u; R = f - P; conv= R^2/(1+f^2); while conv > tol && iter < 20 Kt = 3*u^2+6*u+2; delu = R/Kt; u = u + delu; strain = 0.5*u^2 + u; P = u^3+3*u^2+2*u; R = f - P; conv= R^2/(1+f^2); iter = iter + 1; end fprintf('\n %3d %7.5f %7.5f %7.5f',i,f,u,strain); end

The following table shows the force, displacement, and strain at each force increment.

When displacement is small (e.g., the first increment), the different between displacement and strain is small, while the difference becomes large as displacement increases. This is because of the nonlinear displacement-strain relation.

Increment 1 2 3 4

Force (kN) 1.0 2.0 3.0 4.0

Displacement 0.0467 0.0880 0.1254 0.1597

Strain 0.0478 0.0919 0.1333 0.1725

3‐7

5 6 7 8 9 10

5.0 6.0 7.0 8.0 9.0 10.0

0.1915 0.2222 0.2499 0.2763 0.3013 0.3252

0.2098 0.2469 0.2812 0.3144 0.3467 0.3781

P2.6 Solve Problem P2.5 using the secant method. Do not use the incremental force method. Discuss about the convergence rate.

Solution: For the derivation of nonlinear equation, refer to P2.5. Below is the list of MATLAB programs that solve for the nonlinear equation. % % P2.6 Nonlinear strain bar (secant method) % tol = 1.0e-5; iter = 0; u = 0.0; uold = u; f = 1; P = u^3+3*u^2+2*u; Pold = P; R = f - P; conv= R^2/(1+f^2); strain = 0.5*u^2 + u; fprintf('\n iter u strain conv'); fprintf('\n %3d %7.5f %7.5f %12.3e',iter,u,strain,conv); Ks = 3*u^2+6*u+2; while conv > tol && iter < 20 delu = R/Ks; u = uold + delu; strain = 0.5*u^2 + u; P = u^3+3*u^2+2*u; R = f - P; conv= R^2/(1+f^2); Ks = (P - Pold)/(u - uold); uold = u; Pold = P; iter = iter + 1; fprintf('\n %3d %7.5f %7.5f %12.3e',iter,u,strain,conv); end

The following table shows the convergence history. It converges in the fourth iteration.

The convergence criterion reduces rapidly, but not as fast as the Newton-Raphson method.

Iteration 0 1 2 3 4

Displacement 0.0000 0.5000 0.2667 0.3160 0.3252

Strain 0.0000 0.6250 0.3022 0.3659 0.3781

conv 5.000E-01 3.838E-01 2.746E-02 6.859E-04 2.241E-06

3‐8

P2.7 Consider a uniform bar with cross-sectional area A = 1´10-4m2 and a unit length L = 1m. The bar shows elasto-plastic material behavior as depicted in the figure. The plastic deformation starts at yields stress sY = 400MPa. In the elastic region, the Young’s modulus E = 200GPa, while in the plastic region, the tangent stiffness is ET = 20GPa. When a force F = 50kN is applied at the end, calculate tip displacement and stress of the element using one bar element. Use 10 equal-interval force increments. Plot the force-displacement curve. Assume displacement-strain relation is linear. s ET 1

F

2

E

x L

e

Solution: The discrete weak form of the bar element can be written as L

d T ò BT s A d x = d T F 0

where d = [d1, d2]T is the vector of nodal displacements, d is the vector of virtual nodal displacements, BT = [-1, 1]/L is the displacement-strain matrix, and F = [F1, F2]T is the vector of applied forces. In order to simplify the following steps, the essential boundary condition can be applied in advance; i.e., d1 = d1 = 0 . For simplicity of notation, d = d2 and F = F2 will be used in the following derivations. Then, the above discrete weak form becomes a scalar equation. The residual now becomes

R =F -ò

L sA

L  R = F - s(d )A 0

dx

Note that the residual is nothing but the equilibrium between external and internal forces: P(d) = F. The nonlinearity comes from stress calculation. Initially, the stress increases linearly with strain until it reaches the yield stress. The value of displacement at yield is uY = eY L =

sY L E

= 0.002

Thus, the internal force term is determined based on the magnitude of displacement as ì ï EA ï d if d < dY ïP (d ) = ï L í E A ï ï P (d ) = sY A + T (d - dY ) otherwise ï ï L ï î

The Jacobian relation and stress calculation also have two branches as

3‐9

ì ï EA ï KT (d ) = ï ï L í E A ï ï KT (d ) = T ï ï L ï î

if d < dY otherwise

ì ï E ï s(d ) = d ï ï L í E ï ï s(d ) = sY + T (d - dY ) ï ï L ï î

if d < dY otherwise

Note that actually elasto-plastic material show much more complex behavior that the one explained above. However, when the load is monotonically increasing, the above formulas work. Below is the list of MATLAB program that solves the elasto-plastic bar with ten increments. % % P2.7 Elasto-plastic bar % fprintf('\n inc F u stress'); tol = 1.e-5; u = 0; uY = 0.002; for i=1:10 f = i*5000; iter = 0; if u < uY P = 2E7*u; else P = 2E7*uY + 2E6*(u-uY); end R = f - P; conv= R^2/(1+f^2); while conv > tol && iter < 20 if u < uY Kt = 2E7; else Kt = 2E6; end delu = R/Kt; u = u + delu; if u < uY P = 2E7*u; else P = 2E7*uY + 2E6*(u-uY); end R = f - P; conv= R^2/(1+f^2); iter = iter + 1; end if u < uY stress = 2E11*u; else stress = 4E8 + 2E10*(u-uY); end fprintf('\n %3d %8.1f %7.5f %10.3e',i,f,u,stress); end

The following table and figure show the force-displacement relation at each increment. It is clear that plastic deformation starts at F = 40kN at which stress becomes 400MPa. Since the stress-strain relation is linear in each increment, it only requires one iteration per increment.

Increment 1 2 3 4 5 6

Force (kN) 5.0 10.0 15.0 20.0 25.0 30.0

Displacement 0.00025 0.00050 0.00075 0.00100 0.00125 0.00150

Stress (MPa) 50 100 150 200 250 300

3‐10

7 8 9 10

35.0 40.0 45.0 50.0

0.00175 0.00200 0.00450 0.00700

350 400 450 500

P2.8 Consider three nonlinear springs in the figure. The stiffness of each spring is given by k1 = 500 + 50u, k2 = 200 + 100u, and k3 = 500 + 100u where u is the elongation of the spring. Solve the displacements at Nodes 1 and 2 using the NewtonRaphson method when F = 100. k1 k3 F

k2

u2 u1

Solution: The finite element matrix equation for the three nonlinear springs can be written as ék + k + k -k ù ìïu üï ìï 0 üï 2 3 3 ú ï 1ï ï ï ê 1 í ý=í ý ê ú -k 3 k 3 ú ïïu2 ïï ïïF ïï êë ûî þ î þ

By substituting the stiffness for all spring elements, the following nonlinear equations can be obtained: ì ïï500u12 + 200u1u2 - 100u22 + 1200u1 - 500u2 = 0 í ï 100u12 - 200u1u2 + 100u22 - 500u1 + 500u2 = 100 ï ï î

which is in the form of P(u) = F. The Jacobian matrix can be obtained by differentiating the above relation

3‐11

é100u + 200u + 1200 200u - 200u - 500 ù 1 2 1 2 ú KT = êê ú 200 u 200 u 500 200 u + 200 u + 500 1 2 1 2 êë úû

Below is the list of MATLAB programs that solves for the equilibrium of the three nonlinear springs % % P2.8 Three nonlinear springs % tol = 1.0e-5; iter = 0; u = [0; 0]; uold = u; f = [0; 100]; P = [50*u(1)^2+200*u(1)*u(2)-100*u(2)^2+1200*u(1)-500*u(2) 100*u(1)^2-200*u(1)*u(2)+100*u(2)^2-500*u(1)+500*u(2)]; R = f - P; conv= (R(1)^2+R(2)^2)/(1+f(1)^2+f(2)^2); fprintf('\n iter u1 u2 conv'); fprintf('\n %3d %7.5f %7.5f %12.3e',iter,u(1),u(2),conv); while conv > tol && iter < 20 Kt = [100*u(1)+200*u(2)+1200 200*u(1)-200*u(2)-500 200*u(1)-200*u(2)-500 -200*u(1)+200*u(1)+500]; delu = Kt\R; u = uold + delu; P = [50*u(1)^2+200*u(1)*u(2)-100*u(2)^2+1200*u(1)-500*u(2) 100*u(1)^2-200*u(1)*u(2)+100*u(2)^2-500*u(1)+500*u(2)]; R = f - P; conv= (R(1)^2+R(2)^2)/(1+f(1)^2+f(2)^2); uold = u; iter = iter + 1; fprintf('\n %3d %7.5f %7.5f %12.3e',iter,u(1),u(2),conv); end

The following table shows the convergence history of the Newton-Raphson method. Note that the iteration converges at the third iteration.

Iteration 0 1 2 3

u2 0.0000 0.3429 0.3296 0.3316

u1 0.0000 0.1429 0.1380 0.1388

conv 9.999E-01 1.688E-03 2.722E-05 5.390E-07

P2.9 Consider a uniform bar in the figure. The stress-strain relation and displacementstrain relation are linear. However, the Young’s modulus of the material varies according to the strain.

s = E (u )e(u ),

e(u ) =

du , dx

æ du ö E (u ) = E 0 ççç1 - ÷÷÷ dx ø÷ è

When one element is used to model the bar, formulate the nonlinear equation with the tip displacement being unknown. Solve the tip displacement using the incremental force method with ten equal-interval increments. Use E0 = 1.0GPa, A = 10-4m2, and F =

3‐12

25kN. Plot the force-displacement curve. Test what happens when F = 30kN, and explain why.

1

2

F = 10kN

x L = 1m

Solution: Since Node 1 is fixed for the element, it can be deleted. Thus, the weak form of the nonlinear bar becomes a single DOF equation. By denoting d = d2 and F = F2, the nonlinear equation becomes P(d) = F, where the internal force term is defined by P (d ) = s(d )A = EAe(d ) =

E 0A

(e(d ) - e(d ) ) L 2

Note that since the length of the element is a unit and since a linear bar element is used, displacement is identical to strain. Due to strain-dependent material properties, the Newton-Raphson method is used to find the displacement d. The Jacobian becomes

KT =

EA dP = 0 (1 - 2d ) dd L

The first derivative on the right-hand side can be calculated by differentiating the stressstrain relation, and the second derivative from displacement-strain relation. Using these relations, we have KT =

E A dP = 0 (2d + 1) dd L

Below is the list of MATLAB programs that solves for the nonlinear modulus bar. % % P2.9 Nonlinear modulus bar % fprintf('\n inc iter F tol = 1.e-5; u = 0; for i=1:10 f = i*2500; iter = 0; stress = 1E9*(1-u)*u; P = stress*1E-4; R = f - P; conv= R^2/(1+f^2); while conv > tol && iter < 20 Kt = 1E5*(1-2*u); delu = R/Kt; u = u + delu; stress = 1E9*(1-u)*u; P = stress*1E-4; R = f - P;

u

stress');

3‐13 conv= R^2/(1+f^2); iter = iter + 1; end fprintf('\n %3d %3d end

%7.1f %7.5f %12.3e',i,iter,f,u,stress);

The following table and figure show the force-displacement relation during the force increments. Each increment converges in the second iteration. Note that the slope of the force-displacement curve becomes zero as the force approaches 25kN. In fact, the force will decrease if displacement further increases. However, this cannot be solved using Newton-Raphson method as the tangent stiffness becomes singular.

Increment 1 2 3 4 5 6 7 8 9 10

Force (kN) 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0 22.5 25.0

Displacement 0.0257 0.0528 0.0817 0.1127 0.1464 0.1838 0.2261 0.2764 0.3416 0.4802

Stress (MPa) 25.0 50.0 75.0 100.0 125.0 150.0 175.0 200.0 224.9 249.6

3.9.

Exercise

P3.1

Derive the expression of the Eulerian strain in Eq. (3.17).

Solution: From the definistion of the Eulerian strain, it is necessary to define the inverse of the deformation gradient in terms of the displacement gradient. From the relation X = x u,

F-1 =

¶X ¶(x - u) = = 1 - n u ¶x ¶x

Then, the Eulerian strain in Eq. (4.17) can be written as e = 21 (1 - F-T F-1 )

(

)

= 21 1 - (1 - n u)T (1 - n u) T

T

= (n u + n u - n u n u) 1 2

▄

P3.2 Derive the relation in volume change in Eq. (3.26) for an infinitesimal hexahedron whose edges are initially parallel to the coordinate directions.

Solution: Let E1, E2, and E3 be the unit basis vectors for the coordinate system, and the legths of the hexahedron are dX1, dX2, and dX3. Then, the three edges at the undeformed state can be written as d X1 = d X 1 E 1 ,

dX2 = d X 2 E 2 ,

dX2 = d X 2 E 2

With these three vectors, the volume of the infinitesimal hexahedron can be calculated by dV0 = dX1 ⋅ (dX2 ´ dX3 ) = dX 1dX 2 dX 3 E1 ⋅ (E2 ´ E 3 ) = dX1dX 2 dX 3

Now, after deformation, the three edges are deformed to dx1 = FdX1 = dX 1

¶x , ¶X 1

dx2 = FdX2 = dX 2

¶x , ¶X 2

dx 3 = FdX 3 = dX 3

The deformed volume becomes

dVx = dx1 ⋅ (dx2 ´ dx3 ) = dX1dX 2dX 3

¶x æç ¶x ¶x ö÷÷ ⋅ çç ´ ÷ ¶X1 çè ¶X 2 ¶X 3 ø÷

The determinant of the deformation gradient can be written as

¶x ¶X 3

J = det(F) =

¶x æç ¶x ¶x ö÷÷ ⋅ çç ´ ÷ ¶X1 çè ¶X 2 ¶X 3 ÷ø

Thus, the deformed volume becomes dVx = J dV0

▄

P3.3 Consider a square block under oscillating simple shear deformation. The relation between undeformed and deformed geometry is given as x 1 = X 1 + aX 2 sin wt,

x 2 = X 2,

x3 = X3

Calculate the deformation gradient and the change in volume.

Solution: The deformation gradient can be calculated from its definition as

é1 a sin wt 0ù ê ú 1 0úú F = êê0 ê ú 0 1ú êë0 û The determinant of the deformation gradient is J = det(F) = 1. Thus, the volume of the block does not change due to the given deformation. ▄

P3.4 Many materials often show very different behavior between volume-changing deformation and volume-preserving deformation. The former is called dilatation, while the latter is called distortion. In such a case, it is necessary to separate the dilatational and distortional parts from the deformation gradient. Let F be the deformation gradient, decompose it such that F = Fv⋅Fd, where Fv is the dilatational part and Fd is the distortional part. Calculate Fv and Fd using the third invariant of the deformation gradient.

Solution: The determinant of the product of two tensors satisfies the following relation:

J = det(F) = det(Fv )det(Fd ) From the definition, det(Fv ) = J and det(Fd ) = 1 . In the view of Example 3.1, the dilatational deformation gradient must have the following form:

3

Index él 0 0 ù ê ú Fv = êê 0 l 0 úú ê ú êë 0 0 l úû where the principal stretch can be written in terms of J by det(Fv ) = l 3 = J

1



l =J3

Thus, the following Fv is obtained: 1

Fv = J 3 1 and the distortional part of the deformation gradient can be written as -1

Fd = Fv-1 ⋅ F = J 3 F ▄

P3.5 Repeat Problem P3.4 for the Cauchy-Green deformation tensor; i.e., decompose C = Cv⋅Cd. Solution: From the definition of the Cauchy-Green tensor C = FTF, the determinant of C becomes det(C) = det(FT)det(F) = J2. The volumetric part of the Cauchy-Green tensor comes from the volumetric part of the deformation gradient: Cv = FvT Fv , whose determinant becomes 2

det(Cv ) = det(FvT )det(Fv ) = J 3 Thus, the following Cv is obtained: 2

Cv = J 3 1 and the distortional part of the Cauchy-Green tensor can be written as -2

Cd = Cv-1 ⋅ C = J 3 C ▄

P3.6 Consider a bar with a square cross section in the figure under uniaxial tension loading. The principal stretch in X1 direction is given by l > 1. When material is incompressible, compare X1 component of normal strain using Lagrangian, Eulerian, and engineering strains.

X2 F

X1

X3 Solution: Since the bar will maintain rectangular shape, there is no shear deformation. In addition, since both X2 and X3 directions are unconstrained, and the cross-sectional geometries are identical, the principal stretches in these two directions will be the same. Thus, the relation between undeformed and deformed geometries can be written as x 1 = l X 1,

x 2 = l2X 2 ,

x 3 = l2X 3

Since the material is incompressible, the volume should be preserved:

l1l22 = 1



l2 = l-1/2

Thus, the deformation gradient can be obtained as

él 0 0 ùú ê 0 úú F = êê 0 l-1/2 ê ú l-1/2 ú 0 ê0 ë û And the right Cauchy-Green deformation tensor can be obtained as

él 2 0 0 ùú ê C = êê 0 l-1 0 úú ê ú 0 l-1 ú ê0 ë û The Lagrangian strain becomes

él 2 - 1 0 0 ùú ê 1ê l-1 - 1 E= ê 0 0 úú 2ê ú l-1 - 1ú 0 ê 0 ë û For the Eulerian strain, the inverse of the left Cauchy-Green deformation tensor can be calculated by

5

Index él-2 0 0 ù ê ú G-1 = êê 0 l 0 úú ê ú êë 0 0 l úû And the Eulerian strain becomes

é1 - l-2 0 0 ùú ê 1 1-l 0 úú e = êê 0 2ê ú 0 1 - lú êë 0 û By differentiating the following displacement u1 = (l - 1)X 1,

u2 = (l2 - 1)X 2 ,

u 3 = (l2 - 1)X 3

the engineering strain can be obtained as

él - 1 ù 0 0 ê ú ú l-1/2 - 1 0 e = êê 0 ú ê ú -1/2 l 0 - 1ú ê 0 ë û Below are the normal strains in the X1 direction from the three different strains:

1 E11 = (l 2 - 1) 2

1 e11 = (1 - l-2 ) 2

e11 = l - 1

The figure below shows the difference between these three strain components. When the strain is small; i.e., l » 1, all three strains are similar. However, the difference becomes large as the strain increase.

▄

P3.7 A four node square element undergoes large displacement and rotation in the XY plane, as shown in the figure. The node initially at the origin is moved to (1, 1 – sinp/4) and the element is rotated by 45°. Calculate the deformation gradient. Compute the Lagrangian strain and demonstrate that no strain occurs during rigid body motion. y Current

1 Initial

1

x

Solution: From geometry the coordinates in the initial (X,Y) and the deformed (x,y) configurations are as follows.

Node 1 2 3 4

X 0 1 1 0

Y 0 0 1 1

x 1 1 + sinp/4 1 1 – sinp/4

y 1 – sinp/4 1 1 + sinp/4 1

We first need to develop mapping of the current configuration in terms of the initial configuration. A systematic way to develop this mapping in the finite element context is

Index 7 to use the interpolation functions to map the given configurations into a 2×2 reference element in Chapter 1. The interpolation functions are as follows. ìï N = 1 (1 - s )(1 - t ) ïï 1 4 ïïN = 1 (1 + s )(1 - t ) 4 ïí 2 1 ïïN 3 = 4 (1 + s )(1 + t ) ïï 1 ïïîN 4 = 4 (1 - s )(1 + t )

Using these interpolation functions, the initial configuration is mapped as follows.

s +1 2 t +1 Y = N 1y1 + N 2y2 + N 3y 3 + N 4y 4 = 2 X = N 1x 1 + N 2x 2 + N 3x 3 + N 4x 4 =

By inverting this mapping, the s and t can be written in terms of X and Y as follows. s = 2X - 1 t = 2Y - 1

The current configuration can also be mapped into s, t using the interpolation functions.

1 4 1 y= 4 x=

( 2s - 2t + 4) ( 2s + 2t + 4)

Now the desired mapping between the initial and the current configurations can be written by substituting for s, t in terms of X, Y giving

1 4 1 y= 4 z =Z x=

( 2(2X - 1) - 2(2Y - 1) + 4) ( 2(2X - 1) + 2(2Y - 1) + 4)

The deformation gradient can now be easily computed by direct differentiation.

é1 ê 2 F = êê 12 ê êë 0

- 12 1 2

0

0ùú 0úú ú 1ú û

Using the deformation gradient, the matrix of Lagrangian strains is as follows.

é 0 0 0ù ê ú 1 T ê E = (F F - 1) = ê 0 0 0úú 2 ê ú êë 0 0 0úû This matrix mathematically states the obvious fact that in this example any differential line segment in the original configuration has the same length in the current configuration. This example demonstrates that Lagrangian strains are invariant with respect to rigid body rotations and displacements. ▄

P3.8 A square plane strain element is deformed as shown in the figure. The relation between deformed and undeformed coordinates is given as x 1 = X 1 - aX 1X 2 ,

x 2 = X2,

x3 = X3

Compare the engineering strain and Lagrangian strain. Show that the two strain measures become identical when ‘a’ approach zero. a

2

a

X2 2

X1 a

a

Original element Deformed element

Solution: The displacement vector can be written as

u = [-aX1X2, 0, 0]T For linear elastic model, the engineering strain tensor can be defined as é -aX - 21 aX1 ùú e = sym(0 u) = êê 1 2 0 úú êë - 2 aX1 û

Thus, the strain varies linearly.

For the geometric nonlinear model, the deformation gradient and Lagrangian strain can be calculated by

é 1 - aX2 -aX1 ù ú F = 1 + 0u = êê ú 0 1 êë úû

9

Index é -aX (1 - 1 aX ) 2 T 1 2 E = 2 (F F - 1) = êê 1 2 êë - 2 aX1(1 - aX 2 )

é 1 a 2X 2 - 21 aX1(1 - aX 2 ) ùú 2 ê 2 ú = e + ê 1 a 2X X 1 a 2X 2 êë 2 úû 1 1 2 2

1 a 2X X ù 1 2ú 2 1 a 2X 2 ú 1 úû 2

Note that the the Lagrangian strain is the engineering strain plus nonlinear terms, which will approach zero fast when ‘a’ approaches zero. ▄

P3.9 The relation between deformed and undeformed coordinates for pure bending of a plane strain solid is given as x 1 = X 1 - aX 1X 2 ,

x 2 = X 2 + 21 aX 12 ,

x3 = X3

Compare the engineering strain and Lagrangian strain. Show that the two strain measures become identical when ‘a’ approach zero.

Solution: The displacement vector can be written as

u = [-aX1X2, 21 aX12, 0]T For linear elastic model, the engineering strain tensor can be defined as

é -aX2 0 ù ú e = sym(0u) = êê ú 0 0 úû ëê Thus, e11 is only non-zero component and varies linearly with respect to X 2 .

For the geometric nonlinear model, the deformation gradient and Lagrangian strain can be calculated by

é 1 - aX2 -aX1 ù ú F = 1 + 0u = êê ú aX 1 1 êë úû E = 21 (FT F - 1) é -aX + 1 a 2 (X 2 + X 2 ) 1 a 2X X ù 2 1 2 1 2ú 2 2 = êê 1 a 2X X 1 a 2X 2 ú êë 1 2 1 úû 2 2 éX2 + X2 X X ù 2 1 2ú = e + 21 a 2 êê 1 X1X 2 X12 úú ëê û

Note that the Lagrangian strain is the engineering strain plus nonlinear terms, which will approach zero fast when ‘a’ approaches zero. ▄

P3.10 In the small deformation theory, the volumetric strain (dVx - dV0)/dV0 is approximated by e11 + e22 + e33, while in the large deformation theory, it is given by J 1. Show that when the deformation is small, the latter can be approximated by the former.

Solution: The deformation gradient can be written in terms of displacement gradient as

é ê1 + ¶u1 ê ¶X1 ê ê ¶u 2 F = êê ¶ X 1 ê ê ¶u 3 ê ê ¶X 1 ëê

¶u1 ¶X 2 ¶u2 1+ ¶X 2 ¶u 3 ¶X 2

ù ú ¶X 3 úú ¶u2 úú ¶X 3 úú ¶u 3 úú 1+ ¶X 3 úûú ¶u1

The determinant of the deformation gradient can be written as æ ¶u1 ö÷ éêæç ¶u2 öæ ¶u 3 ÷ö ¶u2 ¶u 3 ùú ÷÷ çç ÷÷ ç1 + ÷÷ + det F = ççç1 + 1 ÷ çè ¶X1 ÷ø êêççè ¶X 2 ÷øèçç ¶X 3 ÷ø ¶X 3 ¶X 2 úú ë û ¶u1 éê ¶u2 ¶u 3 ¶u2 æç ¶u 3 ö÷ùú ÷÷ ç1 + + ¶X 2 êê ¶X 3 ¶X1 ¶X1 çèç ¶X 3 ÷øúú ë û é ù æ ö ¶u ¶u ¶u 3 ¶u 3 ç ¶u2 ÷ú ÷÷ çç1 + + 1 êê 2 ¶X 3 êë ¶X1 ¶X 2 ¶X1 çè ¶X 2 ÷øúú û ¶u1 ¶u2 ¶u 3 = 1+ + + + H.O.T. ¶X1 ¶X 2 ¶X 3

When the deformation is small, the higher-order terms will approach zero quickly. Thus the volumetric strain can be approximated by J -1 =

¶u1 ¶X 1

+

¶u 2 ¶X 2

+

¶u 3 ¶X 3

» e11 + e22 + e33

▄

P3.11 An initially straight beam AB is bent into a circular arc A'B' as shown in the figure. The deformation is specified as

x 1 = g(X 2 )cos

p(1 - X1 ) 2

,

x 2 = g(X 2 )sin

p(1 - X1 ) 2

,

x 3 = X3

Index 11 where g(X2) is a simple function of X2. (a) Find the deformation gradient in terms of g(X2). (b) If the volume of the beam does not change, find g(X2). (c) Using g(X2) in (b), find U, Q, and V. X2 Undeformed beam

1 Deformed beam X1

1

Solution: (a) For the given deformation, the deformation gradient can be obtained as é p ù dg p ê g(X )sin p (1 - X ) cos (1 - X1 ) 0ú 2 1 ê 2 ú 2 dX 2 2 ê ú ê p ú p dg p sin (1 - X1 ) 0ú F = ê- g(X 2 )cos (1 - X1 ) ê 2 ú 2 dX 2 2 ê ú ê 0 0 1ú ê ú êë úû

(b) In order to preserve the volume, the determinant of the deformation gradient must be one. det F =

p dg g(X 2 ) =1 2 dX 2

Using the separation of variables,

gdg =

2 dX 2 p

Integrating both sides

1 2 2 g = X2 p 2 Thus, the expression of g(X2) is obtained as

g(X 2 ) =

4X 2 p

(c) By substituting g(X2) into the deformation gradient,

é ê pX sin p (1 - X ) 2 1 ê 2 ê ê p F = êê- pX 2 cos (1 - X1 ) 2 ê ê 0 ê ê êë

ù p cos (1 - X1 ) 0ú ú 2 pX 2 ú ú p 1 sin (1 - X1 ) 0úú 2 pX 2 ú ú 0 1ú ú úû 1

The left Cauchy-Greed deformation tensor becomes

é pX ê 2 ê C = FT F = êê 0 ê ê 0 êë

0ùú ú 0úú ú 1úú û

0 1 pX 2 0

Note that the left Cauchy-Green deformation tensor has only diagonal components. Thus, the matrices of eigenvectors and eigenvalues become

ìï1 0 0ü ï ïï ï ï ï F = í0 1 0ï ý, ïï ï ïï0 0 1ï ï ï î þ

é pX ê 2 ê L = êê 0 ê ê 0 êë

0ùú ú 0úú ú 1úú û

0 1 pX 2 0

Accordingly, U, Q, and V can be calculated by

é ê pX 2 ê ê T U = F LF = ê 0 ê ê ê 0 ëê

0 1 pX 2 0

é sin cos 0ù ê ú Q = FU-1 = êê- cos sin 0úú ê ú 0 1ú êë 0 û

ù 0ú ú ú 0ú ú ú 1úú û

Index é ê 1 2 cos2 ê pX 2 sin + ê pX 2 ê êæ ö êçç 1 ÷÷ T ÷ V = FQ = êç- pX 2 + ÷÷ sin cos êçç p X ÷ø è 2 ê ê 0 ê ê ê ëê

13 ù æ ö çç ú 1 ÷÷ ÷÷ sin cos 0ú çç- pX 2 + ú pX 2 ÷÷ø çè ú ú ú 1 2 2 pX 2 cos + sin 0ú ú pX 2 ú 0 1úú ú ú ûú

In the above expressions, simplified notations are used such that sin = sin p2 (1 - X1 ) and cos = cos p2 (1 - X1 ) . ▄ P3.12 Consider a square element under pure shear deformation as shown in the figure.

The relation between deformed and undeformed coordinates becomes x 1 = X 1 + kX 2 ,

x 2 = kX 1 + X 2

(a) Calculate deformation gradient F, Lagrangian strain E, Eulerian strain e, and engineering strain e. (b) Calculate principal stretch tensors U and V, and rotation tensor Q. X2

X1 Solution: (a) From the relation between deformed and undeformed coordinates, the deformation gradient and Lagrangian strain can be calculated as é1 k ù ú F = êê ú k 1 êë úû

é1 k2 E = (F F - 1) = êê 2 êë k 1 2

k ùú 1 2ú k ú 2 û

T

The engineering strain becomes é0 k ù ú e = êê ú k 0 êë úû

Note that the Lagrangian strain has normal components in higher order. (b) The relation in the polar decomposition is given as F = QU = VQ . First, the right Cauchy-Green deformation tensor becomes é1 + k 2 2k ùú C = FT F = êê 1 + k 2 úú êë 2k û

The above Cauchy-Green tensor will have two eigenvalues: C - l1 = (1 + k 2 - l)2 - 4k 2 = (1 + k 2 - l - 2k )(1 + k 2 - l + 2k ) = 0

Thus, two eigenvalues are l1 = (1 - k )2 and l2 = (1 + k )2 and é(1 - k )2 0 ùú L = êê (1 + k )2 úú êë 0 û

Two eigenvectors corresponding to the two eigenvalues can be calculated as ì ï1ü ï E1 = 12 íï ýï , ï -1ï ï î ï þ

ì ï1ü ï E2 = 12 íï ýï , ï 1ï ï îï þ

é 1 1ù ú F = 12 êê ú 1 1 ëê ûú

Then, the principal stretch tensor U can be calculated from the following relation: U = F LFT =

0 ùú éê1 -1ùú éê1 k ùú 1 éê 1 1ùú éê1 - k = 1 + k úú êê1 1 úú êêk 1úú 2 êê-1 1úú êê 0 ë ûë ûë û ë û

Note that the principal stretch tensor is identical to the deformation gradient, which means there is no rotation involved in the deformation. Thus, we have é 1 0ù ú Q = FU-1 = 1 = êê ú 0 1 úû ëê

Because Q = 1 , the following relation can also be obtained:

V = FQ-1 = F = U

15

Index ▄

P3.13 A square block of surface area A on all sides is under pure shear deformation due to the uniformly distributed load F on the top surface, as shown in the figure. The deformation of the block is such that the deformed coordinates can be written as x 1 = X 1 + aX 2 ,

x 2 = X 2,

x3 = X3

Calculate Cauchy stress, 1-st and 2-nd Piola-Kirchhoff stresses.

X2

F

X1

Solution: Since the force is uniformly distributed over the area, the Cauchy stress will be

é 0 F / A 0ùú ê s = êêF / A 0 0úú ê ú 0 0ú êë 0 û Using the relation of P = JF-1 s , the first Piola-Kirchhoff stress can be calculated by

é-aF / A F / A 0ù ê ú 0 0úú P = êê F / A ê ú 0 0 0ú êë û And from S = JF-1 sF-T the second Piola-Kirchhoff stress becomes ,

é-2aF / A F / A 0ù ê ú ê 0 0úú S= ê F /A ê ú 0 0 0ú êë û Note that the shear stress components are all the same, but negative normal stress component exists in the case of P and S. However, this component becomes small as the shear deformation becomes small. ▄

P3.14 A force R is applied at the tip of the uniform bar element shown in the figure. The initial length and the cross-sectional area of the bar are, respectively, A0 and L0. The elastic modulus of the material is E. Calculate the tip displacement by solving the total Lagrangian variational equation with St. Vernant-Kirchhoff nonlinear elastic material model. Assume the following numerical values: E = 700MPa, A0 = 1.0´10-4m2, L0 = 1.0m, and R = 10kN. Compare the tip displacement with that from the linear elastic model when (a) E = 700MPa and (b) E = 70GPa.

1

R = 10kN

2 x L0=1m

Solution: If the tip displacement is uNL, the displacement in the bar can be approximated by u(X ) =

X u = lX L0 NL

(1)

where l = uLN/L0 is the stretch ratio. Using Eq. (1), the displacement gradient can be calculated by

0u =

du =l dX

(2)

Since the problem is 1D, the displacement gradient becomes a scalar. Note that the above displacement gradient is in fact engineering strain. The deformation gradient becomes F11 = 1 + 0u = 1 + l =

L L0

(3)

where L = L0 + uLN is the deformed length of the bar. The Lagrangian strain is given as E11 = 21 (F11T F11 - 1) = 21 [(1 + l)2 - 1] = l + 21 l 2

(4)

Index 17 Comparing with the engineering strain, the Lagrangian strain has an additional quadratic term. Assuming that the material is St. Vernant-Kirchhoff nonlinear elastic, the strain energy density becomes W (E) = 12 E : D : E = 21 E ⋅ E112 = 12 E (l + 12 l 2 )2

(5)

By differentiating the above strain energy density with respect to Lagrangian strain, the 2nd Piola-Kirchhoff stress can be calculated as S11 =

dW = E ⋅ E11 = E (l + 21 l 2 ) dE11

(6)

In order to obtain the variational equation, the variation of the strain needs to be calculated. Since the displacement of the bar is expressed in terms of the tip displacement in Eq. (1), the variation of the displacement can also be represented by that of the tip displacement. Let uNL be the variation of the tip displacement, and l = uNL / L0 be the variation of the stretch ratio, the variation of displacement and that of the Lagrangian strain can be obtained as u (X ) =

X u = lX L0 NL

(7)

E11 = 21 (F11T 0u + 0u T F11 ) = (1 + l)l

(8)

Note that E11 is linear with respect to l . Using Eq. (8), the structural energy form for the total Lagrangian formulation becomes a 0 (u, u ) = ò

L0 0

S11E11 A0 dX = ò

L0 0

EA0 (l + 12 l 2 )(1 + l)l dX

(9)

Since the integrand is independent of X, the integral can be evaluated analytically, as a 0 (u, u ) = EA0L0l (l + 23 l 2 + 21 l 3 )

(10)

Since the point load is applied at the tip of the bar, the load form can be evaluated without integration as  0 (u ) = u (L0 )R = l L0R

(11)

By equating Eqs. (10) and (11), the variational equation can be written as

EA0L0l (l + 23 l 2 + 21 l 3 ) = l L0R,

"l Î  h

(12)

In the discrete domain, the space of kinematically admissible displacements will be the space of real numbers. In order to satisfy the above equation for all real number l , the coefficients of l should be equal in the above equation, to yield l + 23 l 2 + 21 l 3 =

R EA0

(13)

For linear elastic material, the above equation becomes l=

R EA0

(14)

Note that both equations have the same right-hand side. Thus, the nonlinear displacement will always be smaller than that of linear displacement. In addition, when the displacement is small; i.e., l  1 , the higher-order terms can be negligible, l 2 » l 3 » 0 . Thus, the nonlinear displacement will approach to the linear one. (a) When E = 700 MPa, Eq. (13) yields l = 0.12028  uNL = 120.28mm , while Eq. (14) yields l = 0.14286  uL = 142.86mm . Thus, the linear elastic model predicts about 19% larger tip displacement. (b) When E = 70 GPa, Eq. (13) yields l = 0.001426  uNL = 1.426mm , while Eq. (14) yields l = 0.001429  uL = 1.429mm . Thus, the linear elastic model predicts about 0.2% larger tip displacement. ▄

P3.15 Solve Problem P3.14 using force equilibrium; i.e., internal force caused by stress is equal to external force.

Solution: From the previous problem, we have S11 = E (l + 12 l 2 )

However, the 2nd Piola-Kirchhoff stress cannot be used for force equilibrium. Thus, the 1st Piola-Kirchhoff stress is calculated using the following relation: P11 = S11 ⋅ F11T = E (l + 12 l 2 )(1 + l) = E (l + 23 l 2 + 12 l 3 )

Note that the 1st Piola-Kirchhoff stress is defined with respect to the initial crosssectional area. Assuming this stress is uniform over the cross-section, the force equilibrium can be obtained by R = P11 ⋅ A0 = EA0 (l + 23 l 2 + 21 l 3 )

Thus, we obtain the same equation with the variational approach, as

19

Index l + 23 l 2 + 21 l 3 =

R EA0

For linear elastic material, the above equation becomes l=

R EA0

Note that both equations have the same right-hand side. Thus, the nonlinear displacement will always be smaller than that of linear displacement. In addition, when the displacement is small; i.e., l  1 , the higher-order terms can be negligible, l 2 » l 3 » 0 . Thus, the nonlinear displacement will approach to the linear one. ▄

P3.16 Consider a plane strain, unit depth, square element as shown in the figure. Use St. Vernant-Kirchhoff isotropic material model with two Lame’s constants l and m. A uniformly distributed force Tx (force per area) is horizontally applied at the top surface.

Assuming it is a simple shear problem, the deformation of the element can be written as ìïx = X + kX 1 2 ï 1 í ïïx 2 = X 2 î

(a) Find the relation between k and Tx , (b) Find the reaction force in X 2 direction at the top surface, and (c) Compare the results with that of the linear elastic model.

X2

Tx

X1 Solution: For the given deformation, the deformation gradient and Lagrangian strain can be calculated as

é1 k ù ú, F = êê ú 0 1 êë úû

1 1 é 0 k ùú E = (FT F - 1) = êê 2 2 2 êk k úú ë û

From the constitutive tensor of St. Vernant-Kirchhoff material C = l1 Ä 1 + 2mI and the Lagrangian strain, the 2nd Piola-Kirchhoff stress can be calculated by é 1 lk 2 ù mk 2 ê ú S = D : E = l tr(E)1 + 2mE = ê 2ú 1 k ( ) k m l m + êë úû 2

Since the 2nd Piola-Kirchhoff stress does not have any physical meaning, it is converted to the 1st Piola-Kirchhoff stress as é ( 1 l + m)k 2 ù mk 2 ê ú P = S⋅F = ê 3 2ú 1 1 êë mk + ( 2 l + m)k ( 2 l + m)k úû T

The unit normal vector of the top surface is N = [0, 1]T. Thus, the surface traction on the top surface becomes ìïmk + ( 1 l + m)k 3 üï ìïT üï 2 ïý = ïí x ïý PT ⋅ N = ï í ïï ( 21 l + m)k 2 ïï ïïTy ïï ïî ïþ î þ

(a) In the above equation, Tx is the horizontal surface traction on the top surface or, equivalently, uniformly distributed force: Tx = mk + ( 21 l + m)k 3 (b) Ty is the uniformly distributed vertical reaction: Ty = ( 21 l + m)k 2 (c) For linear elastic model, the relation between k and Tx is Tx = mk . Thus, the nonlinear model has higher-order terms. In addition, the vertical reaction in linear elastic model is zero, while the nonlinear elastic model yields non-zero vertical reaction. When k  0 , the results from nonlinear model approach that of the linear elastic model. ▄

P3.17 Consider a deformation of a rectangular bar whose deformed geometry is given as x 1 = aX 1,

x 2 = b X 2,

x 3 = bX 3

When the material is incompressible and St. Venant-Kirchhoff material properties are given as E = 600MPa and n = 0.49, write the expression of S11 component of the second Piola-Kirchhoff stress as a function of a. In addition, write the expression of s11 of the Cauchy stress as a function of a. Plot S11 and s11 in the range of a = [0.7 1.5].

Solution: For the given material properties, the Lame’s constants can be calculated from En = 9, 865.8 MPa (1 + n )(1 - 2n ) E m= = 201.3 MPa 2(1 + n )

l=

Index 21 For given deformation, the deformation gradient and Cauchy-Green deformation tensor can be calculated as

éa 0 0 ù ê ú ê F = ê 0 b 0 úú , ê ú êë 0 0 b úû

éa 2 ê T C = F F = êê 0 ê ê0 ë

0 b2 0

0 ùú 0 úú ú b2 ú û

The relation between a and b can be obtained from incompressibility:

det F = ab 2 = 1



b = a-1/2

Thus, the Lagrangian strain can be calculated as

éa 2 - 1 0 0 ùú ê 1 1 0 úú E = (FT F - 1) = êê 0 a-1 - 1 2 2ê ú 0 a-1 - 1ú ê 0 ë û Since all shear components are zero, we can only consider the normal components as a vector. The second Piola-Kirchhoff stress becomes 2 -1 2 ìS ï ü él + 2m ïï ï l l ùú ïïìE11 ïïü ìïï l2 (a + 2a - 3) + m(a - 1) ü ï 11 ï ê ï ï ïï ïï ê ï ï 2 -1 -1 ï ï ï l + 2m l úú ï S = íS 22 ý = ê l íE22 ý = í l2 (a + 2a - 3) + m(a - 1)ý ï ï ï ï ï ïï ê ú ï ï ïl 2 ï ï -1 -1 S E l l l 2 m + ï ï ï ï ï ïï ( a 2 a 3) m ( a 1) + + ê ú ï î 33 ï þ ë û ïî 33 ïþ ï î2 þ

The S11 component of the stress becomes

S11 =

l 2 (a + 2a-1 - 3) + m(a2 - 1) 2

The Cauchy stress can be obtained from the relation

s=

1 F ⋅ S ⋅ FT J

The s11 component of the stress becomes

s11 =

l 4 (a + 2a - 3a2 ) + m(a 4 - a2 ) 2

The following figure shows the two stress components as a function of the principal stretch a. Note that both stresses are highly nonlinear even if the relation between stress and strain is constant.

▄

P3.18 Consider a simple shear deformation of a square whose deformed geometry is given as x 1 = X 1 + aX 2 ,

x 2 = X 2,

x3 = X3

When the material is incompressible and St. Venant-Kirchhoff material properties are given as E = 600MPa and n = 0.49, write the expression of S12 component of the second Piola-Kirchhoff stress as a function of a. In addition, write the expression of s12 of the Cauchy stress as a function of a. Plot S12 and s12 in the range of a = [0.0 1.5].

Solution: For the given material properties, the Lame’s constants can be calculated from En = 9, 865.8 MPa (1 + n )(1 - 2n ) E m= = 201.3 MPa 2(1 + n )

l=

For given deformation with incompressibility, the deformation gradient and Lagrangian strain tensor can be calculated as

é 1 a 0ù ê ú ê F = ê0 1 0úú , ê ú êë0 0 1úû

é0 a ê 1 T 1ê E = (F F - 1) = êa a2 2 2ê êë 0 0

0ùú 0úú ú 1ú û

In the plane strain problem, we can consider only three non-zero stress components. The second Piola-Kirchhoff stress becomes

Index ì ü é ïï l2 a2 üïï ïïS ïï l + 2m ïï E ü ï ì 0 úù ì l 11 11 ï ê ï ï ï ïï ê ï ï 2ï l + 2m 0 úú ïí E22 ïý = ï S=ï íS 22 ý = ê l í( l2 + m)a ý ï ï ïï ï ï ê ú ïï ï ï ï ï S E 0 0 2 m ma ï ï ï ï ï ï ê ú ï 12 þ ï ë ï î û îï 12 þï î þï

23

Thus, S12 is a linear function of a. The Cauchy stress can be obtained from the relation

s=

1 F ⋅ S ⋅ FT J

The s12 component of the stress becomes æl ö s12 = çç + m÷÷÷ a 3 + ma ÷ø çè 2

Different from the hyperelastic material, now s12 is a cubic function, while S12 is a linear function for the shear deformation. The following figure compares the two stresses as a function of shear deformation.

▄

P3.19 Consider the following deformation with ½a½ £ 1:

x1 = X1 + aX 2,

x 2 = 1 - a2 X 2 ,

x 3 = X3

Assume St. Venant-Kirchhoff material with two material parameters l and m. (a) Show that the above deformation is a pure shear deformation in terms of the Lagrangian strain, (b) Calculate the second Piola-Kirchhoff stress and Cauchy stress in terms of a, l, and m.

Solution: (a) For the given deformation, the deformation gradient and the Lagrangian strain become

é ê1 ê F = ê0 ê ê0 ëê

a 1 - a2 0

0úù ú 0ú , ú 1ú ûú

é 0 a 0ù ê ú 1 T 1ê E = (F F - 1) = êa 0 0úú 2 2ê ú êë 0 0 0úû

Thus, the given deformation is a pure shear deformation in terms of the Lgrangian strain. (b) From the St. Venant-Kichhoff material, the stress-strain relation becomes

é 0 a 0ù ê ú S = l(trE)1 + 2mE = m êêa 0 0úú ê ú êë 0 0 0úû Thus, only shear stress component S12 exists. The Cauchy stress becomes

é a2 ê ê 2 ê 1-a ê 1 s = FSFT = m êê a J ê ê ê 0 ê ëê

a

0

0

0

0

0

ù ú ú ú ú ú ú ú ú ú ú ûú

Note that the Cauchy stress has a non-zero normal component s11. The following figure shows the deformation of a square block with a = 0.5.

25

Index

▄

P3.20 A force F is applied at the tip of the uniform bar shown in the figure. The displacement of the bar is given as u(X) = lX where l is the principal stretch. The initial length and the cross-sectional area of the bar are, respectively, A0 and L0. The elastic modulus of the material is E. Calculate the tip displacement by solving the principal stretch using the total Lagrangian formulation with the St. Venant-Kirchhoff material model. Assume the following numerical values: E = 700 MPa, A0 = 1.0´10-4 m2, L0 = 1.0 m, and F = 10 kN. Compare the tip displacement with that from the linear elastic model when (a) E = 700 MPa and (b) E = 70 GPa.

F = 10kN x L0=1m

Figure P3.20

Solution: If the tip displacement is u NL , the displacement in the bar can be approximated by u(X ) =

X u = lX L0 NL

(1)

where l = u NL / L0 is the stretch ratio. Using Eq. (1), the displacement gradient can be calculated by

0u =

du =l dX

(2)

Since the problem is 1D, the displacement gradient becomes a scalar. Note that the above displacement gradient is in fact engineering strain. The deformation gradient becomes F11 = 1 + 0u = 1 + l =

L L0

(3)

where L = L 0 + u NL is the deformed length of the bar. The Lagrangian strain is given as T E11 = 21 (F11 F11 - 1) = 21 [(1 + l)2 - 1] = l + 21 l2

(4)

Comparing with the engineering strain, the Lagrangian strain has an additional quadratic term. Assuming that the material is St. Vernant-Kirchhoff nonlinear elastic, the strain energy density becomes 2 W (E) = 21 E : C : E = 21 E ⋅ E11 = 21 E (l + 21 l2 )2

(5)

By differentiating the above strain energy density with respect to Lagrangian strain, the 2nd Piola-Kirchhoff stress can be calculated as S11 =

dW = E ⋅ E11 = E (l + 21 l 2 ) dE11

(6)

In order to obtain the variational equation, the variation of the strain needs to be calculated. Since the displacement of the bar is expressed in terms of the tip displacement in Eq. (1), the variation of the displacement can also be represented by that of the tip displacement. Let u N L be the variation of the tip displacement, and l = u NL / L0 be the variation of the stretch ratio, the variation of displacement and that of the Lagrangian strain can be obtained as u (X ) =

X u = lX L0 NL

T E11 = 21 (F11 0u + 0u T F11 ) = (1 + l)l

(7) (8)

Note that E11 is linear with respect to l . Using Eq. (8), the structural energy form for the total Lagrangian formulation becomes a 0 (u, u ) = ò

L0

0

S11E11 A0 dX = ò

L0

0

EA0 (l + 21 l2 )(1 + l)l dX

Since the integrand is independent of X , the integral can be evaluated analytically, as

(9)

Index a0 (u, u ) = EA0L0l (l + 23 l2 + 21 l 3 )

27 (10)

Since the point load is applied at the tip of the bar, the load form can be evaluated without integration as (11)

 0 (u ) = u (L 0 )F = l L 0 F

By equating Eqs. (10) and (11), the variational equation can be written as

EA0L0l (l + 23 l2 + 21 l 3 ) = l L0F,

"l Î  h

(12)

In the discrete domain, the space of kinematically admissible displacements will be the space of real numbers. In order to satisfy the above equation for all real number l , the coefficients of l should be equal in the above equation, to yield l + 23 l 2 + 21 l 3 =

F EA0

(13)

For linear elastic material, the above equation becomes l=

F EA0

(14)

Note that both equations have the same right-hand side. Thus, the nonlinear displacement will always be smaller than that of linear displacement. In addition, when the displacement is small; i.e., l  1 , the higher-order terms can be negligible, l2 » l 3 » 0 . Thus, the nonlinear displacement will approach to the linear one. (a) When E = 700 MPa, Eq. (13) yields l = 0.12028  uNL = 120.28mm , while Eq. (14) yields l = 0.14286  uL = 142.86mm . Thus, the linear elastic model predicts about 19% larger tip displacement. (b) When E = 70 GPa, Eq. (13) yields l = 0.001426  uNL = 1.426mm , while Eq. (14) yields l = 0.001429  uL = 1.429mm . Thus, the linear elastic model predicts about 0.2% larger tip displacement. ▄

P3.21 Solve Problem P4.20 using force equilibrium; i.e., internal force caused by stress is equal to external force.

Solution: From the previous problem, we have

S11 = E (l + 21 l2 ) However, the 2nd Piola-Kirchhoff stress cannot be used for force equilibrium. Thus, the 1st Piola-Kirchhoff stress is calculated using the following relation: T P11 = S11 ⋅ F11 = E (l + 21 l2 )(1 + l) = E (l + 23 l2 + 21 l 3 )

Note that the 1st Piola-Kirchhoff stress is defined with respect to the initial crosssectional area. Assuming this stress is uniform over the cross-section, the force equilibrium can be obtained by

F = P11 ⋅ A0 = EA0 (l + 23 l2 + 21 l 3 ) Thus, we obtain the same equation with the variational approach, as l + 23 l 2 + 21 l 3 =

F EA0

For linear elastic material, the above equation becomes l=

F EA0

Note that both equations have the same right-hand side. Thus, the nonlinear displacement will always be smaller than that of linear displacement. In addition, when the displacement is small; i.e., l  1 , the higher-order terms can be negligible, l2 » l 3 » 0 . Thus, the nonlinear displacement will approach to the linear one. (a) When E = 700 MPa, Eq. (13) yields l = 0.12028  uNL = 120.28mm , while Eq. (14) yields l = 0.14286  uL = 142.86mm . Thus, the linear elastic model predicts about 19% larger tip displacement. (b) When E = 70 GPa, Eq. (13) yields l = 0.001426  uNL = 1.426mm , while Eq. (14) yields l = 0.001429  uL = 1.429mm . Thus, the linear elastic model predicts about 0.2% larger tip displacement. ▄

P3.22 Consider two bar elements under a force at the tip. Using the displacementcontrolled method, plot the load-displacement curve (F vs. u2 and u3). Increase the tip displacement u3 up to 1.0m with ten equal increments. Assume St. Venant-Kirchhoff material with E = 100MPa, and cross-sectional areas of A(1) = 1.0´10-4 m2 and A(2) = 0.5´10-4 m2.

29

Index

1

2

3

E, A(1)

E, A(2)

L0=1m

L0=1m

F

x Figure P3.22

Solution: Since this is a one-dimensional problem, only X1-directional component will be used in the following calculations. For Elements 1 and 2, the Lagrangian strains are defined as

E11(1) = u2 + 21 u22 ,

E11(2) = u3 - u2 + 12 (u3 - u2 )2

In the above equation u1 = 0 is used. For the St. Venant-Kirchhoff material, the second Piola-Kirchhoff strsses for the two elements can be calculated by S11(1) = E (u2 + 12 u22 ),

(

S11(2) = E u 3 - u2 + 12 (u 3 - u2 )2

)

Since u1 is fixed, its variation is also equal to zero. The variation of the Lagrangian strans become

E11(1) = (1 + u2 )u2,

E11(2) = (u2 - u3 - 1)(u2 - u3 )

The energy form can be obtained by adding the contributions from two elements as a(u, u ) = ò

L0 0

S11(1)(1 + u2 )u2A(1) dX + ò

2 L0

L0

S11(2)(u2 - u 3 - 1)(u2 - u 3 )A(2) dX

= u2 éêS11(1)(1 + u2 )A(1)L0 + S11(2)(u2 - u 3 - 1)A(2)L0 ùú - u 3 éêS11(2)(u2 - u 3 - 1)A(2)L0 ùú ë û ë û

The load form is simply

(u ) = u3F3 Since the nonlinear variational equation must satisfy for arbitraty u2 and u3 , two nonlinear equation can be obtained:

S11(1)(1 + u2 )A(1)L0 + S11(2)(u2 - u 3 - 1)A(2)L0 = 0 S11(2)(u2 - u3 - 1)A(2)L0 + F = 0 Note that the first equation is the coefficient of u2 and the second is that of u3 . Since u3 is prescribed for the displacement-controll method, its variation is zero, and the applied

force is in fact the reaction force required prescribing the displacement. Thus, for a given u3, the unknown u2 is solved from the first equation. After that, the second equation is used to solve for the reaction force F. In order to solve the above nonlinear equation using the Newton-Raphson method, the increment is stresses are required:

DS11(1) = E (1 + u2 )Du2,

DS11(2) = E (u2 - u3 - 1)Du2

Note that only the increment Du2 is considered because the motion of u3 is prescribed; i.e., the convergence iteration is performed to find Du2 after increasing u3 according to the displacement controlled method. Thus,

a * (u; Du, u ) = ò

L0

+ò

2L0

0

L0

A(1)[E (1 + u2 )2 + S11(1) ]Du2u2 dX A(2)[E (u2 - u 3 - 1)2 + S11(2) ]Du2u2 dX

(

)

(

)

= u2L0 éêA(1) E (1 + u2 )2 + S11(1) + A(2) E (u2 - u 3 - 1)2 + S11(2) ùú Du2 ë û Below is the list of MATLAB programs that solves for the nonlinear variational equation.

The following table and figure show the converged solutions at each increment. % % P3.22 Two bar elements--displacement controlled procedure % tol = 1.0e-5; conv = 0; u2 = 0; u2old = u2; E = 1E8; A1 = 1E-4; A2 = .5E-4; fprintf('\n step u2 u3 F'); % Displacement increment loop for i=1:10 u3 = 0.1*i; S1 = E*(u2+.5*u2^2); S2 = E*(u3-u2+.5*(u3-u2)^2); P = S1*A1*(1+u2)+S2*A2*(u2-u3-1); F = S2*A2*(1+u3-u2); R = -P; conv = R^2; % Convergence loop iter = 0; while conv > tol && iter < 50 Kt = A1*(E*(1+u2)^2+S1) + A2*(E*(u2-u3-1)^2+S2); delu2 = R/Kt; u2 = u2old + delu2; S1 = E*(u2+.5*u2^2); S2 = E*(u3-u2+.5*(u3-u2)^2); P = S1*A1*(1+u2)+S2*A2*(u2-u3-1); R = -P; conv= R^2; u2old = u2; iter = iter + 1; end F = S2*A2*(1+u3-u2); fprintf('\n %3d %7.5f %7.5f %8.3f',i,u2,u3,F); end

Increment

u2

u3

F

31

Index

1 2 3 4 5 6 7 8 9 10

0.0343 0.0704 0.1077 0.1460 0.1851 0.2248 0.2651 0.3058 0.3469 0.3883

0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000

361.3 779.7 1256.9 1795.0 2396.4 3063.4 3798.5 4604.2 5482.9 6437.2

▄

P3.23 Consider a nonlinear elastic uniaxial bar element under tip force F = 100N shown in Figure 3.11. The stress strain relation is given in terms of Cauchy stress and engineering strain in the deformed geometry: s11 = Ee11. Using the updated Lagrangian formulation, solve for displacement at the tip, stress and strain of the uniaxial bar.

Assume E = 200Pa and the cross-sectional area A = 1.0m2.

Solution: It is easy to estimate the stress and strain in order to make equilibrium with the applied load. Since F = 100N and A = 1.0m2, the required Cauchy stress should be s11 = 100Pa. From the stress-strain relation, the required strain should be e11 = 0.5. Since the strain is defined using the deformed geometry, the deformed length of the bar should be

2.0m, which yields the tip displacement of u2 = 1.0m. Now in the following, the updated Lagrangian method is used to solve the problem.

Since the strain is defined in the deformed geometry, e11(u ) =

u2 du du d X , = = dx d X dx 1 + u2

e11(u ) =

u2 1 + u2

,

J = 1 + u2

Thus, since a concentrated force is applied at the tip, the load form is (u ) = Fu2 . The energy form can be wriiten as L

a(u, u ) = ò s11e11(u )A dx =ò

0 L0 0

s11e11(u )AJ dX

= EAL0

u2u2 1 + u2

Since the nonlinear variational equation satisfies for arbitrary u2 , the residual can be defined as R = EAL0

u2 1 + u2

-F = 0

Since the constitutive relation is given in terms of Cauchy stress and engineering strain, it is unnecessary to transform the material description to the spatial descrition. It is more convenient to directly linearlize the energy form in the spatial form. After linearizing the residual, the incremental equation for the Newton-Raphson method becomes EAL0 (1 + u2 )2

Du2 = -R

Below is the list of MATLAB programs that solve the the nonlinear variational equation.

Also the following table shows the convergence interation of the Newton-Raphson method. The solution converges in the fourth iteration. As expected, the tip displacement, strain, and strss converge to u2 = 1.0m., e11 = 0.5, and s11 = 100Pa. % % P3.23 Uniaxial bar--updated Lagrangian formulation % tol = 1.0e-5; iter = 0; E = 200; u = 0; uold = u; f = 100; strain = u/(1+u); stress = E*strain; P = stress; R = f - P; conv= R^2/(1+f^2); fprintf('\n iter u1 Strain Stress conv'); fprintf('\n %3d %7.5f %7.5f %8.3f %12.3e %7.5f',iter,u,strain,stress,conv); while conv > tol && iter < 20 Kt = E/(1+u)^2; delu = R/Kt; u = uold + delu; strain = u/(1+u);

33

Index stress = E*strain; P = stress; R = f - P; conv= R^2/(1+f^2); uold = u; iter = iter + 1; fprintf('\n %3d %7.5f end

Iteration 0 1 2 3 4

%7.5f %8.3f %12.3e %7.5f',iter,u,strain,stress,conv);

u 0.0000 0.5000 0.8750 0.9922 1.0000

Strain 0.0000 0.3333 0.4667 0.4980 0.5000

Stress 0.000 66.667 93.333 99.608 99.998

conv 9.999E-01 1.111E-01 4.444E-03 1.538E-05 2.328E-10

▄

P3.24 Consider a deformation of a rectangular bar whose deformed geometry is given as x 1 = aX 1,

x 2 = b X 2,

x 3 = bX 3

When the material is incompressible, Mooney-Rivlin hyperelastic material with A10 = 80MPa and A01 = 20MPa, write the expression of S11 component of the second PiolaKirchhoff stress as a function of a. In addition, write the expression of s11 of the Cauchy stress as a function of a. Plot S11 and s11 in the range of a = [0.7 1.5].

Solution: For given deformation, the deformation gradient and Cauchy-Green deformation tensor can be calculated as

éa 0 0 ù ê ú ê F = ê 0 b 0 úú , ê ú êë 0 0 b úû

éa 2 ê T C = F F = êê 0 ê ê0 ë

0 b2 0

0 ùú 0 úú ú b2 ú û

The relation between a and b can be obtained from incompressibility:

det F = ab 2 = 1



b = a-1/2

The three invariants of the deformation tensor can be obtained as

I 1 = a2 - 2a-1 I 2 = 2a + a-2 I3 = 1 The reduced invariants become

J 1 = I 1I 3-1/3 = a 2 - 2a-1 J 2 = I 2I 3-2/3 = 2a + a-2 J 3 = I 3-1/2 = 1 In order to calculate stress, we need to differentiate the reduced invariants with respect to strain

é 1 0 0ù ê ú I 1,E = 21 = 2 êê 0 1 0úú ê ú êë 0 0 1úû é2a-1 ù 0 0 ê ú ê ú 2 -1 I 2,E = 2(I 1 1 - C) = 2 ê 0 a +a 0 ú ê 2 -1 ú 0 a +a ú ê 0 ë û éa-2 ê -1 I 3,E = 2I 3C = 2 êê 0 ê êë 0

0 ùú a 0 úú ú 0 aú û 0

The derivatives of the reduced invariants become

é2(1 - a-3 ) 0 0 ùú ê 1 2 J 1,E = I 3-1/3I 1,E - I 1I 3-4/3I 3,E = êê 0 1 - a3 0 úú 3 3ê ú 0 0 1 - a3 ú ê ë û é2(a-1 - a-4 ) ù 0 0 ê ú 2 2 ú -a2 + a-1 J 2,E = I 3-2/3I 2,E - I 2I 3-5/3I 3,E = êê 0 0 ú 3 3ê 2 -1 ú -a + a ú 0 0 ê ë û é a -2 ê 1 J 3,E = I 3-1/2I 3,E = êê 0 2 ê êë 0

0 ùú a 0 úú ú 0 aú û 0

Thus, the second Piola-Kirchhoff stress becomes S = A10J 1,E + A01J 2,E + K (J 3 - 1)J 3,E

Index

35

The S11 component of the stress becomes

4é A10 (1 - a-3 ) + A01(a-1 - a-4 )ùú ê ë û 3 4 = (80 + 20a-1 )(1 - a-3 ) 3

S11 =

The Cauchy stress can be obtained from the relation

s=

1 F ⋅ S ⋅ FT J

The s11 component of the stress becomes

s11 =

4 (80a2 + 20a)(1 - a-3 ) 3

The following figure shows the two stress components as a function of the principal stretch a. Note that the second Piola-Kirchhoff stress is highly nonlinear, but the Cauchy stress is reasonably linear with respect to the principal stretch. Also note that the two stresses are similar when the deformation is small. However, as deformation increases, the difference also increases.

▄

P3.25 Consider a simple shear deformation of a square whose deformed geometry is given as x 1 = X 1 + aX 2 ,

x 2 = X 2,

x3 = X3

When the material is incompressible, Mooney-Rivlin hyperelastic material with A10 = 80MPa and A01 = 20MPa, write the expression of S12 component of the second PiolaKirchhoff stress as a function of a. In addition, write the expression of s12 of the Cauchy stress as a function of a. Plot S12 and s12 in the range of a = [0.0, 1.5].

Solution: For given deformation, the deformation gradient and Cauchy-Green deformation tensor can be calculated as

é 1 a 0ù ê ú F = êê0 1 0úú , ê ú êë0 0 1úû

é1 a 0ùú ê C = FT F = êêa a 2 + 1 0úú ê ú 0 1ú êë 0 û

The three invariants of the deformation tensor can be obtained as

I 1 = a2 + 3,

I 2 = a 2 + 3,

I3 = 1

The reduced invariants become

J 1 = I 1I 3-1/3 = a2 + 3 J 2 = I 2I 3-2/3 = a2 + 3 J 3 = I 3-1/2 = 1 In order to calculate stress, we need to differentiate the reduced invariants with respect to strain

é 1 0 0ù ê ú I 1,E = 21 = 2 êê 0 1 0úú ê ú êë 0 0 1úû é2a2 + 4 -2a 0 ùú ê I 2,E = 2(I 1 1 - C) = êê -2a 4 0 úú ê ú 0 2a2 + 4ú ê 0 ë û é2a2 + 2 -2a 0ù ê ú I 3,E = 2I 3C-1 = êê -2a 2 0úú ê ú 0 2ú êë 0 û The derivatives of the reduced invariants become

Index é-a 4 - 4a2 ê 1 -4/3 2ê 3 -1/3 J 1,E = I 3 I 1,E - I 1I 3 I 3,E = ê a + 3a 3 3ê 0 ê ë

a + 3a 0 ùú -a2 0 úú ú -a2 ú 0 û 3

é-2a 4 - 5a2 2a 3 + 3a 0 ùú ê 2 2 -2a2 J 2,E = I 3-2/3I 2,E - I 2I 3-5/3I 3,E = êê 2a 3 + 3a 0 úú 3 3ê ú -2a2 ú 0 0 ê ë û éa2 + 1 -a 0ù ê ú 1 -1/2 ê J 3,E = I 3 I 3,E = ê -a 1 0úú 2 ê ú 0 1ú êë 0 û Thus, the second Piola-Kirchhoff stress becomes S = A10J 1,E + A01J 2,E + K (J 3 - 1)J 3,E

The S12 component of the stress becomes

2 [A10 (a 3 + 3a) + A01(2a 3 + 3a)] 3 = 80a 3 + 200a

S12 =

The Cauchy stress can be obtained from the relation

s=

1 F ⋅ S ⋅ FT J

The s12 component of the stress becomes s12 = 2(A10 + A01 )a = 200a

Note that the shear stress S12 is a cubic function of a, but s12 is a linear function.

37

▄

P3.26 Derive the energy form and its linearization of a Mooney-Rivlin hyperelastic material using the perturbed Lagrangian method. Use a mixed variable r = [uT, p]T. Solution: Using the distortional strain energy density in Eq. (4.116) and dilatational strain energy density in Eq. (4.123) for the Mooney-Rivlin material, the second Piola-Kirchhoff stress can be obtained by W (J 1, J 2 , J 3 , p) = A10 (J 1 - 1) + A01(J 2 - 3) + p(J 3 - 1) S = W,E = A10J 1,E + A01J 2,E + pJ 3,E .

1 2 p 2K

(1) (2)

Note that the independent pressure p is used. In order to derive the energy form, the first variation of the strain energy density can be written as æ pö W = W,E : E + W,p p = S : E + çççJ 3 - 1 - ÷÷÷ p .

K ø÷ è

(3)

Since both displacement and pressure are independent variables, a new combined variable is introduced as r = [uT, p]T. Then, the energy form can be obtained by integrating Eq. (3) as a(r, r ) º òò 0 S : E dW + òò 0 pH dW W

W

(4)

where H = J3 – 1 – p/K corresponds to the volumetric strain.

The energy form a (r, r ) is nonlinear through the constitutive relation and straindisplacement relation. Linearization of stress can be expressed in terms of displacement and pressure increments as

Index DS = W,E,E : DE + W,E,p Dp = D : DE + J 3,EDp

39 (4)

where D is the fourth-order constitutive tensor defined as D=

¶S = A10J 1,EE + A01J 2,EE + pJ 3,EE ¶E

(5)

and DE and Dp are the incremental strain and pressure. Linearization of energy form can be obtained as a * (r; Dr, r ) º òò 0 éêE : (D : DE + J 3,EDp) + S : DEùú dW û Wë æ ö÷ p D ÷ dW +òò 0 p ççJ 3,E : DE W çè K ÷÷ø

(6)

The pressure term can be condensed on the finite element level by directly solving the terms that contain the pressure variation. This can easily be done if constant pressure approximation is used, which can be done within a finite element. ▄

P3.27 Derive the 6´6 [D] matrix in Eq. (3.147) for two-dimensional Mooney-Rivlin material with three material parameters (A10, A01, and K). Use the penalty method for near-incompressibility Solution: The expression for the constitutive tensor is given in Eq. (3.128). In order to make a matrix expression, it is necessary to define the second-order derivatives of the three invariants in the matrix notation as [I 1,EE ] = [0]6´6

é0 ê ê4 ê ê4 ê [I 2,EE ] = ê ê0 ê ê0 ê ê0 ë

4 4 0 4 4 0 0 0 0 0 0 0

0 ùú 0 0 0 úú 0 0 0 úú -2 0 0 úú ú 0 -2 0 ú ú 0 0 -2ú û 0

0

é ù 4C 3 4C 2 -4C 5 ê ú ê 4C ú 4 C 4 C 3 1 6ú ê ê 4C ú 4C 1 -4C 4 ú 2 [I 3,EE ] = êê ú 4 C 2 C 2 C 2 C ê 4 3 6 5 ú ê ú 2C 6 -2C 1 2C 4 ú ê-4C 5 ê ú -4C 6 2C 5 2C 4 -2C 2 ú ê ë û

From Eq. (3.129), the second-order derivatives of the three reduced invariants can be written in the matrix notation as

1 -4 4 -7 1 -4 [J 1,EE ] = - I 3 3 ({I 1,E }{I 3,E }T + {I 3,E }{I 1,E }T ) + I 1I 3 3 {I 3,E }{I 3,E }T - I 1I 3 3 [I 3,EE ] 3 9 3 2 - 53 10 - 83 2 -5 - 23 T T [J 2,EE ] = I 3 [I 2,EE ] - I 3 ({I 2,E }{I 3,E } + {I 3,E }{I 2,E } ) + I 2I 3 {I 3,E }{I 3,E }T - I 2I 3 3 [I 3,EE ] 3 9 3 1 1 - 23 1 [J 3,EE ] = - I 3 {I 3,E }{I 3,E }T + I 3 2 [I 3,EE ] 4 2 Note that the expressions of {I1,E}, {I2,E}, and {I3,E} are available in Section 3.5.2. From Eq. (3.128), the constitutive matrix can be obtained as

[D] = A10 [J 1,EE ] + A01[J 2,EE ] + K (J 3 - 1)[J 3,EE ] + K {J 3,E }{J 3,E }T ▄

P3.28 Derive the 6´6 [D] matrix in Eq. (3.147) for two-dimensional Mooney-Rivlin material with three material parameters (A10, A01, and K). Use the perturbed Lagrangian method for near-incompressibility Solution: For the perturbed Lagrangian formulation, the expression of stress is given in Example 3.15, and the constitutive tensor can be written as

D=

¶S = A10J 1,EE + A01J 2,EE + pJ 3,EE ¶E

In order to make a matrix expression, it is necessary to define the second-order derivatives of the three invariants in the matrix notation as [I 1,EE ] = [0]6´6

é0 ê ê4 ê ê4 ê [I 2,EE ] = ê ê0 ê ê0 ê ê0 ë

Index 4 4 0 0 4 4 0 0 0 0 0 0 0

41

0 0 ùú 0 0 0 úú 0 0 0 úú -2 0 0 úú ú 0 -2 0 ú ú 0 0 -2ú û

é ù 4C 3 4C 2 -4C 5 ê ú ê 4C ú 4 C 4 C 3 1 6ú ê ê 4C ú 4C 1 -4C 4 ú 2 [I 3,EE ] = êê ú 4 C 2 C 2 C 2 C ê 4 3 6 5 ú ê ú 2C 6 -2C 1 2C 4 ú ê-4C 5 ê ú 2C 5 2C 4 -2C 2 ú -4C 6 ê ë û

From Eq. (3.129), the second-order derivatives of the three reduced invariants can be written in the matrix notation as

1 -4 4 -7 1 -4 [J 1,EE ] = - I 3 3 ({I 1,E }{I 3,E }T + {I 3,E }{I 1,E }T ) + I 1I 3 3 {I 3,E }{I 3,E }T - I 1I 3 3 [I 3,EE ] 3 9 3 8 2 - 53 10 2 -5 - 23 [J 2,EE ] = I 3 [I 2,EE ] - I 3 ({I 2,E }{I 3,E }T + {I 3,E }{I 2,E }T ) + I 2I 3 3 {I 3,E }{I 3,E }T - I 2I 3 3 [I 3,EE ] 3 9 3 1 1 - 23 1 [J 3,EE ] = - I 3 {I 3,E }{I 3,E }T + I 3 2 [I 3,EE ] 4 2 Note that the expressions of {I1,E}, {I2,E}, and {I3,E} are available in Section 3.5.2. Thus, the constitutive matrix can be obtained as [D] = A10 [J 1,EE ] + A01[J 2,EE ] + p[J 3,EE ]

▄

P3.29 A nearly incompressible rubber block is confined between two frictionless rigid walls as shown in the figure. When uniform pressure P is applied on the right end, the length of the block is changed by x1 = (1 - a)X1. When a = 0.1, (a) calculate the value of strain energy density and (b) the magnitude of applied pressure P on the right end.

Assume plane strain problem and use Mooney-Rivlin material with A10 = 80MPa, A01 = 20MPa, and K = 1,000MPa.

Rigid wall

X1

Rubber

P Figure P3.29

Solution: Since both X2 and X3 directions are fixed, the deformation of the rubber block can be written as x 1 = 0.9X 1,

x 2 = X 2,

x3 = X3

from which the deformation gradient and the right Cauchy-Green deformation tensor can be calculated as é.9 0 0ù ê ú ê F = ê 0 1 0úú , ê ú êë 0 0 1úû

é.81 0 0ù ê ú ê C = ê 0 1 0úú ê ú êë 0 0 1úû

The three eigenvalues of the right Cauchy-Green deformation tensor can be obtained as

l12 = l22 = 1,

l32 = 0.81

Using the three eigenvalues, the three invariants can be obtained as I 1 = 2.81,

I 2 = 2.62,

I 3 = 0.81

In addition, the reduced invariants are J 1 = 3.0145,

J 2 = 3.0151,

J 3 = 0.9

(a) Strain energy density K (J 3 - 1)2 2 = 80 ´ 0.0145 + 20 ´ 0.0151 + 500 ´ (-0.1)2 = 6.462

W = A10 (J 1 - 3) + A01(J 2 - 3) +

(b) Hydrostatic pressure p = K (J 3 - 1) = -100MPa

▄

P3.30 Consider a unit cube shown in Figure 3.15. Using an eight-node solid element, perform biaxial extension analysis using ABAQUS. Apply uniform extensions in both X1

Index 43 and X2 directions so that deformed shape will be 5´5´t3. Plot stress s11 and thickness t3 as a function of stretch.

Solution: The following program list shows the ABAQUS input file for the biaxial loading: *HEADING - INCOMPRESSIBLE HYPERELASTICITY (MOONEY-RIVLIN), BIAXIAL TENSION *NODE,NSET=ALL 1, 2,1.

3,1.,1., 4,0.,1., 5,0.,0.,1.

6,1.,0.,1.

7,1.,1.,1.

8,0.,1.,1. *NSET,NSET=FACE1 1,2,3,4 *NSET,NSET=FACE2 5,6,7,8 *NSET,NSET=FACE3 1,2,5,6 *NSET,NSET=FACE4 2,3,6,7 *NSET,NSET=FACE5 3,4,7,8 *NSET,NSET=FACE6 4,1,8,5 *ELEMENT,TYPE=C3D8RH,ELSET=ONE 1,1,2,3,4,5,6,7,8

*SOLID SECTION,ELSET=ONE,MATERIAL=MNEY *MATERIAL,NAME=MNEY *HYPERELASTIC, MOONEY-RIVLIN 80., 20., *STEP,NLGEOM,INC=20 BIAXIAL TENSION *STATIC,DIRECT 1.,20. *BOUNDARY,OP=NEW FACE1,3 FACE3,2 FACE6,1 FACE4,1,1,5.

FACE5,2,2,5. *EL PRINT,F=1 S, E, *NODE PRINT,F=1 U,RF *OUTPUT,FIELD,FREQ=1 *ELEMENT OUTPUT S,E *OUTPUT,FIELD,FREQ=1 *NODE OUTPUT U,RF *END STEP

The analysis results are shown in the following figure:

▄

4.10.

Exercise

P4.1 A force is gradually applied at the end of an elastoplastic bar such that it is in the plastic phase. When the total magnitude of strain is e = 0.003, calculated the applied force, axial stress, elastic strain, and plastic strain. Use the following material properties: E = 100GPa, H = 10GPa, and sY = 100MPa. The cross-sectional area of the bar is A = 1.0´10-4m2.

Solution: If the total strain is pure elastic, then stress will become s = Ee = 300MPa, which is larger than the yield stress. Thus, the material is in the plastic phase. It is convenient to separate the initial elastic deformation, e(1), until the yield stress from the elastoplastic deformation, e(2) such that e = e(1) + e(2). The initial elastic strain can be calculated from

e(1) = ee(1) = eY =

sY = 0.001, E

s(1) = sY = 100MPa

After the initial elastic deformation, the remaining deformation is elastoplastic deformation. In this phase, the strain increment is De(2) = 0.002. Since the total strain increment is given, Eq. (4.9) can be used to calculate the plastic strain increment as De(2) p =

De(2) = 0.00182 1+H /E

Thus, the plastic strain is ep = Dep(2) = 0.00182 , and elastic strain is ee = e - ep = 0.00118 . The axial stress can be calculated using the elastic strain, s = E ee = 118MPa . The applied force can be calculated from the assumption that the axial stress is uniformly distributed over the cross section: F = sA = 11.8kN . Below is the MATLAB program that solves for the problem: % % P4.1 elastoplastic bar (MPa, mm) % delE=0.003; A=100; mp = [1E5, 0, 1E4, 100]; nS=0; nA=0; nep=0; [Snew, Anew, epnew]=combHard1D(mp,delE,nS,nA,nep) eelast=delE - epnew Force = Snew*A

▄

P4.2 A force 12kN is gradually applied and then removed at the end of an elastoplastic bar. When the yield stress of the material is 100MPa, calculate plastic strains and tip displacement after removing the applied force. Use the following material properties: E = 100GPa and H = 10GPa. The cross-sectional area of the bar is A = 1.0´10-4m2 and the length of the bar is L0 = 1m.

Solution:

4-2

In a one-dimensional bar, it is assumed that the force is uniformly distributed over the cross section. During the loading process, since the total stress, s = F/A = 120MPa, is larger than the yield stress, it can be concluded that the material is under plastic deformation. It is convenient to divide the entire deformation into elastic and plastic phases. The material is initially elastic until it reaches yield stress. Thus, when stress reaches yield stress s(1) = sY = 100MPa, strain is purely elastic:

Dee(1) =

sY = 0.001 E

After yielding, the remaining stress increment, Ds = 20MPa, is in the plastic phase. The elastic and plastic strain increments can be calculated from

Dee(2) =

Ds(2) = 0.0002 E

Dep(2) =

Ds(2) = 0.002 H

Thus, the total elastic and plastic strains become ee = Dee(1) + Dee(2) = 0.0012 ep = De(2) p = 0.002

Now, during the unloading, elastic strain is removed, but the plastic strain remains. Thus, ep = 0.002 and u = epL = 2mm . ▄ P4.3 A uniaxial bar is under tensile force F = 12kN at load step tn. (a) When the plastic strain is epn = 0.002 , determine the yield status of the material. (b) If the applied force is increased to F = 15kN at load step tn+1, calculate plastic strain and tip displacement. Assume the initial yield stress sY = 100MPa, E = 100GPa and H = 10GPa. The cross-sectional area of the bar is A = 1.0´10-4m2 and the length of the bar is L0 = 1m. Assume isotropic hardening model.

Solution: (a) At load step tn, the stress in the bar is

sn =

F = 120MPa A

The yield stress is

sYn = sY0 + H epn = 100 + 10000 ´ 0.002 = 120MPa Since s n = sYn the material is in the loading stage. (b) During the loading stage up to tn, the elastic and plastic strains can be calculated by

4-3

een =

sn = 0.0012, E

epn = 0.002

At load step tn+1, the elastic and plastic strain increments can be calculated from stress increment Dsn +1 = 30MPa as

Deen +1 =

Dsn +1 = 0.0004, E

Depn +1 =

Dsn +1 = 0.004 H

Thus, the total elastic and plastic strains become een +1 = een + Deen +1 = 0.0016,

epn +1 = epn + Depn +1 = 0.006

The tip displacement is

u = (een +1 + epn +1 )L = 7.6mm ▄

P4.4 An elastoplastic bar is under variable load history. At load step tn, the stress and plastic strain are sn = 200MPa and epn = 1.0 ´ 10-4 , respectively. (a) Is the material in elastic or plastic state? (b) When strain increment is De = -0.003, calculate stress and plastic strain. Assume isotropic hardening with E = 200GPa, H = 25GPa, and sY = 250MPa. Solution: (a) At a given plastic strain epn = 1.0 ´ 10-4 , the yield stress is

sY = sY0 + H epn = 252.5MPa Since sn < sY, the material is in the elastic state. (b) For given strain increment, the trial stress can be obtained as Ds = E De = -600MPa,

s tr = s n + Ds = -400MPa

Since f tr =| str | -sY = 147.5 > 0 , the material is yielded in the compression side.

From Eq. (4.26), the plastic strain increment becomes Dep =

tr

f = 6.5556 ´ 10-4 E +H

Therefore, the stress and plastic strain are updated as

sn +1 = tr s - sgn( tr s)E Dep = -268.9MPa enp +1 = enp + Dep = 7.5556 ´ 10-4

4-4

Below is the MATLAB program that solves for the problem: % % P4.4 Elastoplastic bar (isotropic hardening) % delE=-0.003; nS=200; nA=0; nep=1E-4; mp=[2E5, 0, 2.5E4, 250]; [Snew, Anew, epnew]=combHard1D(mp,delE,nS,nA,nep)

▄

P4.5 Repeat Problem P4.4 using the kinematic hardening model. For back stress, use an = 2.5MPa. Solution: (a) At a given stress sn = 200MPa and back stress an = 2.5MPa , the trial shifted stress is hn = sn - an = 197.5MPa

Since hn < sY, the material is in the elastic state. (b) For given strain increment, the trial stress can be obtained as Ds = E De = -600MPa,

s tr = s n + Ds = -400MPa

Since f tr =| str - an | -sY = 152.5 > 0 , the material is yielded in the compression side. From Eq. (4.41), the plastic strain increment becomes Dep =

tr

f = 6.7778 ´ 10-4 E +H

Therefore, the stress, back stress and plastic strain are updated as

sn +1 = tr s - sgn( tr h)E Dep = -264.4MPa

an +1 = n a - sgn( tr h)H Dep = -14.444MPa enp +1 = enp + Dep = 7.7778 ´ 10-4 Below is the MATLAB program that solves for the problem: % % P4.5 Elastoplastic bar (kinematic hardening % delE=-0.003; nS=200; nA=0; nep=1E-4; mp=[2E5, 1, 2.5E4, 250]; [Snew, Anew, epnew]=combHard1D(mp,delE,nS,nA,nep)

▄

P4.6

Repeat Problem 4.5 using the combined hardening model with b = 0.5.

4-5 Solution: (a) At a given stress sn = 200MPa, back stress an = 2.5MPa , and plastic strain epn = 1.0 ´ 10-4 , the trial shifted stress and the yield stress are hn = sn - an = 197.5MPa

sY = sY0 + (1 - b )H enp = 251.25MPa Since hn < sY, the material is in the elastic state. (b) For given strain increment, the trial stress can be obtained as Ds = E De = -600MPa,

s tr = s n + Ds = -400MPa

Since f tr =| str - an | -sY = 151.25 > 0 , the material is yielded in the compression side. From Eq. (4.41), the plastic strain increment becomes Dep =

tr

f = 6.2222 ´ 10-4 E +H

Therefore, the stress, back stress and plastic strain are updated as

sn +1 = tr s - sgn( tr h)E Dep = -265.6MPa an +1 = n a - sgn( tr h)bH Dep = -5.9028MPa

enp +1 = enp + Dep = 7.2222 ´ 10-4 Below is the MATLAB program that solves for the problem: % % P4.6 Elastoplastic bar (combined hardening) % delE=-0.003; nS=200; nA=0; nep=1E-4; mp=[2E5, 0.5, 2.5E4, 250]; [Snew, Anew, epnew]=combHard1D(mp,delE,nS,nA,nep)

▄ P4.7 For the combined isotropic/kinematic hardening model, derive the expression of plastic strain increment from the plastic consistency condition.

Solution: The plastic consistency condition means that the yield function at the current time step, tn+1, remains zero; that is, n +1

f =

n +1

h - n + 1 sY = 0

The following stress update formulas and yield stress are used for the plastic consistency condition: n +1

s = tr s - sgn( tr h)E Dep

4-6 n +1

a = n a + sgn( tr h)bH Dep

n +1

sY = n sY + (1 - b )H Dep

By substituting the above 3 formulas into the plastic consistency condition, The above consistency condition can be expanded in terms of plastic strain increment as



tr

s - sgn(tr h)E Dep - tr a - sgn( tr h)bH Dep - ( n sY + (1 - b)H Dep ) = 0

tr

s - tr a - n sY - (E + H )Dep = 0

Note that the above formula works for both positive and negative trial stress. Therefore, the plastic strain increment can be obtained as D ep =

tr

f E +H

Note that the formula for plastic strain increment is identical to that of isotropic hardening in Eq. (4.26) and kinematic hardening in Eq. (4.41). ▄ P4.8 An elastoplastic bar is clamped at the left end, and variable loads are applied at the right end, as shown in the table. Plot the stress-strain curve by changing the applied forces by 5kN increments. Assume the following material properties with isotropic hardening: E = 70GPa, H = 10GPa, sY = 250MPa. The length of the bar is L = 1m, and the cross-sectional area is A = 1.0´10-4m2.

Load step Force (kN)

1 30

2 20

3 35

4 20

Solution: Since the given condition is applied force at the tip, the convergence iteration must be performed to find equilibrium. Below is the MATLAB program that solves for the problem: % % P4.8 Variable loadings % E=70000; H=10000;Et=1000; sYield=250; mp = [E, 0, H, sYield]; Et=E*H/(E+H); nS=0; nep=0; nA=0; A = 100; L = 1000; tol = 1.0E-5; u = 0; iter=0; Res=0; Force = 1000*[5:5:30 25 20 25 30 35 30 25 20]; N = size(Force',1); X=zeros(N,1);Y=zeros(N,1); fprintf('\nstep iter u S ep Residual'); fprintf('\n %3d %3d %7.4f %7.3f %8.6f %10.3e',i,iter,u,nS,nep,Res); for i=1:N P = Force(i); iter = 0; Res = P - nS*A; conv = Res^2/(1+P^2);

4-7 du=0; while conv > tol && iter < 20 Eep = E; if epnew>nep; Eep = Et; end delu = Res / (Eep*A/L); du = du + delu; delE = du / L; [Snew, Anew, epnew]=combHard1D(mp,delE,nS,nA,nep); Res = P - Snew*A; conv = Res^2/(1+P^2); iter = iter + 1; end u=u+du; nS = Snew; nep = epnew; X(i) = u; Y(i) = nS; fprintf('\n %3d %3d %7.4f %7.3f %8.6f %10.3e',i,iter,u,nS,nep,Res); end X=[0;X];Y=[0;Y];plot(X,Y);

▄

P4.9 An elastoplastic bar is clamped at the left end, and variable displacements are applied at the right end, as shown in the table. Plot the stress-strain curve by changing the tip displacement by 1mm increments. Assume the following material properties with isotropic hardening: E = 70GPa, H = 10GPa, sY = 250MPa. The length of the bar is L = 1m, and the cross-sectional area is A = 1.0´10-4m2.

Load step Displacement (mm)

1 5.0

2 3.0

3 7.0

4 6.0

Solution: Since the tip displacements are given, there is no need to perform convergence iteration.

At each step, displacement increment is calculated by the difference between current and previous load increments. After that, the strain increment is calculated from the small deformation assumption. The program combHard1D will calculate stress and plastic

4-8

strain for the given strain increment. Below is the MATLAB program that solves for the problem: % % P4.9 Variable displacement % E=70000; H=10000;Et=1000; sYield=250; mp = [E, 0, H, sYield]; Et=E*H/(E+H); nS=0; nep=0; nA=0; A=100; L=1000; tol = 1.0E-5; u=0; iter=0; Res=0; disp=[0 1 2 3 4 5 4 3 4 5 6 7 6]; N = size(disp',1); X=zeros(N,1);Y=zeros(N,1); fprintf('\nstep u S ep'); fprintf('\n %3d %7.4f %7.3f %8.6f',i,u,nS,nep); for i=2:N delu = disp(i) - disp(i-1); delE = delu / L; [Snew, Anew, epnew]=combHard1D(mp, delE, nS, nA, nep); nS = Snew; nep = epnew; X(i) = disp(i); Y(i) = nS; fprintf('\n %3d %7.4f %7.3f %8.6f',i,u,nS,nep); end plot(X,Y);

▄

P4.10 A force of P = 15 is applied to the two parallel bars in Example 4.2 and then removed. Using combHard1D program, calculate tip displacement and residual stresses for the two bars after unloading. Use 15 load increments for each loading and unloading cycle. Plot stresses vs. tips displacement in the XY graph.

Solution: Below is the MATLAB program to solve loading/unloading cycle for two parallel bars in Example 4.2. After unloading, the tip displacement is u = 0.1227, and stresses are s1 =

4-9

-4.091, s2 = 2.455. Since the area of bar2 is twice than that of bar1, the stress s2 is half of s1. % % P4.10 Two-bar in parallel - unloading % E1=10000; Et1=1000; sYield1=5; E2=5000; Et2=500; sYield2=7.5; mp1 = [E1, 0, E1*Et1/(E1-Et1), sYield1]; mp2 = [E2, 1, E2*Et2/(E2-Et2), sYield2]; nS1 = 0; nA1=0; nep1 = 0; epnew1=0; nS2 = 0; nA2=0; nep2 = 0; epnew2=0; A1 = 0.75; L1 = 100; A2 = 1.25; L2 = 100; tol = 1.0E-5; u = 0; Force = [1:15 14:-1:0]; N = size(Force',1); X=zeros(N,1);Y1=zeros(N,1);Y2=zeros(N,1); fprintf('\nstep iter u S1 S2 ep1 ep2 fprintf('\n %3d %3d %7.4f %7.3f %7.3f %8.6f %8.6f %10.3e',...

0,0,u,nS1,nS2,nep1,nep2,0); for i=1:N P = Force(i); iter = 0; Res = P - nS1*A1 - nS2*A2; conv = Res^2/(1+P^2); while conv > tol && iter < 20 Eep1 = E1; if epnew1>nep1; Eep1 = Et1; end Eep2 = E2; if epnew2>nep2; Eep2 = Et2; end delu = Res / (Eep1*A1/L1 + Eep2*A2/L2); u = u + delu; delE = delu / L1; [Snew1, Anew1, epnew1]=combHard1D(mp1, delE, nS1, nA1, nep1); [Snew2, Anew2, epnew2]=combHard1D(mp2, delE, nS2, nA2, nep2); Res = P - Snew1*A1 - Snew2*A2; conv = Res^2/(1+P^2); iter = iter + 1; nS1 = Snew1; nep1 = epnew1; nA1 = Anew1; nS2 = Snew2; nep2 = epnew2; nA2 = Anew2; end X(i) = u; Y1(i) = nS1; Y2(i) = nS2; fprintf('\n %3d %3d %7.4f %7.3f %7.3f %8.6f %8.6f %10.3e',... i,iter,u,nS1,nS2,nep1,nep2,Res); end X=[0;X];Y1=[0;Y1];Y2=[0;Y2];plot(X,Y1,X,Y2);

Residual');

4-10

▄

P4.11 A force 12kN is gradually applied at the end of an elastoplastic bar. When the yield stress of the material is 100MPa, calculate displacement at the tip. Use the following material properties: E = 100GPa and H = 10GPa. The cross-sectional areas of the bars are A(1) = 1.0´10-4m2 and A(2) = 0.5´10-4m2.

1

2

3

E, A(1)

E, A(2)

L0=1m

L0=1m

F

x Figure P4.11

Solution: Since the two bars are connected in parallel, the element forces of the two bars are the same as the applied force at the tip. By considering the cross-sectional areas of the bars, the stresses of the bars can be calculated by

s(1) =

F A(1)

= 120MPa,

s(2) =

F A(2)

= 240MPa

Since both bars are in the plastic state, it is possible to separate the entire deformation into the initial yielding stage (at s = 100MPa) and followed by the elastoplastic state.

For bar1, the elastic strain at the initial yielding stage can be calculated from

ee(1) =

sY = 0.001 E

After that, the remaining Ds = 20MPa is in the elastoplastic stage:

4-11

ee(1) =

Ds(1) = 0.0002, E

e(1) p =

Ds(1) = 0.002, H

e(1) = ee(1) + ep(1) = 0.0022

Thus the total strain becomes e(1) = 0.001 + 0.0002 + 0.002 = 0.0032 , and the displacement at Node 2 becomes

u2 = e(1)L0 = 0.32mm For bar2, the strain at the initial yielding stage can be calculated from

ee(2) =

sY = 0.001 E

After that, the remaining Ds = 140MPa is in the elastoplastic stage:

ee(2) =

Ds(2) = 0.0014, E

e(2) p =

Ds(2) = 0.014, H

e(2) = ee(2) + e(2) p = 0.0154

the total strain becomes e(1) = 0.001 + 0.0014 + 0.014 = 0.0164 , and the displacement at Node 3 becomes

u3 = u2 + e(2)L0 = 1.96mm The same problem can be solved using combHard1D program with convergence iteration. The following MATLAB program solves for the displacement and stress for the two-bar problem: % % P4.11 Two-bar in serial % E=100000; H=10000; sYield=100;Et=E*H/(E+H); mp = [E, 0, H, sYield]; nS1 = 0; nA1=0; nep1=0; epnew1=0; Eep1=E; nS2 = 0; nA2=0; nep2=0; epnew2=0; Eep2=E; A1 = 100; L1 = 100; A2 = 50; L2 = 100; tol = 1.0E-8; iter = 0; u = [0 0]'; F = [0 12000]'; Res = F - [nS1*A1-nS2*A2;nS2*A2]; conv = norm(Res)^2/(1+norm(F)^2); fprintf('\n iter u1 u2 S1 S2 ep1 ep2 Residual'); fprintf('\n %3d %7.4f %7.4f %7.3f %7.3f %8.6f %8.6f %10.2e %10.2e',... iter,u(1),u(2),nS1,nS2,nep1,nep2,Res); while conv > tol && iter < 20 Kt = [Eep1*A1/L1+Eep2*A2/L2,-Eep2*A2/L2;-Eep2*A2/L2,Eep2*A2/L2]; delu = Kt\Res; u = u + delu; delE1 = delu(1) / L1; delE2 = (delu(2)-delu(1)) / L2; [Snew1, Anew1, epnew1]=combHard1D(mp, delE1, nS1, nA1, nep1); [Snew2, Anew2, epnew2]=combHard1D(mp, delE2, nS2, nA2, nep2); Eep1 = E; if epnew1>nep1; Eep1 = Et; end Eep2 = E; if epnew2>nep2; Eep2 = Et; end nS1 = Snew1; nep1 = epnew1; nA1=Anew1; nS2 = Snew2; nep2 = epnew2; nA2=Anew2; Res = F - [nS1*A1-nS2*A2;nS2*A2]; conv = norm(Res)^2/(1+norm(F)^2);

4-12 iter = iter + 1; fprintf('\n %3d %7.4f %7.4f %7.3f %7.3f %8.6f %8.6f %10.2e %10.2e',... iter,u(1),u(2),nS1,nS2,nep1,nep2,Res); end

As shown in the output below, the Newton-Raphson iteration converges in the second iteration. iter 0 1 2

u1 0.0000 0.1200 0.3200

u2 S1 S2 ep1 ep2 Residual 0.0000 0.000 0.000 0.000000 0.000000 0.00e+000 0.3600 101.818 112.727 0.000182 0.001273 -4.55e+003 1.9600 120.000 240.000 0.002000 0.014000 0.00e+000

1.20e+004 6.36e+003 0.00e+000

▄

P4.12 Two one-dimensional bars are connected serially as shown in the figure. At load step n, bar1 was plastic and bar2 was elastic. At load step n+1, the increments of nodal displacements are given as Du = [Du1, Du2, Du3] = [0.0, -0.01, 0.0]. Calculate stresses and plastic strains of both bars at load step n+1.

1

3

2 bar1 L=100

bar2 L=100

Figure P4.12

Young modulus (E)

bar1 10,000

bar2 5,000

Tangent modulus (Et)

1,000

500

Previous stress (sn)

6.0

7.4

Initial yield stress (sY)

5.0

7.5

9E-4 Plastic Isotropic

0.0 Elastic Isotropic

Plastic strain (ep) Yield status Hardening

Solution: Bar1: From the given nodal displacements, the strain of the element can be calculated by De =

Du2 - Du1 = -10-4 L1

First, assume that the strain increment is purely elastic to obtain the following trial state:

str = sn + E De = 5.0,

sYn = sY0 + H epn = 6.0

Since the material is initially in the plastic state and the stress is positive at load step n, the bar yielded in tension. However, since the incremental strain is negative, the bar is under unloading. Thus, the material becomes elastic and

4-13

sn +1 = str = 5.0,

epn +1 = epn = 9 ´ 10-4

Bar2: From the given nodal displacements, the strain of the element can be calculated by De =

Du3 - Du2 = 10-4 L1

First, assume that the strain increment is purely elastic to obtain the following trial state:

str = sn + E De = 7.9,

sYn = sY0 + H epn = 7.5

Since str > sYn , the material becomes plastic in this load increment. Since the element is elastic at load step n, the ratio R needs to be calculated, which can be written as

R = 1-

str - sYn Ds

= 0.2

Thus, the updated stress and plastic strain become

sn +1 = sn + RDs + Et (1 - R)De = 7.54 epn +1 = epn +

1-R De = 7.2 ´ 10-5 1+H /E

▄

P4.13 Write the expression of the rank-4 unit symmetric tensor and unit deviatoric tensor in the 6´6 matrix notation.

Solution: The index of rank-2 tensor can be converted into a vector via [11 22 33 12 23 13]. From the definition of I ijkl = (dik djl + dil djk ) / 2 , the matrix version of the fourth-order unit symmetric tensor becomes é1 ê ê0 ê ê0 ê I=ê ê0 ê ê0 ê ê0 ë

0 0 0 0 0 ùú 1 0 0 0 0 úú 0 1 0 0 0 úú ú 0 0 21 0 0 ú ú 0 0 0 21 0 ú ú 0 0 0 0 21 ú û

In addition, from the definition of rank-4 unit deviatoric tensor Idev = I - 13 1 Ä 1 , its matrix version becomes

4-14

é 2 -1 -1 0 0 ê 3 3 3 ê- 1 2 -1 0 0 ê 3 3 3 ê- 1 - 1 2 0 0 ê 3 3 IDEV = ê 3 0 0 21 0 ê 0 ê 0 0 0 21 ê 0 ê ê 0 0 0 0 0 ë

0 ùú 0 úú 0 úú ú 0ú ú 0ú ú 1ú 2û

▄

P4.14 A solid shaft as shown in the figure is subjected to tensile force P and a torque T. The force and torque are such that the normal stress sxx = s and shear stress t = s. The shear stress is along the circumference of the shaft. Using the von Mises criterion, determine the values of s when the material yields first time. The yield stress from the uniaxial tension test is sY. X2

P X1 T

X3

Figure P4.14 Solution: In the case of P, the stress component is fixed to be s11. In the case of T, the shear stress component varies at different location on the boundary of the cross-section. For the simplicity, let’s consider the case in which t12 = s. Thus, the stress matrix and deviatoric stress become é2s s 0 ùú ê3 s = êê s - 13 s 0 úú ê 0 - 13 s úú 0 êë û

és s 0ù ê ú s = êê s 0 0 úú , ê 0 0 0ú úû ëê

The norm of deviatoric stress can be calculated by s =s

4 + 1 + 1 +1+1 = s 9 9 9

8 3

The yield criterion is f = s -

2s 3 Y

=

8s 3

Thus, the first yielding starts when

s=

1 s 2 Y

2s 3 Y

=0

4-15 ▄

P4.15 A plane stress plate is under biaxial stress state in which sxx = -syy = s. When the applied load is proportional, determine s when the material yields first time. The yield stress from the uniaxial tension test is sY. Solution: In the case of biaxial loading, the stress matrix and deviatoric stress become és 0 0ù ê ú s = s = êê 0 -s 0 úú ê 0 0 0ú êë úû

The norm of deviatoric stress can be calculated by s =s 2

The yield criterion is f = s -

2s 3 Y

=

2s -

2s 3 Y

=0

Thus, the first yielding starts when s=

1s 3 Y

▄

P4.16 A square is under proportional loading with shear stress t12 = t21 = t. When the effective plastic strain is ep = 0.1, calculate the value of shear stress. Consider three different hardening models: (a) isotropic, (b) kinematic, and (c) combined hardening with b = 0.5. Assume that the initial yield stress is 400MPa and the plastic modulus is H = 200MPa. Solution: Since the applied stress is proportional loading, it is expected that the material is in the plastic phase, and all three models provide the same stress value. The difference occurs only when the direction of loading changes. In the case of pure shear loading, the stress and deviatoric stress become é 0 t 0ù ê ú s = s = êê t 0 0 úú ê 0 0 0ú êë úû

Thus, the norm of the deviatoric stress becomes

4-16

s = 2t

(a) Isotropic hardening: from the definition of yield function, s -

2 (s 0 3 Y

2 (400 + 200 ´ 0.1) = 0 3

+ He p ) = 2t -

s = 171.5MPa (b) Kinematic hardening: from the definition of yield function, s-a -

2s 0 3 Y

= 0

Note that Da is parallel to h and the loading direction remains fixed, and thus, a is parallel to s. In that case, the norm of the shifted stress can be written as s - a = s - a . Thus, the yield function can be rewritten as s-a -

2s 0 3 Y

= s - a -

t = 21

2 (s 0 3 Y

2s 0 3 Y

2 He p 3

= 2t -

2s 0 3 Y

= 0

+ He p ) = 171.5MPa

(c) Combined hardening: Similar to the kinematic hardening model, a is parallel to s.

Thus, the yield function can be written as 2 é s 0 + (1 - b )He ù p úû 3 êë Y 0 2 é a - 3 ê sY + (1 - b )Hep ùú ë û 2 b He - 2 s 0 - 2 (1 - b )He p p 3 3 Y 3

s-a = s = 2t =0

Thus, the applied stress can be solved for t = 21

2 (s 0 3 Y

+ He p ) = 171.5MPa

Note that all three models provide the same stress value.

▄

P4.17 A pure shear deformation is applied to the square element as shown in the figure such that s12 = s21 is only non-zero stress component. At load step n, the stress value was s12 = 50, and there was no plastic deformation. At load step n+1, incremental strain De12 = De21 = 0.005 is applied. Calculate stress components and effective plastic strain at load step n+1. Use the following material properties: shear modulus m = 1,000, plastic modulus H = 100, initial yield stress sY = 100.

4-17 X2

Figure P4.17 X1 Solution: Since s12 is only non-zero component, it is convenient to work with it as a scalar rather n = 50 and De12 = 0.005 , the trial than stress matrix. With given information of s12 state can be calculated as tr n s12 = s12 + 2mDe12 = 60

The yield state can be tested using the yield function as f = htr -

2sn 3 Y

=

tr 2s12 -

2s 0 3 Y

= 3.2032 > 0

Thus, the material becomes plastic in this load increment. The plastic consistency parameter can be calculated by

Dg =

f = 0.00155 2m + 23 H

In addition, the deviatoric unit tensor for the trial state becomes

N=

htr htr

é0 1 0ù ú 1 êê ú 1 0 0 = ê ú 2ê ú 0 0 0 êë úû

Thus, only s12 will be updated due to plastic deformation. The updated stress and plastic strain become n +1 tr s12 = s12 -

epn +1 = epn +

2mDg 2

= 57.808

2Dg = 0.00127 3

Below is the MATLAB program that solves the above problem: % % P4.17 shear deformation of a square % lambda=1000; mu=1000; mp = [lambda, mu, 0, 100, 100];

4-18 Iden=[1 1 1 0 0 0]'; deps=[0 0 0 0.01 0 0]'; stressN=[0 0 0 50 0 0]'; alphaN=[0 0 0 0 0 0]'; epN=0; D=2*mu*eye(6) + lambda*Iden*Iden'; D(4,4) = mu; D(5,5) = mu; D(6,6) = mu; [stress, alpha, ep]=combHard(mp,D,deps,stressN,alphaN,epN)

▄

P4.18 Displacements of a simple shear deformation in the figure can be expressed by u1 = kx2, u2 = 0. At load step n, k = 0.016 and the material is elastic. At load step n+1, Dk = 0.008. Calculate stress and plastic strain. Check if the updated state is on the yield function; i.e., f (s n +1, epn +1 ) = 0 . Use the following material properties: shear modulus m = 100, plastic modulus H = 10, initial yield stress sY = 12 . x2

F

x1 Figure P4.18 Solution: From the infinitesimal deformation assumption, the strain at load step n can be calculated by é 0 en = êê 1 êë 2 k

1k ù 2 ú

0 úú û



n e12 = 0.008,

De12 = 0.004

Since e12 is only non-zero component, strain will be considered as a scalar in the following calculations. For the given strain at load step n, the stress becomes

é 0 sn = sn = 2me = 2me = êê êë mk

é 0 1.6 ù mk ù ú=ê ú ê 1.6 0 ú , 0 úú ê ú û ë û



n s12 = 1.6

Again, since s12 is only non-zero component, stress will be considered as a scalar. For a given increment De12, the trial stress becomes tr n s12 = s12 + 2mDe = 2.4

The yield function can be used to check the status of the material, as

4-19 f = str 12 -

2s 0 3 Y

tr 2 2(S12 ) - 8 = 0.566 > 0

=

Thus, the material becomes plastic in this increment, yieldStatus = 1. The unit deviatoric tensor and the plastic consistency parameter can be calculated by

N=

Dg =

str str

=

1 éê 0 2.4 ùú 3.394 êêë 2.4 0 úúû

f 0.5657 = = 0.002737 2 200 + 6.667 2m + 3 H

Using the unit deviatoric tensor and the plastic consistency parameter, the stress and the effective plastic strain can be updated by

é 0 2.4 ù 0.5473 é 0 2.4 ù é 0 2.013 ù úê ú=ê ú sn +1 = str - 2mDg N = êê ú ê ú ê ú 2.4 0 2.4 0 2.013 0 3.394 êë êë úû úû êë úû

epn +1 = epn +

2 Dg = 0.002234 3

After the plastic deformation, the yield surface is expanded by

k(epn +1 ) = sY0 + Hepn +1 = 3.4864 Thus, the yield function after update becomes f (sn +1, e pn +1 ) = sn +1 -

2k(e n +1 ) = 0 p 3

Thus, the updated state of the material is on the surface of the yield function.

Below is the MATLAB program that solves the above problem: % % P4.18 shear deformation of a square % lambda=100; mu=100; mp = [lambda, mu, 0, 10, sqrt(12)]; Iden=[1 1 1 0 0 0]'; stressN=[0 0 0 0 0 0]'; alphaN=[0 0 0 0 0 0]'; epN=0; D=2*mu*eye(6) + lambda*Iden*Iden'; D(4,4) = mu; D(5,5) = mu; D(6,6) = mu; deps=[0 0 0 0.016 0 0]'; [stress, alpha, ep]=combHard(mp,D,deps,stressN,alphaN,epN); stressN=stress; alphaN=alpha;epN=ep; deps=[0 0 0 0.008 0 0]'; [stress, alpha, ep]=combHard(mp,D,deps,stressN,alphaN,epN)

▄ P4.19 At load step tn, a unit cube is under unaxial stress state with s11 = 100MPa, and all other stress components and plastic variables are zero. At load step tn+1, additional shear stress is applied such that Dg12 = 0.002. Determine stress, back stress, and

4-20

effective plastic strain. Assume the following material properties: l = m = 100GPa, H = 10GPa, sY = 100MPa, combined isotropic/kinematic hardening with b = 0.5.

Solution: At load step tn, the material reaches the initial yield stress. At load step tn+1, the trial stress becomes tr

s = n s + D : De = [100 0 0 200 0 0]T

Since back stress is zero at load step tn, the deviatoric stress is the same with the shifted stress as tr

h = tr s = tr s - tr ( tr s)1 = [66.67 -33.33 -33.33 200 0 0]T

From the shifted stress, the unit deviatoric tensor that is normal to the yield surface can be obtained as N=

tr

h

tr

h

= [.2265 -.1132 -.1132 .6794 0 0]T

Using the norm of the shifted stress, the yield function can be evaluated as

f =

tr

h -

2s 0 3 Y

= 294.4 - 100

2 = 212.7 > 0 3

It is clear that the material is in the plastic state. The plastic consistency parameter can be calculated from g =

f = 0.001 2m + 23 H

Using  and N, the stress, back stress, and effective plastic strain can be updated as n +1

n +1

s = tr s - 2mg N = [53.4 23.3 23.3 60.1 0 0]T

a = n a + 23 b H g N = [.777 -.389 -.389 2.331 0 0]T n +1

e p = ne p +

2 g = 8.405 ´ 10-4 3

Below is the MATLAB program that solves for the above problem: lambda=100000; mu=100000; mp = [lambda, mu, 0.5, 10000, 100]; Iden=[1 1 1 0 0 0]'; stressN=[100 0 0 0 0 0]'; alphaN=[0 0 0 0 0 0]'; epN=0; D=2*mu*eye(6) + lambda*Iden*Iden'; D(4,4) = mu; D(5,5) = mu; D(6,6) = mu; deps=[0 0 0 0.002 0 0]'; [stress, alpha, ep]=combHard(mp,D,deps,stressN,alphaN,epN)

4-21 ▄

P4.20 Using Abaqus perform a uniaxial tension test of a unit cube (C3D8) in x3direction. Assume elastoplastic material with linear isotropic hardening (E = 2.0E5,  = 0.3, Y = 200, H = 2.0E4). Displace at x3 = 1 surface is controlled as shown in the figure with three steps. Use 10 increments in each step. Plot stress-strain curve for all 30 increments.

0.006 0.004

0.001 t1

t2

t3

Figure P4.20

Solution: The following program list shows the ABAQUS input file for the elastoplastic loading:

4-22

*HEADING MISES PLASTICITY/LINEAR ELASTICITY, UNIAXIAL TENSION TEST, C3D8 *NODE,NSET=ALLN 1,0.,0.,0.

2,1.,0.,0.

3,1.,1.,0.

4,0.,1.,0.

5,0.,0.,1.

6,1.,0.,1.

7,1.,1.,1.

8,0.,1.,1. *ELEMENT,TYPE=C3D8,ELSET=ALLE 1,1,2,3,4,5,6,7,8 *SOLID SECTION,ELSET=ALLE,MATERIAL=ALLE *MATERIAL,NAME=ALLE *ELASTIC 200.E3,.3 *PLASTIC 200.,0.

400.,.01 *BOUNDARY 1,PINNED 2,2 5,2 6,2 4,1 5,1 8,1 2,3 3,3 4,3 *STEP,INC=10 *STATIC,DIRECT

*BOUNDARY 7,3,,.001 5,3,,.001 6,3,,.001 8,3,,.001 *EL PRINT,FREQ=1 S, E, EP, *NODE PRINT U,RF *EL FILE,FREQ=1 S, E,EP *END STEP *STEP,INC=10 *STATIC,DIRECT 1.,10. *BOUNDARY 7,3,,.006 5,3,,.006 6,3,,.006 8,3,,.006 *EL PRINT,FREQ=1 S, E, EP, *NODE PRINT U,RF *EL FILE,FREQ=1 S, E,EP *END STEP

The analysis results are shown in the following figure:

▄

4-23 P4.21 Calculate Dep and Dalg for one-dimensional elastoplasticity problem using the von Mises yield criterion and linear combined isotropic/kinematic hardening. Assume material properties: (E , H , sY0 , b ) .

Solution: (a) In one-dimensional problems, the lateral strains are ignored, and stress s and strain e are scalar. In addition, there is no need to calculate deviatoric stress and strain. The yield function becomes

f (h,ep ) = s - a - (sY0 + (1 - b )Hep ) £ 0 Note that the above yield function is slightly different from the multi-dimensional yield function in Eq. (4.84) because equivalent stress and effective strain are not used. The constitutive relation is s = E (e - e p ) with e p = ep = g , and the hardening models are

a = bH g There is no difference between the plastic strain and effective plastic strain. From the consistency condition:

f = E (e - e p ) - a - (1 - b )Hep = 0 The plastic consistency parameter can be calculated as

g =

E e E +H

Thus, æ E ÷ö EH s = E e - E e p = E çç 1 e ÷÷ e = E +H ø E +H èç

Thus, the elastoplastic tangent modulus can be obtained as

Dep =

EH E +H

Note that the above Dep is the same as the tangent modulus in Eq. (4.8). (b) In numerical integration, the trial stress is obtained from tr

s = n s + E De,

tr

a = n a,

tr

e p = ne p

Using the property of De p = Dep = Dg , the plastic return mapping becomes n +1

s = tr s - E Dg,

The yield function becomes

n +1

a = tr a + b H Dg,

n +1

ep = trep + Dg

4-24

f ( n +1h, n +1ep ) = tr h - (E + b H )Dg - (sY0 + (1 - b )H ( tre p + Dg )) = 0

from which the plastic consistency parameter can be calculated by tr

Dg =

h - (sY0 + (1 - b )H trep ) (E + H )

For the consistent tangent stiffness, it is clear that N is fixed in one-dimensional problems. The derivative of the plastic consistency parameter with respect to strain increment becomes

¶Dg E = ¶De E +H Thus, the consistent tangent stiffness becomes

D alg = E -

E2 EH = E +H E +H

Note that Dalg = Dep because in the case of one-dimensional case, the direction N is fixed. ▄

P4.22 In the saturated isotropic hardening model, the yield stress starts from initial value of sY0 and approaches sY¥ as the plastic strain increases. é

æ

öù

ç ep ÷÷ ú k(ep ) = sY0 + (sY¥ - sY0 ) êê 1 - exp çç ÷ú ççè ep¥ ÷÷ø úû êë

Since the hardening model is nonlinear, it is required to have a local Newton-Raphson method to find the plastic consistency parameter. Modify MATLAB program combHard so that it can solve for the above saturated isotropic hardening model. Test the program by solving the pure shear problem in P4.15. Assume the following material properties: shear modulus m = 1,000, initial yield stress sY = 100, asymptotic yield stress sY¥ = 200, and asymptotic effective plastic strain e p¥ = 0.05.

Solution: The return-mapping algorithm will be similar to the linear isotropic hardening model except for the yield function and the local Newton-Raphson method to find the plastic consistency parameter.

1. Yield function f

k

=

tr

2. Jacobian relation

h -

æ

2ç çç sY0 3ç

çè

k öù ö é æ çç e p ÷÷ ú ÷÷ ¥ 0 ê + (sY - sY ) ê 1 - exp ç ÷ú ÷ ççè e p¥ ÷ø÷ úû ø÷÷ êë

4-25

æ e k öù ¶f 2 (sY¥ - sY0 ) éê ç p ÷ ç 1 exp = 2m + - ÷÷÷ úú ç ê ççè e ¥ ÷ø ú ¶g 3 ep¥ êë p û 3. Update the plastic consistency parameter and effective plastic strain

g k +1 = g k +

fk , ¶f ¶g

epk +1 = nep +

2 g k +1 3

4. Check convergence If (½fk½ > fTOL) k = k + 1 and go to Step 1 If (k > kMAX) stop with error message Below is MATLAB program, combHardSat, which can solve for stress and effective plastic strain from the saturated hardening model: % % Saturated isotropic hardening model % function [stress, ep]=combHardSat(mp,D,deps,stressN,epN) % Inputs: % mp = [lambda, mu, epinf, Y0, Yinf]; % D = elastic stiffness matrix % stressN = [s11, s22, s33, t12, t23, t13]; % Iden = [1 1 1 0 0 0]'; two3 = 2/3; stwo3=sqrt(two3); %constants mu=mp(2);epinf=mp(3);Y0=mp(4);Yinf=mp(5); %material properties ftol = Y0*1E-6; %tolerance for yield stresstr = stressN + D*deps; %trial stress I1 = sum(stresstr(1:3)); %trace(sigmatr) eta = stresstr - I1*Iden/3; %deviatoric stress etat = sqrt(eta(1)^2 + eta(2)^2 + eta(3)^2 ... + 2*(eta(4)^2 + eta(5)^2 + eta(6)^2));%norm of eta fyld = etat - stwo3*(Y0+(Yinf-Y0)*(1-exp(-epN/epinf))) if fyld < ftol %yield test stress = stresstr; ep = epN; %trial states are final return; else iter =0; gamma = 0; ep = epN; while fyld > ftol %local N-R iteration iter=iter+1; if iter > 40; error('Fail to compute gamma'); end; dfdg = 2*mu+two3*((Yinf-Y0)*exp(-ep/epinf)/epinf) gamma = gamma + fyld/dfdg %update plast cons param ep = epN + stwo3*gamma; fyld = etat -2*mu*gamma - stwo3*(Y0+(Yinf-Y0)*(1-exp(-ep/epinf))) end end N = eta/etat; %unit vector normal to f stress = stresstr - 2*mu*gamma*N; %updated stress

Below is the MATLAB program that can solve for P4.15 with the saturated isotropic hardening model:

4-26 % % P4.22 shear deformation of a square (saturated isotropic hardening) % lambda=1000; mu=1000;epinf=0.05;Y0=100;Yinf=200; mp = [lambda, mu, epinf, Y0, Yinf]; Iden=[1 1 1 0 0 0]'; deps=[0 0 0 0.01 0 0]'; stressN=[0 0 0 50 0 0]'; alphaN=[0 0 0 0 0 0]'; epN=0; D=2*mu*eye(6) + lambda*Iden*Iden'; D(4,4) = mu; D(5,5) = mu; D(6,6) = mu; [stress, ep]=combHardSat(mp,D,deps,stressN,epN)

The local Newton-Raphson converged after the second iteration.

Iteration 1: f = 3.2032, df/dg = 3.33E3, g = 9.61E-4 Iteration 2: f = 0.01, df/dg = 3.31E3, g = 9.64E-4 Iteration 3: f = 9.77E-8 After convergence, the stress and effective plastic strain are updated to n +1

s12 = 58.63,

n +1

ep = 7.87 ´ 10-4

▄

P4.23 An plane strain square undergoes the following elastic deformation: x 1 = X1 + kX 2 ,

x 2 = X 2,

x3 = X3

Using the linear relationship between principal Kirchhoff stress and logarithmic stretch, find the Kirchhoff stress tensor when k = 0.02. Use the following material properties: l = m = 100GPa. Solution: Since the cube is elastic, there is no need to separate elastic and plastic part of deformation. Thus, the superscript ‘e’ will be omitted in the following derivation. For given deformation, the deformation gradient and left Cauchy-Green deformation tensor becomes é 1 0.02 0 ù ê ú 1 0 úú , F = êê 0 ê0 0 1 úú ëê û

é 1.0004 0.02 0 ù ê ú 1 0 úú b = FFT = êê 0.02 ê 0 0 1 úú ëê û

The three eigenvalues and eigenvectors of b become l1 = 1.02,

n1 = [-0.711 -0.704 0]T

l2 = 0.98,

n 2 = [0.704 -0.711 0]T

l3 = 1,

n 3 = [0 0 1]T

4-27 Then, the logarithmic stretch can be obtained by T

e = { 0.02 -0.02 0 }

The stress-strain relation in the principal space, tp = c⋅e, can be written as é3 1 1ùì ìï 4 üï ï 0.02 ü ï ï ï ïï ê úï ï ï ï 11 9 ï ï t p = êê 1 3 1 úú ï ´ = 0.02 10 4 í ý í ïý ´ 10 Pa ï ï ï ï ê1 1 3ú ï ï ï 0 ïï úû ï ï 0 ï ï ï þï ëê î þ îï

Then, the Kirchhoff stress can be obtained using é 4 0 ùú ê 0.04 t = å tip ni Ä ni = êê 4 -0.04 0 úú GPa ê 0 i =1 0 0 úú êë û 3

▄

P4.24 A history of biaxial loadings is applied to a 1mm´1mm square, as shown in the figure. The square is constrained in the Y-direction along the bottom edge and in the Xdirection along the left edge. The model is displaced in the X and Y directions at the right and top edges by R = 2.5´10-5mm, respectively. Calculate sxx, syy, szz, and vonMises stress at each load step. Use the following material properties: E = 250GPa, n = 0.25, sY = 5MPa, and ET = 50GPa.

Y DY

DX

1m

X 1m

Load step 1 2 3 4

DX R R 0 0

Figure P4.24

DY 0 0 R R

Description First yield Plastic flow Elastic unloading Plastic reloading

4-28

-R -R 0 0

5 6 7 8

0 0 -R -R

Plastic flow Plastic flow Elastic unloading Plastic flow

Solution: The problem is modeled using one square element in ANSYS. Since the strain in Zdirection is constrained, stress in Z-direction. Below is the table of stresses at each load step.

Load step 1 2 3 4 5 6 7 8

sxx 7.50000 11.7 14.2 16.4 9.9 5.1 2.6 0.2

syy 2.5 6.7 14.2 19.7 15.6 10.7 3.2 -3.0

szz 2.5 6.7 9.2 13.9 12.0 9.2 6.7 2.8

von Mises stress 5.0 5.0 5.0 5.0 5.0 5.0 3.8 5.0

Below is ANSYS script to model and solve the problem: /COM,ANSYS MEDIA REL. 11.0 (10/27/2006) REF. VERIF. MANUAL: REL. 11.0 /VERIFY,VMR049-PL1A-182 /TITLE, VMR049-PL1A-182, 2D PLANE STRAIN PLASTICITY BENCHMARK /COM, REFERENCE: NAFEMS REPORT-R0049 /PREP7 R = 2.5E-5 ET,1,182,,, KEYOPT,1,3,2 N,1,,,, N,2,0,1,, N,3,1,0,, N,4,1,1,, E, 1,3,4,2 MP,EX,1,250E3, MP,NUXY,1,0.25, TB,BISO,1,1, , , TBMODIF,2,1,5 TBMODIF,3,1,0.0 TB,HILL,1 TBDATA,1,1.0,1.0,1.0,1.0,1.0,1.0 NSEL,S,LOC,X D,ALL,UX NSEL,S,LOC,Y D,ALL,UY NSEL,S,LOC,Y,1.0 D,ALL,UY FINISH /SOLU NLGEOM,ON NSEL,S,LOC,X,1.0 D,ALL,UX,R

4-29 NSEL,ALL NSUBST,10,10,10, OUTRES,ALL,5 SOLVE NSEL,S,LOC,X,1.0 D,ALL,UX,2*R NSEL,ALL NSUBST,10,10,10, OUTRES,ALL,5 SOLVE NSEL,S,LOC,Y,1.0 D,ALL,UY,R NSEL,ALL NSUBST,10,10,10, OUTRES,ALL,5 SOLVE NSEL,S,LOC,Y,1.0 D,ALL,UY,2*R NSEL,ALL NSUBST,10,10,10, OUTRES,ALL,5 SOLVE NSEL,S,LOC,X,1.0 D,ALL,UX,R NSEL,ALL NSUBST,10,10,10, OUTRES,ALL,5 SOLVE NSEL,S,LOC,X,1.0 D,ALL,UX,0.0 NSEL,ALL NSUBST,10,10,10, OUTRES,ALL,5 SOLVE NSEL,S,LOC,Y,1.0 D,ALL,UY,R NSEL,ALL NSUBST,10,10,10, OUTRES,ALL,5 SOLVE NSEL,S,LOC,Y,1.0 D,ALL,UY,0.0 NSEL,ALL NSUBST,10,10,10, OUTRES,ALL,5 SOLVE FINISH /POST26 /GROPT,VIEW,0 /GTHK,CURVE,1 /GROPT,FILL,OFF /GRID,1 /GTHK,GRID,1 /GROPT,CGRID,1 /AXLAB,X,STEP /AXLAB,Y,STRESS /GTHK,AXIS,1 /GRTYP,0 /XRANGE,0,8 /YRANGE,-10,25 ESOL,2,1,4,S,X, ESOL,3,1,4,S,Y,

4-30 ESOL,4,1,4,S,Z, ESOL,5,1,4,S,EQV, PLVAR,2,3,4,5 PRVAR,2,3,4,5 *DIM,VALUEX,ARRAY,8,1 *DO,JJ,1,8,1 *GET,VALUEX(JJ,1),VARI,2,RTIME,JJ *ENDDO R1=VALUEX(1,1)/7.500 R2=VALUEX(2,1)/11.666 R3=VALUEX(3,1)/14.166 R4=VALUEX(4,1)/16.418 R5=VALUEX(5,1)/9.927 R6=VALUEX(6,1)/5.134 R7=VALUEX(7,1)/2.635 R8=VALUEX(8,1)/1.218 *DIM,VALUEY,ARRAY,8,1 *DO,JJ,1,8,1 *GET,VALUEY(JJ,1),VARI,3,RTIME,JJ *ENDDO R9=VALUEY(1,1)/2.500 R10=VALUEY(2,1)/6.666 R11=VALUEY(3,1)/14.166 R12=VALUEY(4,1)/19.669 R13=VALUEY(5,1)/15.622 R14=VALUEY(6,1)/10.745 R15=VALUEY(7,1)/3.245 R16=VALUEY(8,1)/(-3.715) *DIM,VALUEZ,ARRAY,8,1 *DO,JJ,1,8,1 *GET,VALUEZ(JJ,1),VARI,4,RTIME,JJ *ENDDO R17=VALUEZ(1,1)/2.500 R18=VALUEZ(2,1)/6.666 R19=VALUEZ(3,1)/9.166 R20=VALUEZ(4,1)/13.912 R21=VALUEZ(5,1)/11.951 R22=VALUEZ(6,1)/9.120 R23=VALUEZ(7,1)/6.620 R24=VALUEZ(8,1)/3.521 *DIM,VALUEEF,ARRAY,8,1 *DO,JJ,1,8,1 *GET,VALUEEF(JJ,1),VARI,5,RTIME,JJ *ENDDO R25=VALUEEF(1,1)/5.000 R26=VALUEEF(2,1)/5.000 R27=VALUEEF(3,1)/5.000 R28=VALUEEF(4,1)/5.000 R29=VALUEEF(5,1)/5.000 R30=VALUEEF(6,1)/5.000 R31=VALUEEF(7,1)/3.719 R32=VALUEEF(8,1)/5.000 *DIM,STEP,CHAR,10 *DIM,TARGETX,CHAR,10 *DIM,TARGETY,CHAR,10 *DIM,TARGETZ,CHAR,10 *DIM,TARGETEF,CHAR,10 *DIM,RATIOX,,8,1

4-31 *DIM,RATIOY,,8,1 *DIM,RATIOZ,,8,1 *DIM,RATIOEF,,8,1 STEP(1)='1.0','2.0','3.0','4.0','5.0','6.0','7.0','8.0' TARGETX(1)='7.500','11.666','14.166','16.418','9.927','5.134','2.635','1.218' TARGETY(1)='2.500','6.666','14.166','19.669','15.622','10.745','3.245','-3.715' TARGETZ(1)='2.500','6.666','9.166','13.914','11.951','9.120','6.620','3.521' TARGETEF(1)='5.000','5.000','5.000','5.000','5.000','5.000','3.719','5.000' *VFILL,RATIOX,DATA,R1,R2,R3,R4,R5,R6,R7,R8 *VFILL,RATIOY,DATA,R9,R10,R11,R12,R13,R14,R15,R16 *VFILL,RATIOZ,DATA,R17,R18,R19,R20,R21,R22,R23,R24 *VFILL,RATIOEF,DATA,R25,R26,R27,R28,R29,R30,R31,R32 /COM, /COM, ----------- VMR049-PL1A-182 RESULTS COMPARISON -------------/COM, /COM, vmr049-pl1a-182.jpeg RESULTS SHOULD MATCH R0049 NAFEMS MANUAL /COM, GRAPH RESULTS ON PAGE 49, FIGURE 2.14(A). THE RESULTS /COM, DISPLAYED ARE INCREMENTED FOR THIS PURPOSE. /COM, /COM, /COM, ------ VMR049-PL1A-182 STRESS RESULTS IN X DIRECTION ------/COM, /COM, | STEP | TARGET | ANSYS | RATIO /COM, *VWRITE,STEP(1),TARGETX(1),VALUEX(1,1),RATIOX(1,1) (1X,A8,' ',1x,A8,' ',F8.3,' ',F8.3,' ') /COM, /COM, ------ VMR049-PL1A-182 STRESS RESULTS IN Y DIRECTION ------/COM, /COM, | STEP | TARGET | ANSYS | RATIO /COM, *VWRITE,STEP(1),TARGETY(1),VALUEY(1,1),RATIOY(1,1) (1X,A8,' ',1x,A8,' ',F8.3,' ',F8.3,' ') /COM, /COM, /COM, ------ VMR049-PL1A-182 STRESS RESULTS IN Z DIRECTION ------/COM, /COM, | STEP | TARGET | ANSYS | RATIO /COM, *VWRITE,STEP(1),TARGETZ(1),VALUEZ(1,1),RATIOZ(1,1) (1X,A8,' ',1x,A8,' ',F8.3,' ',F8.3,' ') /COM, /COM, /COM, ------ VMR049-PL1A-182 EFFECTIVE STRESS RESULTS ------/COM, /COM, | STEP | TARGET | ANSYS | RATIO /COM, *VWRITE,STEP(1),TARGETEF(1),VALUEEF(1,1),RATIOEF(1,1) (1X,A8,' ',1x,A8,' ',F8.3,' ',F8.3,' ') /COM, /COM, FINISH

5.7.

Exercise

P5.1 In the beam contact problem in Section 5.2.1, determine the contact force and tip deflection using the Lagrange multiplier method. Choose the gap g as a Lagrange multiplier.

Solution: The contact consistency condition is given as

lg = l(vtip - d) = 0 Since the gap is considered as a Lagrange multiplier, the contact force, l, needs to be written in terms of the gap, or equivalently, in terms of vtip. From the tip deflection formula in Eq. (5.1),

l=

3 ´ 103 - 3 ´ 105 ´ v tip 8

By substituting vtip into the consistency condition in Eq. (5.3), we have

æ3 ö 3 5 -3 ççç ´ 10 - 3 ´ 10 ´ v tip ÷÷÷(v tip - 1 ´ 10 ) = 0 è8 ø The above equation has two solutions: v tip = 1.25 ´ 10-3 and v tip = 1 ´ 10-3 . The former yields l = 0 but g = 0.25 ´ 10-3 > 0 , which violates the contact requirement.

Therefore, it is invalid. The latter, however, yields l = 75N > 0 and g = 0 , which satisfies the contact requirement. This means that the beam contacts with the rigid block with the contact force of 75N. ▄

P5.2 For the beam contact problem in Section 5.2.1, determine the contact force and tip deflection using the Lagrange multiplier method. Model the beam using a two-node Euler beam element. Compare the results with the results in Section 5.2.1, and explain the reason for different results.

Solution: The finite element equation for a two-node Euler beam element with a distributed load and clamped condition at Node 1 becomes é 12 6L -12 6L ù ì v1 ü ïï ïìï -qL / 2 + R1 ïüï ê ú ïï ï 2 2 ê ú q ïï ïïï -qL2 / 12 + C ïïï EI ê 6L 4L -6L 2L ú ïï 1ï ï 1ï ï ý=ï í ý úí 3 ê -12 -6L ï ï ï ï 12 6 L v qL / 2 L ê úï 2ï ï ï ï ï ï ï ê 2 2 ú ïq ï 2 ï ïï qL / 12 êë 6L 2L -6L 4L úû ï ï ï ï 2þ ï ïî î ïþ

In addition, since the first two degrees-of-freedom are fixed, we can delete the first two rows and columns, yielding the following reduced equation:

4-2 ì v2 ï ü ìï -qL / 2 üï EI éê 12 -6L ùú ï ï ï ï=ï í í 2 ý 2 ú q ý ê ï ï ï qL / 12 L3 êë -6L 4L ûú ï 2 ï þ ï îï î þï

In the Lagrange multiple method, it is considered that the Lagrange multiplier is the contact force in the interface. Therefore, the above equation is modified as é 12 -6 ù ìï v2 üï ìï -500 + l ïü úï ï ï 105 êê ýï úí ý = í êë -6 4 úû ïïî q2 ïïþ ïïî 1000 / 12 ïïþ

Solving the above equation yields the vertical tip displacement as a function of Lagrange multiplier as v2 =

10-5 l - 0.00125 3

Using this relation, the contact consistency condition in Eq. (5.3) can be written as

æ 10-5 ö lg = l(v2 + d) = l ççç l - 0.00025 ÷÷÷ = 0 ÷ø çè 3 Different from Eq. (5.2), g = v2 + d is used because the nodal displacement is positive in the positive y-direction. The above quadratic equations has two solutions: l = 0N and l = 75N . When l = 0N , the gap becomes g = 0.00025 > 0 , which violates the condition of no penetration. Therefore, this cannot be a possible configuration. On the other hand, when l = 75N , the gap becomes g = 0 . Since this solution satisfies all requirements, this is the solution. In fact, the solution is consistent with the solution from Section 5.2.1. ▄

P5.3 For the frictional contact problem in Section 5.2.2, determine the frictional force and slip displacement using the Lagrange multiplier method. Choose the slip utip as a Lagrange multiplier.

Solution: In the Lagrange multiplier method, the consistency condition is used to find the unknown slip displacement. Using the tip displacement formula, the consistency condition can be written as

(P - t )L (t - ml) = 0 EA The above consistency condition has two solutions: t = P and t = ml . The condition of t = P is equivalent to the requirement of u tip = 0 . Therefore, the other condition of t - ml < 0 should be checked. This condition is satisfied if P < ml , which corresponds to the stick condition. For the given load conditions, however, P > ml , and this cannot satisfy the requirement.

The condition of t = ml corresponds to the slip condition, and needs to show that u tip > 0 . The tip displacement can then be expressed by

4-3

u tip =

(P - ml)L = 0.625mm > 0 EA

Since the tip displacement is greater than zero, it satisfies the requirement of the slip condition, and it is a valid state. ▄

P5.4 During a Newton-Raphson iteration, a rectangular plane element is in contact with a rigid surface as shown in the figure. Due to the penalty method, the penetration of g = –1×10–4m is observed with penalty parameter wn = 106. In the two-dimensional problem, the element has eight degrees-of-freedom {u1x, u1y, u2x, u2y, u3x, u3y, u4x, u4y}T. Calculate the contact force and contact stiffness matrix in terms of 8x1 vector and 8x8 matrix, respectively.

4

3

1

2 Rigid surface Figure P5.4 Contact of a rectangular block

Solution: Since the element boundary is straight and the rigid surface is flat, the contact boundary is constant, the edge between Nodes 1 and 2. The contact force term in Eq. (5.59) can be applicable for both Nodes 1 and 2. Using en = [0,1]T , the contact forces are T

fnc1 = fnc 2 = -wn gn en = { 0, 102 }

Therefore, through the assembly process, the contact force vector of the element becomes

Fnc = { 0 102

0 102

T

0 0 0 0}

This is equivalent to applying vertical forces at Nodes 1 and 2.

The contact stiffness is the same for both nodes, as

é0 0 ù ú kcn 1 = kcn 1 = wn en enT = êê ú w 0 n úû êë Therefore, after assembly, the element contact stiffness matrix becomes

4-4

é0 ê ê wn ê ê ê ê ê Kcn = ê ê ê ê ê 0 ê ê ê ë

0 wn

ù ú ú 0 ú ú ú ú ú ú 0 ú ú 0 ú ú 0 ú ú 0ú û

▄

P5.5 A sphere of radius r = 8mm is pressed against a rigid flat plane. Using a commercial program, determine the contact radius, a, for a given load F = (30×2p)N. Assume a linear elastic material with Young’s modulus E = 1000 N/mm2 and Poisson’s ratio n = 0.3. Use an axi-symmetric model. Compare the finite element result with the analytical contact radius of a = 1.010mm.

F

y

r x

Figure P5.5 Contact of a sphere

Solution: The problem can be solved using different commercial programs, but ANSYS is used to solve it. An axisymmetric model is used. A node is placed near the expected radius of contact. Midside nodes are removed along the surface where contact is likely to occur.

The model is comprised of both PLANE82 and PLANE183. The model is solved using 3D node-to-node contact elements (CONTA178). The following figure shows finite element mesh along with boundary conditions:

As shown in the following table, the contact radius 1.011mm is close to the target value of 1.010mm.

Target ANSYS Ratio A, mm 1.010 1.011 1.001

4-5 The following program is the ANSYS command script to solve the problem: /PREP7 SMRT,OFF /TITLE, STATIC HERTZ CONTACT PROBLEM SOLVED USING CONTAC178 ELEMENTS ET,1,PLANE82,,,1 ! AXISYMMETRIC ELEMENTS ET,2,PLANE183,,,1 ET,3,CONTA178,,4 ! NODAL CONTACT R,1 RMOD,1,7,1 !CONTACT NORMAL ALONG UY MP,EX,1,1E3 MP,NUXY,1,.3 LOCAL,11,1,0,8,0 ! LOCAL CYLINDRICAL C.S. AT CENTERLINE K,1,8,-90 ! DEFINE KEYPOINTS K,2,8 K,3,7.5,-90 K,4,7.5 K,5 K,6,8,-82.65 ! PLACE KEYPOINT AND NODE AT EXPECTED CONTACT RADIUS K,7,7.5,-82.65 L,1,3 ! DEFINE LINES L,2,4 L,6,7 LESIZE,ALL,,,1 ! DEFINE ELEMENT DIVISIONS ON ALL EXISTING LINES A,1,6,7,3 ! DEFINE AREAS A,6,2,4,7 A,3,7,4,5 LOCAL,12,0,0,8,0 ARSYM,Y,1,3,1 ! CREATE HALF-SYMMETRY MODEL NUMMRG,KPOI ESIZE,,4 ! DEFINE ELEMENT DIVISIONS ON REMAINING LINES LESIZE,4,,,5 *REPEAT,2,1 LESIZE,6,,,8,8 LESIZE,7,,,8,(1/8) LESIZE,10,,,1 *REPEAT,2,2 LESIZE,9,,,6,.2 TYPE,1 ! CREATE NODES AND ELEMENTS AMESH,1,2,1 AMESH,4,5,1 TYPE,2 MSHAPE,1,2D MSHKEY,0 AMESH,3,6,3 CSYS,0 N,1001,-1,1E-8 !NODE 1001 IS THE GROUND D,1001,ALL !X POSITION DOES NOT MATTER IN THIS CASE BECAUSE !THE CONTACT NORMAL IS ONLY ALONG UY TYPE,3 REAL,1 EN,205,1001,2 !USE THE SAME ELEMENT NUMBERS AS VM63 FOR POST-PROC EN,201,1001,4 EN,202,1001,6 EN,203,1001,8 EN,204,1001,10 EN,206,1001,31 MODMSH,NOCHECK TYPE,1 EMODIF,7,7,0 *REPEAT,6,1 MODMSH,CHECK

! REMOVE MIDSIDE NODES ALONG CONTACT SURFACE

4-6 FINISH /SOLU NSEL,S,LOC,X,-.01,.01 ! BOUNDARY CONDITIONS AND LOADING D,ALL,UX,0 NSEL,R,LOC,Y,0 D,ALL,UY,0 NSEL,ALL LOAD=0 *CREATE,LOADSTEP ! MACRO TO INCREMENTALLY APPLY LOAD FK,8,FY,ARG1 SOLVE *END *DO,I,1,3 LOAD=LOAD-10 *USE,LOADSTEP,LOAD*6.2831853 *ENDDO FINISH /POST1 ! POSTPROCESS /OUT, SET,3 ESEL,,TYPE,,3 ETABLE,RFOR,SMISC,1 NSLE PRETAB,RFOR ! PRINT REACTION FORCE TO DETERMINE CONTACT AREA SSUM ! SUM OF REACTION FORCE NLIST ! LIST COORDINATES OF NODES OF CONTACT SURFACE PRNSOL,U,COMP ! LIST DISPLACEMENTS OF NODES /COM CALCULATE RATIO OF A - ACTUAL TO A - TARGET PI=(4*ATAN(1)) LOAD=-(LOAD)*(2*PI) ATAR=(0.88*((LOAD*0.008)**(1/3))) ! A - TARGET *GET,EMAX,ELEM,,NUM,MAX *DO,ENUM,201,EMAX ! START SEARCH FROM ELEM 201 *GET,GRFR,ELEM,ENUM,ETAB,RFOR ! FIND LAST ELEMENT IN CONTACT *IF,GRFR,EQ,0.0,EXIT *ENDDO ESEL,,ELEM,,(ENUM-1) ! SELECT LAST CONTACTING ELEMENT NSLE ! SELECT NODES ATTACHED TO SELECTED ELEMENTS *GET,NMIN,NODE,0,NUM,MIN NODX=NX(NMIN) NODY=NY(NMIN) NUX =UX(NMIN) NUY =UY(NMIN) AACT=NODX+NUX ! A - ACTUAL YCHK=NODY+NUY RATA=(AACT/ATAR) ! RATIO *status,parm *DIM,LABEL,CHAR,1,2 *DIM,VALUE,,1,3 LABEL(1,1) = ' A,' LABEL(1,2) = ' mm ' *VFILL,VALUE(1,1),DATA,1.010 *VFILL,VALUE(1,2),DATA,AACT *VFILL,VALUE(1,3),DATA,ABS(AACT/1.010) /COM /COM,----------------------------------------------------------------------/COM,---------- RESULTS COMPARISON (OBTAINED USING CONTACT178 ELEMENTS)----/COM, /COM, | TARGET | ANSYS | RATIO /COM, *VWRITE,LABEL(1,1),LABEL(1,2),VALUE(1,1),VALUE(1,2),VALUE(1,3) (1X,A8,A8,' ',F10.3,' ',F10.3,' ',1F5.3) /COM,-----------------------------------------------------------------------

4-7 /OUT FINISH

▄

P5.6 A long rubber cylinder with radius r = 200mm is pressed between two rigid plates using a maximum imposed displacement of dmax = 200mm. Determine the forcedeflection response. Use Mooney-Rivlin material with A10 = 0.293MPa and A01 = 0.177MPa. Assume a plane strain condition and symmetry. Compare the results with the target results of F = 250N at d = 100mm and F = 1400N at d = 200mm. dmax r Figure P5.6 Rubber cylinder contact problem Solution: The problem can be solved using different commercial programs, but ANSYS is used to solve it. Since the rubber is in the cylindrical shape, a plane strain solution is assumed.

Due to geometric and loading symmetry, the analysis can be performed using one quarter of the cross section, as shown in the figure below.

All nodes on the left edge (x = 0) are constrained, UX = 0. All nodes on the top edge (y = 0) are coupled in UY. An imposed displacement of -0.1m acts upon the coupled nodes.

Using ANSYS, the problem can be solved in several ways: 1) A 2-D model using PLANE182 with CONTA175 elements 2) A 3-D model using SOLID185 with CONTA175 elements 3) A 2-D model using PLANE182 with CONTA175 element and solved using Lagrange Multipliers method.

4) A 3-D model using SOLID185 with CONTA175 element and solved using Lagrange Multipliers method.

In the 3-D case, a MESH200 element is used as the target face for the automatic generation of contact elements. The target surface is given a high contact stiffness (KN = 2000 MPa) to model a rigid surface and no contact stiffness is required to be specified while performing the solution using Lagrange Multipliers method. The following two figures show deformed geometry and force-displacement curve.

4-8

The force results at d = 100mm and d = 200mm are compared with the values in the literature (T. Tussman, K-J Bathe, "A Finite Element Formulation for Nonlinear Incompressible Elastic and Inelastic Analysis", pg. 385, fig. 6.14). The following table compares the results with different combinations of elements. It is found that the numerical simulation has an error less than 6.6%.

2-D with augmented Lagrangian method Force at d = 100mm Force at d = 200mm 2-D with augmented Lagrangian method Force at d = 100mm Force at d = 200mm 2-D with Lagrange multiplier method Force at d = 100mm Force at d = 200mm 3-D with Lagrange multiplier method Force at d = 100mm Force at d = 200mm

Target

ANSYS

Ratio

250.00 1400.00

266.05 1397.06

1.064 0.998

250.00 1400.00

258.77 1398.79

1.035 0.999

250.00 1400.00

266.20 1400.42

1.065 1.000

250.00 1400.00

266.40 1400.48

1.066 1.000

4-9 The following program is the ANSYS command script to solve the problem: R = 200 ! RADIUS OF CYLINDER (mm) /PREP7 smrt,off /TITLE, RUBBER CYLINDER PRESSED BETWEEN TWO PLATES ET,1,PLANE182, , ,2 ! 2-D PLANE-STRAIN 4-NODE STRUCTURAL SOLID KEYOPT,1,6,1 ET,2,CONTA175 ! 2-D 1-NODE NODE-TO-SURFACE CONTACT ELEMENT R,2, , ,-2000 ! SET SURFACE STIFFNESS ET,3,TARGE169 ! 2-D TARGET ELEMENT MP,EX,1,2.82 ! YOUNG'S MODULUS [MPA] MP,NUXY,1,0.49967 ! POISSON'S RATIO C10 = 0.293 C01 = 0.177 NU1 = 0.49967 DD = (1-2*NU1)/(C10+C01) TB,HYPER,1,1,2,MOONEY TBDATA,1,C10,C01,DD CSYS,1 ! SWITCH TO CYLINDRICAL C.S. K,1 ! DEFINE KEYPOINTS K,2,R,-90 K,3,R K,4,(0.5*R),-90 K,5,(0.6*R),-45 K,6,(0.5*R) K,7,R,-45 L,2,7 L,7,3 CSYS,0 ! SWITCH TO CARTESIAN C.S. A,2,7,5,4 A,7,3,6,5 A,4,5,6,1 ESIZE,,4 ! SET ELEMENT DIVISION SIZE AMESH,ALL ! MESH ALL AREAS SAVE ! SAVE MODEL FOR MORE ANALYSIS N,1001,(-2*R),-R ! TARGET SURFACE NODES N,1002,(2*R) ,-R NSEL,S,NODE,,1001,1002 TYPE,3 REAL,2 TSHAP,LINE ! SET TARGET SHAPE TO LINE E,1002,1001 ! GENERATE RIGID TARGET D,ALL,ALL,0 ! FIX TARGET NSEL,S,LOC,X ! SELECT LEFT EDGE D,ALL,UX ! CONSTRAIN LEFT EDGE IN UX NSEL,S,LOC,Y ! SELECT TOP EDGE CP,1,UY,ALL ! COUPLE TOP EDGE IN UY *GET,NCEN,NODE,,NUM,MIN ! GET MINIMUM NODE NUMBER FROM SELECTED SET NSEL,ALL CSYS,1 ! SWITCH TO CYLINDRICAL C.S. ESEL,S,TYPE,,1 NSLE NSEL,R,LOC,X,R TYPE,2 REAL,2 ESURF ! DEFINE CONTACT ELEMENTS ALLSEL,ALL CSYS,0 SAVE,CONT2D ! SAVE 2D CONTACT MODEL FOR SECOND ANALYSIS FINISH *CREATE,SOLVIT,MAC

! MACRO TO SOLVE MODEL

4-10 /SOLU ANTYPE,STATIC CNVTOL,F,,,,-1 NLGEOM,ON NSUBST,6 OUTRES,,1 D,NCEN,UY,-100 SOLVE FINISH *END SOLVIT

! INCLUDE LARGE DEFORMATION EFFECTS ! SPECIFY NUMBER OF SUBSTEPS IN LOAD STEP ! WRITE SOLUTION FOR EVERY SUBSTEP ! APPLY DISPLACEMENT UY = -100 TO COUPLED NODES

! USE MACRO SOLVIT

*CREATE,PLOTS,MAC ! MACRO FOR POST-PROCESSING /POST1 /DSCALE,1,1 PLDISP,1 ! PLOT DISPLACED SHAPE FINISH /POST26 /AXLAB,Y,FORCE /AXLAB,X,DISPLACEMENT NSOL,2,NCEN,U,Y RFORCE,3,NCEN,F,Y PROD,2,2,,,,,,-2 PROD,3,3,,,,,,-2 XVAR,2 PLVAR,3 ! PLOT DISPLACEMENT VS FORCE PRVAR,2,3 ! PRINT DISPLACEMENT, FORCE *GET,F1,VARI,3,RTIME,.5 *GET,F2,VARI,3,RTIME,1 *DIM,LABEL,CHAR,2,2 *DIM,VALUE,,2,3 LABEL(1,1) = 'F (N) @ ','F (N) @ ' LABEL(1,2) = '.1','.2' *VFILL,VALUE(1,1),DATA,250,1400 *VFILL,VALUE(1,2),DATA,F1,F2 *VFILL,VALUE(1,3),DATA,ABS(F1/250),ABS(F2/1400) FINISH *END PLOTS ! USE MACRO PLOTS SAVE,TABLE_1 RESUME /PREP7 SMRT,OFF ET,5,SOLID185 KEYOPT,5,6,1 ET,6,CONTA175 R,6, , ,-2000,-0.1, ET,7,TARGE170 ET,8,MESH200,6 R,8,0.05 ALLSEL TYPE,5 ESIZE,,1 VEXT,ALL,,,,,1 N,1001,,-R N,1002,2*R,-R N,1003,2*R,-R,8*R N,1004,,-R,8*R TYPE,8 REAL,8 E,1002,1001,1004,1003 NSEL,S,NODE,,1001,1004

! 3-D 8-NODE STRUCTURAL SOLID ! 3-D 1-NODE NODE-TO-SURFACE CONTACT ELEMENT ! 3-D TARGET ELEMENT ! 2-D 4-NODED QUAD

! CREATE TARGET PLANE OF NODES

4-11 TYPE,7 REAL,6 ESURF D,ALL,ALL,0 CSYS,1 ESEL,S,TYPE,,5 NSLE NSEL,R,LOC,X,R ESEL,S,TYPE,,5,7 TYPE,6 REAL,6 ESURF CSYS,0 NSEL,ALL D,ALL,UZ NSEL,S,LOC,X D,ALL,UX NSEL,S,LOC,Y CP,1,UY,ALL *GET,NCEN,NODE,,NUM,MIN ESEL,S,TYPE,,5,7 NSLE SAVE,CONT3D FINISH SOLVIT PLOTS SAVE,TABLE_2 RESUME,CONT2D /PREP7 R,2 KEYOPT,2,2,3 FINISH SOLVIT PLOTS SAVE,TABLE_3 RESUME,CONT3D /PREP7 R,6 KEYOPT,6,2,4 FINISH SOLVIT PLOTS SAVE,TABLE_4

! GENERATE TARGET ELEMENTS ! SWITCH TO CYLINDRICAL C.S.

! SWITCH TO CARTESIAN C.S. ! CONSTRAIN ALL NODES IN Z (PLANE STRAIN)

! COUPLE TOP NODES IN Y

! SAVE 3D CONTACT MODEL FOR SECOND ANALYSIS ! USE MACRO TO OBTAIN SOLUTION ! USE MACRO TO POSTPROCESS

! RESUME CONT175 -2D MODEL ! CONTACT STIFFNESS IS NOT REQUIRED ! LAGRANGE MULTIPLIER ON CONTACT NORMAL AND ! PENALTY ON TANGENT ! USE MACRO TO OBTAIN SOLUTION ! USE MACRO TO POSTPROCESS

! RESUME CONT175 -3D MODEL ! CONTACT STIFFNESS IS NOT REQUIRED ! PURE LAGRANGE MULTIPLIER ON CONTACT NORMAL AND ! TANGENT ! USE MACRO TO OBTAIN SOLUTION ! USE MACRO TO POSTPROCESS

RESUME,TABLE_1 /COM /OUT,results,vrt /COM,------------------- RESULTS COMPARISON -------------/COM, /COM, | TARGET | ANSYS | RATIO /COM, /COM,RESULTS USING PLANE182 AND 2D-CONTA175: *VWRITE,LABEL(1,1),LABEL(1,2),VALUE(1,1),VALUE(1,2),VALUE(1,3) (1X,A8,A8,' ',F10.2,' ',F10.2,' ',1F5.3) /NOPR RESUME,TABLE_2 /GOPR /COM, /COM,RESULTS USING SOLID185 AND 3D-CONTA175:

4-12 *VWRITE,LABEL(1,1),LABEL(1,2),VALUE(1,1),VALUE(1,2),VALUE(1,3) (1X,A8,A8,' ',F10.2,' ',F10.2,' ',1F5.3) /NOPR RESUME,TABLE_3 /GOPR /COM, /COM,RESULTS USING PLANE182 AND 2D-CONTA175 WITH K(2)=3: *VWRITE,LABEL(1,1),LABEL(1,2),VALUE(1,1),VALUE(1,2),VALUE(1,3) (1X,A8,A8,' ',F10.2,' ',F10.2,' ',1F5.3) /NOPR RESUME,TABLE_4 /GOPR /COM, /COM,RESULTS USING SOLID185 AND 3D-CONTA175 WITH K(2)=4: *VWRITE,LABEL(1,1),LABEL(1,2),VALUE(1,1),VALUE(1,2),VALUE(1,3) (1X,A8,A8,' ',F10.2,' ',F10.2,' ',1F5.3) /COM,----------------------------------------------------------/COM, /OUT FINISH

▄

P5.7 Two long cylinders of radii R1 = 10mm and R2 = 13mm, in frictionless contact with their axes parallel to each other, are pressed together with a force per unit length, F = 3200N/mm. Determine the semi-contact length b and the approach distance d. Both materials are linear elastic with E1 = 30000N/mm2 and n1 = 0.25 for Cylinder 1, and E2 = 29120N/mm2 and n2 = 0.3 for Cylinder 2. Assume a plane stress condition with a unit thickness and symmetry. Compare the results with the target results of d = – 0.4181mm and b = 1.20mm.

Symmetric model

E1, n1

d

R1

b

y R2

F

x E2, n2

Figure P5.7 Hertzian contact problem Solution: The problem can be solved using different commercial programs, but ANSYS is used to solve it. Each analysis uses two load steps; in the first load step a small imposed displacement is used on the upper cylinder to engage contact, whereas in the second load step the imposed displacement is deleted and the force load is applied. The problem is solved in four different ways:

4-13 1) Contact Algorithm: Augmented Lagrangian - KEYOPT(2) = 0 2-D analysis with PLANE182 and CONTA175 3-D analysis with SOLID185 and CONTA175 2) Contact Algorithm: Lagrange Multiplier - KEYOPT(2) = 3 2-D analysis with PLANE182 and CONTA175 3-D analysis with SOLID185 and CONTA175 Plane stress condition is modeled using a unit thickness slice through the cylinders. The region modeled is shown shaded in the problem sketch. The ESURF command is used to automatically generate the contact and target elements between "contactor" nodes on the upper cylinder and "target" nodes on the lower cylinder. The default value of contact stiffness FKN is chosen while performing a solution using Augmented Lagrangian contact algorithm (KEYOPT(2) = 0) whereas no contact stiffness input is required to be specified while performing a solution using Lagrange Multiplier contact algorithm (KEYOPT(2) = 3).

The approach distance, d = -0.4181, and the radius of contact region, b = 1.20mm are compared with the values in the literature (N. Chandrasekaran, W. E. Haisler, R. E. Goforth, "Finite Element Analysis of Hertz Contact Problem with Friction", Finite Elements in Analysis and Design, Vol. 3, 1987, pp. 39-56.). The following table compares the results with different combinations of elements. It is found that the numerical simulation has an error less than 3.3%.

Augmented Lagrangian method PLANE182 SOLID185 Lagrange multiplier method PLANE182 SOLID185

Target

ANSYS

Ratio

d (mm) b (mm) d (mm) b (mm)

-0.4181 1.20 -0.481 1.20

-0.4183 1.1609 -0.4191 1.1609

1.000 0.967 1.002 0.967

d (mm) b (mm) d (mm) b (mm)

-0.4181 1.20 -0.4181 1.20

-0.4181 1.1609 -0.4190 1.1609

1.000 0.967 1.002 0.967

The following program is the ANSYS command script to solve the problem: /TITLE, HERTZ CONTACT BETWEEN TWO CYLINDERS /COM /COM 2-D ANALYSIS USING PLANE182 AND CONTA175 /COM CONTACT ALGORITHM: AUGMENTED LAGRANGIAN - KEYOPT(2) = 0 /COM /PREP7 SMRT,OFF ANTYPE,STATIC ET,1,PLANE182 ! 2-D SOLID ELEMENTS ET,2,TARGE169 ! 2-D TARGET ELEMENTS ET,3,CONTA175 ! 2-D CONTACT ELEMENTS MP,EX,1,30000 ! SMALLER CYLINDER PROPERTIES MP,NUXY,1,0.25 MP,EX,2,29120 ! LARGER CYLINDER PROPERTIES MP,NUXY,2,0.30

4-14 CSYS,1 K,1 ! CREATE BIGGER CYLINDER K,2,13 K,3,13,82 K,4,13,90 K,5,11,90 L,1,5 L,2,3 LESIZE,ALL,,,7 L,3,4 ! TARGET SURFACE (LINE 3) LOCAL,11,1,,13 L,3,5 CSYS,1 A,1,2,3,5 A,5,3,4,4 MAT,2 MSHK,1 ! MAPPED AREA MESH MSHA,0,2D ! USING QUADS ESIZE,,4 AMESH,1,2 LOCAL,12,1,,23-1E-5,,-90 ! INTRODUCE SLIGHT INTERFERENCE K,11 ! CREATE SMALLER CYLINDER K,12,10 K,13,10,8 K,14,10,90 K,15,8 L,11,15 L,13,14 LESIZE,7,,,6 LESIZE,8,,,6 L,12,13 ! CONTACT SURFACE (LINE 9) CSYS,11 L,13,15 CSYS,12 MAT,1 A,12,13,15,15 A,15,13,14,11 ESIZE,,6 AMESH,3,4 LSEL,S,LINE,,9 ! SELECT CONTACT NODES ON SMALLER CYLINDER NSLL,,1 CM,CYL1,NODE REAL,1 TYPE,3 ESURF ! GENERATE COTAC175 ELEMENTS LSEL,S,LINE,,3 NSLL,,1 ! SELECT TARGET NODES ON BIGGER CYLINDER REAL,1 TYPE,2 ESURF ! GENERATE TARGE169 ELEMENTS NSEL,ALL CSYS,0 NSEL,S,LOC,Y,23 ! SELECT TOP EDGE OF MODEL CP,1,UY,ALL ! COUPLE NODES ON TOP EDGE *GET,NC,NODE,,NUM,MIN ! GET LOWEST NODE NUMBER (MASTER) NSEL,S,LOC,X ! SYMMETRY CONSTRAINTS D,ALL,UX NSEL,S,LOC,Y D,ALL,UY NSEL,ALL FINISH SAVE,MODEL2D

4-15 *CREATE,SOLV2D,MAC /SOLU D,NC,UY,-0.005 SOLVE DDELE,NC,UY F,NC,FY,-1600 nsub,2,10,1 SOLVE FINISH *END

! CREATE SOLUTION MACRO FOR 2-D CASE

SOLV2D

! EXECUTE SOLUTION MACRO FOR 2-D CASE

! APPLY SMALL DISPLACEMENT TO ENGAGE CONTACT ! SOLVE FIRST LOAD STEP ! DELETE IMPOSED DISPLACEMENT ! APPLY HALF LOAD ON (SYMMETRY) MODEL ! SOLVE SECOND LOAD STEP

*CREATE,RES2D,MAC ! CREATE RESULTS MACRO FOR 2-D CASE /POST1 NSEL,,LOC,Y,23 ! SELECT TOP EDGE OF SMALLER CYLINDER *GET,D,NODE,NC,U,Y ! GET APPROACH DISTANCE (D) ESEL,S,TYPE,,3 ! SELECT CONTACT ELEMENTS ETABLE,NSTAT,CONT,STAT ! STORE CONTACT STATUS ESEL,R,ETAB,NSTAT,2,2 ! SELECT ELEMENTS WITH CONTACT (STAT=2) CMSEL,S,CYL1 ! SELECT CONTACT COMPONENT NODES NSLE,R ! RESELECT NODES WITH CONTACT NSORT,LOC,X,1 ! SORT CONTACT NODES BY ASCENDING X LOCATION *GET,B,SORT,,MAX ! GET SEMI-CONTACT LENGTH (B) *STATUS,PARM *DIM,LABEL,CHAR,2,2 *DIM,VALUE,,2,3 LABEL(1,1) = 'AP DIS ','S-CON LEN ' LABEL(1,2) = ' mm',' mm' *VFILL,VALUE(1,1),DATA,-.4181,1.2 *VFILL,VALUE(1,2),DATA,D,B *VFILL,VALUE(1,3),DATA,ABS(D/.4181),ABS(B/1.2) FINISH *END RES2D SAVE,TABLE_1

! EXECUTE POSTPROCESSING MACRO FOR 2-D CASE

/CLEAR, NOSTART /COM /COM 3-D ANALYSIS USING SOLID185 AND CONTA175 /COM CONTACT ALGORITHM: AUGMENTED LAGRANGIAN - KEYOPT(2) = 0 /COM /PREP7 $SMRT,OFF ANTYPE,STATIC ET,1,SOLID185 ! 3-D SOLID ELEMENTS ET,2,170 ! 3-D TARGET ELEMENTS ET,3,175 ! 3-D CONTACT ELEMENTS MP,EX,1,30000 ! SMALLER CYLINDER PROPERTIES MP,NUXY,1,0.25 MP,EX,2,29120 ! LARGER CYLINDER PROPERTIES MP,NUXY,2,0.30 CSYS,1 K,1 ! CREATE LOWER BIGGER CYLINDER K,2,13 K,3,13,82 K,4,13,90 K,5,11,90 KGEN,2,1,5,1,,,1,100 ! UNIT THICKNESS SLICE L,1,5 L,2,3 L,101,105 L,102,103

4-16 LESIZE,ALL,,,7 L,1,101 *REPEAT,5,1,1 LESIZE,5,,,1 *REPEAT,5,1 LOCAL,11,1,,13 L,3,5 L,103,105 CSYS,1 MAT,2 MSHK,1 ! MAPPED VOLUME MESH MSHA,0,3D ! USING HEX ESIZE,,4 V,1,2,3,5,101,102,103,105 V,5,3,4,4,105,103,104,104 VMESH,ALL LOCAL,12,1,,23-1E-5,,-90 ! INTRODUCE SLIGHT INTERFERENCE K,11 ! CREATE UPPER SMALLER CYLINDER K,12,10 K,13,10,8 K,14,10,90 K,15,8 KGEN,2,11,15,1,,,1,100 L,11,15 L,13,14 LESIZE,18,,,6 LESIZE,19,,,6 L,11,111 *REPEAT,5,1,1 LESIZE,20,,,1 *REPEAT,5,1 CSYS,11 L,13,15 L,113,115 CSYS,12 MAT,1 ESIZE,,6 V,12,13,15,15,112,113,115,115 V,15,13,14,11,115,113,114,111 VMESH,3,4 ASEL,S,AREA,,12 NSLA,,1 ! SELECT CONTACT NODES ON SMALLER CYLINDER CM,CYL1,NODE ! CONTACT NODES COMPONENT REAL,1 TYPE,3 ESURF ! GENERATE 3-D CONTA175 ELEMENTS ASEL,S,AREA,,8 NSLA,,1 ! SELECT TARGET NODES ON BIGGER CYLINDER CM,CYL2,NODE ! TARGET NODES COMPONENT REAL,1 TYPE,2 ESURF ! GENERATE 3-D TARGE170 ELEMENTS NSEL,ALL CSYS,0 NSEL,S,LOC,Y,23 ! SELECT TOP EDGE OF MODEL CP,1,UY,ALL ! COUPLE NODES ON TOP EDGE *GET,NC,NODE,,NUM,MIN ! GET LOWEST NODE NUMBER (MASTER) NSEL,S,LOC,X ! SYMMETRY CONSTRAINTS D,ALL,UX NSEL,S,LOC,Y D,ALL,UY NSEL,S,LOC,Z D,ALL,UZ

4-17 NSEL,ALL FINISH SAVE,MODEL3D *CREATE,SOLV3D,MAC /SOLU D,NC,UY,-0.001 SOLVE DDELE,NC,UY F,NC,FY,-1600 nsub,2,10,1 SOLVE FINISH *END

! CREATE SOLUTION MACRO FOR 3-D CASE

SOLV3D

! EXECUTE SOLUTION MACRO FOR 3-D CASE

! APPLY SMALL DISPLACEMENT TO ENGAGE CONTACT ! SOLVE FIRST LOAD STEP ! DELETE IMPOSED DISPLACEMENT ! APPLY HALF LOAD ON (SYMMETRY) MODEL ! SOLVE SECOND LOAD STEP

*CREATE,RES3D,MAC ! CREATE RESULTS MACRO FOR 3D CASE /POST1 NSEL,,LOC,Y,23 ! SELECT TOP EDGE OF SMALLER CYLINDER *GET,D,NODE,NC,U,Y ! GET APPROACH DISTANCE (D) ESEL,S,TYPE,,3 ! SELECT CONTACT ELEMENTS ETABLE,NSTAT,CONT,STAT ! STORE CONTACT STATUS ESEL,R,ETAB,NSTAT,2,2 ! SELECT ELEMENTS WITH CONTACT (STAT=2) CMSEL,S,CYL1 ! SELECT CONTACT COMPONENT NODES NSLE,R ! RESELECT NODES WITH CONTACT NSORT,LOC,X,1 ! SORT CONTACT NODES BY ASCENDING X LOCATION *GET,B,SORT,,MAX ! GET SEMI-CONTACT LENGTH (B) *STATUS,PARM *DIM,LABEL,CHAR,2,2 *DIM,VALUE,,2,3 LABEL(1,1) = 'AP DIS ','S-CON LEN ' LABEL(1,2) = ' mm',' mm' *VFILL,VALUE(1,1),DATA,-.4181,1.2 *VFILL,VALUE(1,2),DATA,D,B *VFILL,VALUE(1,3),DATA,ABS(D/.4181),ABS(B/1.2) FINISH *END RES3D SAVE,TABLE_2

! EXECUTE POSTPROCESSING MACRO FOR 3-D CASE

/CLEAR, NOSTART /COM /COM 2-D ANALYSIS USING PLANE182 AND CONTA175 /COM CONTACT ALGORITHM: LAGRANGE MULTIPLIER - KEYOPT(2) = 3 /COM RESUME,MODEL2D /PREP7 KEYOPT,3,2,3 ! LAGRANGE MULTIPLIER ON CONTACT NORMAL AND PENALTY ON TANGENT FINISH SOLV2D RES2D SAVE,TABLE_3 /CLEAR, NOSTART /COM /COM 3-D ANALYSIS USING SOLID185 AND CONTA175 /COM CONTACT ALGORITHM: LAGRANGE MULTIPLIER - KEYOPT(2) = 3 /COM RESUME,MODEL3D /PREP7

4-18 KEYOPT,3,2,3 TANGENT FINISH SOLV3D RES3D SAVE,TABLE_4

! LAGRANGE MULTIPLIER ON CONTACT NORMAL AND PENALTY ON

RESUME,TABLE_1 /COM /OUT,results,txt /COM,=========== VM191 RESULTS COMPARISON ====================== /COM, /COM, | TARGET | ANSYS | RATIO /COM, /COM, 2-D ANALYSIS USING PLANE182 AND CONTA175: /COM, CONTACT ALGORITHM: AUGMENTED LAGRANGIAN /COM, *VWRITE,LABEL(1,1),LABEL(1,2),VALUE(1,1),VALUE(1,2),VALUE(1,3) (1X,A8,A8,' ',F10.4,' ',F10.4,' ',1F5.3) /NOPR RESUME,TABLE_2 /GOPR /COM, /COM, 3-D ANALYSIS USING SOLID185 AND CONTA175: /COM, CONTACT ALGORITHM: AUGMENTED LAGRANGIAN *VWRITE,LABEL(1,1),LABEL(1,2),VALUE(1,1),VALUE(1,2),VALUE(1,3) (1X,A8,A8,' ',F10.4,' ',F10.4,' ',1F5.3) /NOPR RESUME,TABLE_3 /GOPR /COM, /COM, 2-D ANALYSIS USING PLANE182 AND CONTA175: /COM, CONTACT ALGORITHM: LAGRANGE MULTIPLIER *VWRITE,LABEL(1,1),LABEL(1,2),VALUE(1,1),VALUE(1,2),VALUE(1,3) (1X,A8,A8,' ',F10.4,' ',F10.4,' ',1F5.3) /NOPR RESUME,TABLE_4 /GOPR /COM, /COM, 3-D ANALYSIS USING SOLID185 AND CONTA175: /COM, CONTACT ALGORITHM: LAGRANGE MULTIPLIER *VWRITE,LABEL(1,1),LABEL(1,2),VALUE(1,1),VALUE(1,2),VALUE(1,3) (1X,A8,A8,' ',F10.4,' ',F10.4,' ',1F5.3) /COM,=========================================================== /OUT FINISH

▄

P5.8 Deepdrawing is a manufacturing process that can create a complex shape out of a simply shaped plate (blank). The deepdrawing configuration is shown in the figure, which is composed of a blank, punch, die and blank holder. The thickness of the initial blank is 0.78mm. The die is fixed throughout the entire process, while the punch moves down by 30mm to shape the blank. The holder controls the slip of the blank by applying friction force. The fillet radii of both punch and die are 5mm. After the maximum downstroke of the punch, both the punch and holder are removed. Then, the blank will experience elastic springback. The objective of this project is to simulate the final geometry of the blank after springback.

4-19 Model the process using an axi-symmetric problem. You many use CAX4R elements.

The whole simulation is divided by three steps. (1) The blank holder is pushed (displacement control) to provide about 100kN of holding force. (2) While the blank holder is fixed at the location of step (1), the punch is moved down by 30mm. (3) Punch, die, and blank holder are removed so that the blank is elastically deformed by springback.

It is possible to change processes.

The following results need to be submitted: (1) deformed shape plots of five different steps, (2) graph of radial position vs. radial strain, and (3) graph of radial position vs. thickness change, (4) graph of punch displacement vs. punch force, and (5) comparison of deformed shapes at the maximum stroke and after springback.

25 mm

u2 Punch

u6

u1 u3

u5 5 mm

Blank 26 mm Plane of Symmetry

u4

Die

Blank Holder E = 206.9 GPa  = 0.29 y = 167 MPa H = 129 MPa f = 0.144 Isotropic Hardening

Figure P5.8 Deepdrawing problem Solution: The geometry of the problem is shown in Figure P5.8. The circular blank being drawn has an initial radius of 100 mm and an initial thickness of 0.78 mm. The punch has a radius of 25 mm and is rounded off at the corner with a radius of u3 = 13 mm. The die has an internal radius of 56 mm and is rounded off at the corner with a radius of u4 = 5 mm.

The blank holder has an internal radius of 26 mm.

The blank is modeled using 40 elements of type CAX4R. The contact between the blank and the rigid punch, the rigid die, and the rigid blank holder is modeled with the *CONTACT PAIR option. The top and bottom surfaces of the blank are defined by means of the *SURFACE option. The rigid punch, the die, and the blank holder are modeled as analytical rigid surfaces with the *RIGID BODY option in conjunction with the *SURFACE option. The mechanical interaction between the contact surfaces is assumed to be frictional contact. Therefore, the *FRICTION option is used in conjunction with the various *SURFACE INTERACTION property options to specify coefficients of friction.

At the start of the analysis, the blank is positioned precisely on top of the die and the blank holder is precisely in touch with the top surface of the blank. The punch is positioned 0.18 mm above the top surface of the blank.

The entire analysis is carried out in five steps. In the first step the blank holder is pushed onto the blank with a prescribed displacement to establish contact. In the shell models this displacement roughly corresponds to zero clearance across the interface.

In the second step the boundary condition is removed and replaced by the applied force of 100 kN on the blank holder. This force is kept constant during Steps 2 and 3.

This technique of simulating the clamping process is used to avoid potential problems

4-20

with rigid body modes of the blank holder, since there is no firm contact between the blank holder, the blank, and the die at the start of the process. The two-step procedure creates contact before the blank holder is allowed to move freely.

In the third step the punch is moved toward the blank through a total distance of 30 mm. This step models the actual drawing process. During this step the option *CONTROLS, ANALYSIS=DISCONTINUOUS is included since contact with friction tends to create a severely discontinuous nonlinearity and we wish to avoid premature cutbacks of the automatic time incrementation scheme.

The last two steps are used to simulate springback. In the fourth step all the nodes in the model are fixed in their current positions and the contact pairs are removed from the model with the *MODEL CHANGE, TYPE=CONTACT PAIR, REMOVE option. This is the most reliable method for releasing contact conditions. In the fifth, and final, step the regular set of boundary conditions is reinstated and the springback is allowed to take place. This part of the analysis with the CAX4R elements is included to demonstrate the feasibility of the unloading procedure only and is not expected to produce realistic results, since the reduced-integration elements have a purely elastic bending behavior.

The springback is modeled with more accuracy in the shell element models.

The following figure shows deformed shapes that are predicted at various stages of the drawing process. The profiles show that the metal initially bends and stretches and is then drawn in over the surface of the die.

The distributions of radial and circumferential strain for all three models and thickness strain are shown in the following figure. The thickness does not change very much: the change ranges from approximately –12% in the cylindrical part to approximately +16% at the edge of the formed cup. Relatively small thickness changes are usually desired in deep drawing processes and are achieved because the radial tensile strain and the circumferential compressive strain balance each other.

4-21

The drawing force as a function of punch displacement is shown in the following figure, where the results are compared with different types of element, and all curves are seen to match closely.

The deformed shape after complete unloading is shown in the following figure, superimposed on the deformed shape under complete loading. The analysis shows the lip of the cup springing back strongly after the blank holder is removed. The springback in the CAX4R model is not physically realistic: in the first-order reduced-integration elements an elastic “hourglass control” stiffness is associated with the “bending” mode, since this mode is identical to the “hourglass” mode exhibited by this element in continuum situations. In reality the bending of the element is an elastic-plastic process, so that the springback is likely to be much less. A better simulation of this aspect would be achieved by using several elements through the thickness of the blank, which would also increase the cost of the analysis. The springback results for the shell models do not exhibit this problem and are clearly more representative of the actual elastic-plastic process.

4-22

The following program is the ANSYS command script to solve the problem: *HEADING DEEP DRAWING OF CYLINDRICAL CUP WITH CAX4R *RESTART,WRITE,FREQUENCY=25 *NODE 101,0.0,0.0 181,100.,0.0 301,0.0,0.78 381,100.,0.78 *NGEN,NSET=BOT 101,181,2 *NGEN,NSET=TOP 301,381,2 *NSET,NSET=WRKPC BOT,TOP *NODE,NSET=DIE 100,0.1,-50. *NODE,NSET=PUNCH 200,0.,50. *NODE,NSET=HOLDER 300,0.1,50. *NSET,NSET=TOOLS PUNCH,DIE,HOLDER *NSET,NSET=CENTER 101,301 *ELEMENT,TYPE=CAX4R,ELSET=BLANK 201,101,103,303,301 *ELGEN,ELSET=BLANK 201,40,2,2 *ELSET,ELSET=ALL BLANK, *SOLID SECTION,MATERIAL=STEEL,ORIENTATION=LOCAL, ELSET=BLANK *ORIENTATION,NAME=LOCAL 1.,0.,0.,0.,1.,0.

0,0., *MATERIAL,NAME=STEEL *ELASTIC 2.06E5,0.29 *PLASTIC,HARDENING=ISOTROPIC 0.167E+03, 0.00000E+00 0.425E+03, 2.0 *ELSET,ELSET=EDIE,GENERATE 231,279,2 *ELSET,ELSET=EHOLDER,GENERATE 241,279,2 *ELSET,ELSET=EPUNCH,GENERATE

4-23 201,249,2 *RIGID BODY,ANALYTICAL SURFACE=BSURF,REF NODE=100 *SURFACE,TYPE=SEGMENTS,NAME=BSURF START,26,-60.

LINE,26,-5.

CIRCL,31,0.,31,-5.

LINE,100,0. *RIGID BODY,ANALYTICAL SURFACE=DSURF,REF NODE=300 *SURFACE,TYPE=SEGMENTS,NAME=DSURF START,100,0.78 LINE,31,0.78 CIRCL,26,5.78,31,5.78 LINE,26,60.78 *RIGID BODY,ANALYTICAL SURFACE=FSURF,REF NODE=200 *SURFACE,TYPE=SEGMENTS,FILLET RADIUS=.013, NAME=FSURF START,25,60.78 LINE,25,5.78 CIRCL,20,0.78,20,5.78 LINE,0.0,0.78 *SURFACE,NAME=ASURF EDIE,S1 *SURFACE,NAME=CSURF EHOLDER,S3 *SURFACE,NAME=ESURF EPUNCH,S3 *CONTACT PAIR,INTERACTION=ROUGH1 ASURF,BSURF *CONTACT PAIR,INTERACTION=ROUGH2 CSURF,DSURF *CONTACT PAIR,INTERACTION=ROUGH3 ESURF,FSURF *SURFACE INTERACTION,NAME=ROUGH1 *FRICTION 0.144, *SURFACE INTERACTION,NAME=ROUGH2 *FRICTION 0.144, *SURFACE INTERACTION,NAME=ROUGH3 *FRICTION 0.144, *STEP,NLGEOM, UNSYMM=YES PUSH THE BLANKHOLDER DOWN BY A PRESCRIBED DISPLACEMENT *STATIC 1.,1. *BOUNDARY CENTER,1,1 DIE,1,1 DIE,2,2 DIE,6,6 PUNCH,1,1 PUNCH,2,2 PUNCH,6,6 HOLDER,1,1 HOLDER,2,2,-1.75E-5 HOLDER,6,6 *MONITOR,NODE=200,DOF=2 *CONTACT CONTROLS,FRICTION ONSET=DELAY *PRINT,CONTACT=YES *NODE PRINT,NSET=TOOLS,FREQUENCY=100 COORD,U,RF *EL PRINT,ELSET=ALL,FREQUENCY=500

4-24 S,E *NODE FILE,NSET=TOOLS,FREQUENCY=10 U,RF *CONTACT FILE,SLAVE=ASURF,FREQUENCY=10 *CONTACT FILE,SLAVE=CSURF,FREQUENCY=10 *CONTACT FILE,SLAVE=ESURF,FREQUENCY=10 *END STEP *STEP,NLGEOM APPLY PRESCRIBED FORCE ON BLANKHOLDER AND RELEASE DISPLACEMENT *STATIC 1.,1. *BOUNDARY,OP=NEW CENTER,1,1 DIE,1,1 DIE,2,2 DIE,6,6 PUNCH,1,1 PUNCH,2,2 PUNCH,6,6 HOLDER,1,1 HOLDER,6,6 *CLOAD HOLDER,2,-1000. *END STEP *STEP,INC=500,NLGEOM MOVE THE PUNCH DOWN *STATIC .01,1.,1.E-6 *CONTROLS,ANALYSIS=DISCONTINUOUS *BOUNDARY,OP=NEW CENTER,1,1 DIE,1,1 DIE,2,2 DIE,6,6 PUNCH,1,1 PUNCH,2,2,-30.0 PUNCH,6,6 HOLDER,1,1 HOLDER,6,6 *CLOAD HOLDER,2,-1000. *END STEP *STEP,NLGEOM FIX ALL NODES AND REMOVE THE CONTACT SURFACES *STATIC 1.,1.,1.,1. *BOUNDARY,FIXED WRKPC,1,2 *MODEL CHANGE,TYPE=CONTACT PAIR,REMOVE ASURF,BSURF CSURF,DSURF ESURF,FSURF *CLOAD,OP=NEW HOLDER,2,0. *END STEP *STEP,NLGEOM, UNSYMM=NO REPLACE THE BOUNDARY CONDITIONS BY THE REGULAR SET *STATIC .1,1.,1.E-6 *BOUNDARY,OP=NEW 181,2 CENTER,1,1

4-25 *END STEP

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