From
Mechanics
Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
NOTES TO THE INSTRUCTOR
The Solutions Manual to accompany the text AdvancedMechanicsofMaterialsand AppliedElasticity supplements the study of stress and deformation analyses developed in the book. The main objective of the manual is to provide efficient solutions for problems dealing with variously loaded members. This manual can also serve to guide the instructor in the assignments of problems, in grading these problems, and in preparing lecture materials as well as examination questions. Every effort has been made to have a solutions manual that can cut through the clutter and is self - explanatory as possible thus reducing the work on the instructor. It is written and class tested by the author, Ansel Ugural.
As indicated in its preface, the text is designed for the senior and/or first year graduate level courses in stress analysis. In order to accommodate courses of varying emphasis, considerably more material has been presented in the book than can be covered effectively in a single three-credit course. The instructor has the choice of assigning a variety of problems in each chapter. Answers to selected problems are given at the end of the text. A description of the topics covered is given in the introduction of each chapter throughout the text. It is hoped that the foregoing materials will help instructor in organizing his or her course to best fit the needs of his or her students.
Holmdel, NJ
Ansel C. Ugural
CHAPTER 1
SOLUTION (1.1)
We have 50753.75(10)m32 A =×= , o 50 = θ , and AP x = σ
Equations (1.11), with o 50 = θ : P x o xx 18.110 413.050 cos )10(700 2 3 ' == = = σ σ σ or 6.35kN P = and 3 '' 560(10)sin50cos500.492131.2 oo xyxx P τσσ == ==
Solving 4.27kN P = allP =
SOLUTION (1.2)
Normal stress is 3 125(10) 0.050.05 50MPa P xAσ × ===
( a ) Equations (1.11), with o 20 = θ :
2 ' 50cos2044.15MPa o x σ == '' 50sin20cos2016.08MPa oo xy τ = −= 2 ' 50cos(2090)5.849MPa oo y σ =+=
MPa
( b ) Equations (1.11), with o45 = θ : 2 ' 50cos4525MPa o x σ == '' 50sin45cos4525MPa oo xy τ = −= 2 ' 50cos(4590)25MPa oo y σ =+=
SOLUTION (1.3)
From Eq. (1.11a), ' 22 75 coscos30 100MPa x xo σ θ σ ===−
For o 50 = θ , Eqs. (1.11) give then 2 ' 100cos5041.32MPa o x σ =−=− oo yx 50 cos 50 sin )100( '' −−= τ 49.24MPa =
Similarly, for o 140 = θ : 2 ' 100cos14058.68MPa o x σ = −= '' 49.24MPa xy τ =−
SOLUTION (1.4)
Refer to Fig. 1.6c. Equations (1.11) by substituting the double angle-trigonometric relations, or Eqs. (1.18) with 0 = y σ and 0 = xy τ , become θσσσ 2 cos 2 1 2 1 ' xxx += and θστ 2 sin 2 1 '' yxx = or cos)2 1(20 2 θ += A P and θ2 sin 10 2 A P =
The foregoing lead to 12 cos 2 sin 2 θθ=− (a)
By introducing trigonometric identities, Eq. (a) becomes 0 cos 2 cossin 4 2 θθθ=− or 21 tan = θ . Hence o 56.26 = θ
Thus, 2(1300) 20(10.6) P =+ gives 32.5kN P = It can be shown that use of Mohr’s circle yields readily the same result.
SOLUTION (1.5)
Equations (1.12):
SOLUTION (1.6)
Shaded transverse area: 22(10)(75)1.5(10)mm32Aat===
Metal is capable of supporting the load 63 90(10)(1.510)135kN PA σ ==×=
Apply Eqs. (1.11): 62 ' 3 25(10)(cos55) 1.5(10)
Thus, 38.3kN allP =
SOLUTION (1.7)
Use Eqs. (1.11): 62 ' 3 20(10)(cos40) 1.5(10) o x P σ == , 51.1kN P = 6 '' 3 8(10) sin40cos40 1.5(10)
Thus, 24.4kN allP =
SOLUTION (1.8) 2 1530450mm A =×=
Apply Eqs. (1.11): 3 2 ' 6 120(10) (cos40)156MPa 45010 o x σ = = × 3 '' 6 120(10) sin40cos40131MPa 45010 oo xy τ = −= ×
SOLUTION (1.9)
We have 450(10)m62 A = . Use Eqs. (1.11): 3 2 ' 6 100(10) (cos60)55.6MPa 45010 o x σ = = × 3
'' 6 100(10) sin60cos6096.2MPa 45010 oo xy τ = −= ×
SOLUTION (1.10)
ooo1309040=+= θ 3
Equations (1.11): 2 ' 31.83cos13013.15MPa o x σ = −= '' 31.83sin130cos13015.67MPa oo xy τ = =−
SOLUTION (1.11)
Use Eqs. (1.14), 0)()2()2(=++−+ x xxyFx
0)0()2()( 2 =+++−+− y Fxyyz
0)2()0()4(=+−++− z zxyFz
Solving, we have (in 3 MNm ): xFxy x 23+−= 2 2 y Fxyyz =−++ Fxyz z += 4 (a)
Substituting x=-0.01 m, y=0.03 m, and z=0.06 m, Eqs. (a) yield the following values 333 29.4kNm14.5kNm58.8kNm xyz FFF = = =
Resultant body force is thus 222 3 67.32kNm xyz FFFF=++=
SOLUTION (1.12)
Equations (1.14): 04,00022 1 1 1 ≠ =++−−ycycyc 0 ,0000 3 3 ≠ =+++ zczc 00000=+++
No. Eqs. (1.14) are not satisfied.
SOLUTION (1.13)
( a ) No. Eqs. (1.14) are not satisfied.
( b ) Yes. Eqs. (1.14) are satisfied.
SOLUTION (1.14)
Eqs. (1.14) for the given stress field yield:
SOLUTION (1.15)
1.16 (CONT.)
' '' 0:50sin25cos25 oo yxy FAA τ =∆−∆
2 90sin25cos2515cos25 oooAA−∆ −∆ 2 15sin250 Ao +∆= '' 19.1534.4712.322.6863.3MPa xy τ =++−=
SOLUTION (1.17)
' 11 22(4060)(4060)cos4050sin40 oo x σ =−++−−+ 1038.332.13.8MPa =−+=
'' 1 (4060)sin4050cos40 2 oo xy τ =−−−+
32.1438.370.4MPa=+=
SOLUTION (1.18)
' 11 22(9050)(9050)cos23015sin230 oo x σ =−++− 204511.513.5MPa=−+=−
'' 1 (9050)sin23015cos230 2 oo xy τ =−+− 53.629.6463.3MPa=+=
SOLUTION (1.19)
Transform from 40 o θ = to 0 θ = . For convenience in computations, Let 160MPa,80MPa,40MPaxyxy σστ = −= −= and 40 o θ =−
Then ' 11 22()()cos2sin2xxyxyxy σσσσσθτθ =++−+ 11
22(16080)(16080)cos(80)40sin(80) oo=−−+−+−+− 166.3MPa=− '' 1 ()sin2cos2 2 xyxyxy τσσθτθ =−−+ 1 (16080)sin(80)40cos(80) 2 oo=−−+−+− 32.4MPa=−
So '' 16080166.373.7MPa yxyx σσσσ=+−=−−+=−
For 0 o θ = :
SOLUTION (1.21)
070 o xy τθ==
(a) '' 60 30sin140 2 o xy σ τ =−=− 153.3MPa σ =
(b) ' 6060 80 cos140 22 o x σσ σ +− ==+ 231MPa σ =
SOLUTION (1.22)
Equations(1.18) with 60 o θ = , 110MPa x σ = , 0 y σ = , 50MPa xy τ = give 11 ' 22(110)(110)cos12050sin12070.8MPa oo x σ =++=
1 '' 2 (110)sin12050cos12072.6MPa oo xy τ =−+=− 11 ' 22(110)(110)cos12050sin12039.2MPa oo y σ =−−=
SOLUTION (1.23)
Equations(1.18) with 30 o θ = , 110MPa x σ = , 0 y σ = , 50MPa xy τ = result in
1 ' 2 (110)55cos6050sin60125.8MPa oo x σ =++= 1 '' 2 (110)sin6050cos6022.6MPa oo xy τ =−+=−
1 ' 2 (110)55cos6050sin6015.8MPa oo y σ =−−=−
SOLUTION (1.24)
We have 2590115o θ =+= 10MPa x σ =− 30MPa y σ = 0 xy τ = (a) ' 11 22()()cos2xxyxy σσσσσθ =++− 11 22(1030)(1030)cos23022.86MPa o =−++−−=
Thus, ' 22.86MPa wx σσ==
(CONT.) θ x’ 10 MPa ' x σ ''xy τ 25o 30 MPa x y’
1.24 (CONT.)
(b) '' 1 ()sin2 2 xyxy τσσθ =−− 1 (1030)sin23015.32MPa 2 o =−−−=− So '' 15.32MPa wxy ττ==−
SOLUTION (1.25)
(a) 22 1 050050 80 ( ) 22 στ +− ==++ 49MPa τ =
(b) 22 max 50 ()4955MPa 2 τ =+= 50 '25MPa 2 σ == 1 050 2tan[]27 2(49) o s θ =−= '' 50 sin2749cos2755MPa 2 oo xy τ =+= Thus, '13.5o s θ =
(1.26)
1.26 (CONT.)
40804080 cos(60)601050MPa 22 o x σ +− =+−=−= 601070MPa y σ =+= 4080 sin(60)17.32MPa 2 o xy τ =−−=−
' 5070 60cos(15)2.68sin(15) 2 oo x σ =+−+− 609.660.69449.65MPa =−−= Thus, "7.5o p θ =−
1.27 (CONT.)
1 50 tan45 50 o α ==
1 2 22 (5050)70.7 R =+=
'' sin85(70.7)70.4MPa o xy τ = = ' 10cos85(70.7)3.84MPa o x σ =−=
SOLUTION (1.28)
1 15 tan12.1 70 o α ==
1 2 22 (1570)71.6 R =+=
'' 71.6sin62.163.3MPa o xy τ = = ' 71.6cos62.120 o x σ = −+ 13.5MPa=−
1.29 (CONT.)
'' 22.5sin73.821.6MPa o xy τ = = ' 67.522.5cos73.873.8MPa o x σ =+=
Sketch of results is as shown in solution of Prob. 1.20.
SOLUTION (1.30)
(a)
(b) ' 60 8060[1cos(40)] 2 o x σ σ ==+−− 231MPa σ =
SOLUTION (1.31)
( a ) From Mohr’s circle, Fig. (a): 1 2 max 121MPa71MPa96MPaσστ == −= o s o p 7.25'3.19' = −= θ θ By applying Eq. (1.20):
Eq. (1.19):
(CONT.)
1.31 (CONT.)
( b ) From Mohr’s circle, Fig. (b): 1 2 max 200MPa50MPa125MPaσστ == −= o s o p 55.71'55.26' = = θ θ
(b)
Through the use of Eq. (1.20), [ ] 12575000,10 75 2 1 4 500,22 2,1 ±=+±= σ or 12200MPa50MPa σσ = =−
Using Eq. (1.19),342 tan = p θ : '26.57'71.57 oo ps θθ = =
SOLUTION (1.32)
Referring to Mohr’s circle, Fig. 1.15:
From Eqs. (a), 21'' σσσσ+=+ yx
By using sin12 2 cos 2 2 θθ=+ , and Eqs. (a) and (b), we have const yxyx =⋅=−⋅ 21 2 ''''σστσσ . SOLUTION (1.33)
We have
(a)
(b)
Figure
1.33 (CONT.)
Equations (1.18):
SOLUTION (1.34)
Substituting the given
SOLUTION (1.35)
Transform from o 60 = θ to o 0 = θ with ''20MPa,60MPa xy σσ = −= , '' 22MPa xy τ =− , and o 60 θ−= . Use Eqs. (1.18): 20602060 22 cos2(60)22sin2(60)59MPa oo x σ −+−−
y x 19 MPa
MPa
MPa
SOLUTION (1.36)
Figure (a)
( a ) Figure (a): 14sin6012.12MPa o y σ == 14cos607MPa o xy τ ==
Figure (b): 060 sin 60 cos 12.12 = =
o xy o y F τ or 7MPa xy τ = (as before)
= ++ −= 060 cos 73060 sin oo xx F σ or 38.68MPa x σ =
( b ) Equation (1.20) is therefore:
2 12.1268.38 2,1 7)( + ±= + σ or 1240.41MPa,10.39MPa σσ = =
Also, o p 9.13 tan12.1268.38 )7(2 1 2 1 = = θ
Note: Eq. (1.18a) gives, ' 40.41MPa x σ = Thus, o p 9.13' = θ
xsin60o
Figure (b)
SOLUTION (1.37)
Figure (a):
100cos4570.7MPa o x σ ==
100sin4570.7MPa o y σ ==
100cos4570.7MPa o xy τ ==
Now, Eqs. (1.18) give (Fig. b):
' 70.7070.7sin2409.47MPa o x σ =++=
'' 070.7cos24035.35MPa o xy τ =−+ =−
' 70.7070.7sin240131.9MPa o y σ =−−=
SOLUTION (1.38)
70sin3035MPa o y σ =−=−
70cos3060.6MPa o xy τ ==
( a ) Figure (a): 0)866.0(6.605.0150 = ++−= ∑ xx F σ or 195MPa x σ =
150 MPa
(a)
xsin30o
Area=1
Figure (a)
Figure (b)
Figure
1.38 (CONT.)
( b ) Equation (1.20): 1 2 22 1953519535 1,2 22()60.6 σ =±+−+
or 12210MPa50MPa σσ = =−
Also, o p 89.13 tan 35195 )6.60(2 1 2 1 = = + θ
Equation (1.18a): ' 80115cos2(13.89)60.6sin2(13.89)210MPa oo x σ = ++=
Thus, o p 89.13' = θ
SOLUTION (1.39)
For pure shear, 2 1 σσ −= : rt P t pr t pr π 22 +−= from which 2 3 prP π =
SOLUTION (1.40)
Table D.4: Art π2 = trJ 3 2π =
Stresses are (Fig. a):
1.40 (CONT.)
Hence,
Therefore,
Also
Equation (1.18b) with
Thus,
SOLUTION (1.42)
At
Hence
Equations(1.18):
SOLUTION (1.44)
Equation
From Mohr’s circle (Fig. a):
Results are shown in Fig. (b).
SOLUTION (1.47)
Figure (a)
( a ) At o 60 θ−= (Fig. a): )60(2 sin )60(2 cos 0 0 2 2 yxyxoo −+− += + τ σσσσ or 0 732.15.15.00−+=τσσ xy (a)
We also have 0 2 sin2(60)cos2(60) xyoo o σσ ττ=−−+− or xy στσ+= 0 464.3 (b)
Substituting Eq. (b) into (a), we obtain 0 = y σ . Results are shown in Fig. b.
Alternatively, using an element ABC (Fig. c):
ττσ xx F or ,464.3 0 τσ = x as before.
Figure (c)
Stresses on planes at o 20 , taking o
(Fig. b):
Figure (b)
1.47 (CONT.)
( b ) Principal stresses:
The maximum principal stress is on plane inclined at
SOLUTION (1.48)
At a critical point on the shaft surface, the state of stress of stress is as shown in Fig. (a). We have
Figure (a)
Results are shown in Fig. (b).
SOLUTION (1.49)
Apply Eqs. (1.20) to Fig. P1.49b, for o 30 θ−= : 40sin2(30)203MPa o xbσ =−−= 203MPa yb σ =− (b) 40cos2(30)20MPa o xybτ =−−=− (CONT.) xy τ x σ y x
Figure (b)
1.49 (CONT.)
Now apply Eqs. (1.18) to Fig. P1.49c, for o 60 θ−= : 10sin2(60)53MPa o xc σ =−=− 53MPa yc σ = (c) 10cos2(60)5MPa o xyc τ =−=−
Superposing stresses in Eqs. (b) and (c) and those in Fig. P1.49a, we obtain Fig. (a).
Figure (a) 153MPa x x’ ' p θ
Referring to Fig. (a): [ ]2 1 2 2 2,1 )45()315(0 −+±= σ or 1251.96MPa51.96MPa σσ = =− When o p 30 tan ' )315(2 )45(2 1 2 1 −= = θ is substituted into Eq. (1.18a), we have 51.96 MPa (Fig. b).
SOLUTION (1.50)
Apply Eqs. (1.18) to Fig. P1.50a, for o15 θ−= , to obtain stresses in Fig. (a): 3030 22 cos2(15)27.99MPa o xa σ =−−−=− 1515cos2(15)2.01MPa o ya σ =−+−=− 15sin2(15)7.5MPa o xya τ =−=− (CONT.) y x 153MPa 45MPa
51.96 MPa
51.96 MPa
Figure (b)
1.50 (CONT.)
Superposition of stresses in Figs. (a) and P1.50b gives Fig. (b).
Figure (a)
Apply Eq. (1.20) to Fig. (b): [
or
When o p 58.5 tan )99.4799.27( 1)5.7(2 2 1 = = +− θ is substituted into Eq. (1.18a), we obtain –28.72 MPa (Fig. c).
SOLUTION (1.51)
Equations (1.18) are applied to Fig. P1.51a, for o 30 θ−= : 20302030 22 cos2(30)22.5MPa o
Figure (b)
These stresses and that of Fig. P1.51b are superimposed to yield Fig. (a).
(CONT.)
1.51 (CONT.)
Principal stresses are thus
We have
Equation (1.18a) results in 37.522.537.522.5 '
Therefore o p 2.31' = θ
Results are shown in a properly oriented element in Fig. (b).
SOLUTION (1.52)
Figure (b)
State of stress is represented by Mohr’s circle in Fig. (a).
From this circle, we determine
Results are shown in Fig. (b).
SOLUTION (1.53)
'' yxyx σσσσ+=+
453027;42MPa σσ −=−+= ' 45304530 42 cos215sin2 22 x σ θθ −+ ==++ or 34.537.5cos215sin2 =+θθ (1)
'' 2137.5sin215cos2 xy τθθ =−=−+
Multiply this by :5.2 52.593.75sin237.5cos2θθ = (2)
Add Eqs. (1) and (2), 87108.75sin2,253.13o θθ = = or 26.6 o θ =
SOLUTION (1.54)
State of stress is represented by Mohr’s circle in Fig. (a).
Referring to this circle, we obtain the results (Fig. b).
SOLUTION (1.55)
' 60MPa18MPa15MPa30MPa xyxyx σστσ == −= −=
From Equation (1.18a):
We have '' 12MPa yxyx σσσσ=+−=
Equation (1.18b) gives 6018 '' 2 sin56.5215cos56.5240.8MPa oo xy τ + = −−=
SOLUTION (1.56)
We have
State of stress is represented by Mohr’s circle in Fig. (a).
SOLUTION (1.57)
1.57 (CONT.)
From Mohr’s circle,
Solving 2.771MPa p =
SOLUTION (1.58)
Mohr’s circle representing stress at point A is shown in Fig. (a).
Figure (a)
From this circle: pp 90 42 005.0 )45.0( == or 467kPa p =
Then 467(0.45) 3 2(0.005)2(0.45)(0.005) 98(10) P π =+ gives 1088kN P = SOLUTION (1.59)
1.59 (CONT.)
( a ) 36.62sin3521MPa o xy τ = −=
( b ) Because of symmetry: '' 21MPa xyxy ττ=−= and '' 140MPa xyxy σσσσ+=+= gives ' 40MPa y σ =
SOLUTION (1.60)
State of stress is represented by Mohr’s circle in Fig. (a).
Figure (a) O τ (MPa) σ (MPa) C (σx,-20) -14 (-12,20)
max −+=+=
( a ) Using this circle, we write [ ]2 1 22 2 12 max 20)(+= + x σ τ and )12(14 14 2
OC
τ Solving, 186MPa x σ =
Note that, alternately, [ ]2 1 22 2 12 2 12 20)( 14 +−=− + xx σ σ yields 186MPa x σ = , as before.
( b ) We have [ ]2 1
or 12188MPa14MPa σσ = =− and 1
2 ()101MPaτσσ=−= Also
(CONT.)
1.60 (CONT.)
Results are shown in Fig. (b).
SOLUTION (1.61)
Figure (b)
( a ) 123 96.05MPa23.95MPa0σσσ = = =
( b ) 1 12max12 2 ()()36.05MPa τσσ=−= 1 13max13 2 ()()48.03MPa τσσ=−= 1 23max23 2 ()()11.98MPa τσσ=−=
Plane of max 12 )( τ is shown in Fig. (a). Other maximum shear planes are sketched similarly.
(a) x
1.62 (CONT.) 1 2 '(3011.684)20.84MPa σ =+= 1 2 (3011.684)9.158MPa r =−=
( a ) ' 'cos6416.82MPa o x r σσ=−= ( b ) '' sin648.231MPa o xy r τ ==
SOLUTION (1.63)
Equilibrium of ' x and ' y directed forces results in (Fig. a): 0))(5.225()(5021
3 =−−pp or 494kPa allp = and 0)(50))(5.225(7
3 =−−+ pp from which 547kPa p =
SOLUTION (1.64) Direction cosines are:
Equation (1.28a) is thus 3 3
Similarly, applying Eqs. (1.28b) through (1.28e), we obtain][ ''ji τ : (CONT.)
1.64 (CONT.) 25.3922.667.99 2.665.39216.16MPa
Then,
(1.34)
Equation (1.28a) is therefore 00))()(40(200)(60
Similarly,
Then,
Referring to Appendix B:
SOLUTION (1.67)
Referring to Appendix B: 123 66.016MPa28.418MPa44.479MPaσσσ = = =− and
SOLUTION (1.68)
Referring to Appendix B: 123 30.493MPa12.485MPa16.979MPaσσσ = = =−
SOLUTION (1.69)
Referring to Appendix B: 123 24.747MPa8.480MPa2.773MPaσσσ = =
and
SOLUTION (1.70)
( a ) Equation (1.32) becomes
Expanding, 0]400)30([
Thus
SOLUTION (1.71)
( a ) At point (3,1,5) with respect to xyz axis, we have][ ijτ :
Then, Eqs. (1.34) result in
Direction cosines of x’ y’ z’, referring to Fig. (a) are
Now Eqs. (1.28) and (a) give][ ''ji τ :
Thus, Eqs. (1.34) yield
as before.
( b ) Direction cosines are (Fig. b):
Figure (b)
1.71 (CONT.)
With these and Eq. (a), Eqs. (1.28) yield][ ''ji τ : 7.25.60
Thus, Eqs. (1.34) result in
The I’s are thus invariants.
SOLUTION (1.72)
Introducing the given data into Eq. (1.28a), we obtain 00)])((6[2)0(14)(10)(12
Remaining stress components are determined in a like manner. The result,][ ''ji τ , is 15.6963.8667.089 3.8666.3046.294MPa 7.0896.29414.
SOLUTION (1.73)
Equations (1.34) become
Equation (1.33) is then
Solution of this quadratic equation is
SOLUTION (1.74)
Referring to Appendix B, we obtain the following values.
SOLUTION (1.75)
( a ) Direction cosines are:
8.054 == l 6.053 == m 0 = n
Equation (1.40) is thus 22
100(0.8)60(0.6)2(40)(0.8)(0.6)124MPa σ =++=
Equations (1.26) yield
100(0.8)40(0.6)104MPa x p =+=
40(0.8)60(0.6)68MPa py =+=
80(0.6)48MPa z p ==
Equation (1.41) is then 1 2 2222 [1046848124]48.66MPa τ =++−=
( b ) Direction cosines are: 0 = l
447.0202 == m
894.0204 == n
Equation (1.40) results in 22
60(0.447)20(0.894)2(80)(0.447)(0.894)91.912MPa σ =++=
Equations (1.26) yield 40(0.447)17.88MPa x p ==
60(0.447)80(0.894)98.34MPa py =+=
80(0.447)20(0.894)53.64MPa z p =+=
Equation (1.41) leads to 1 2 2222 [17.8898.3453.6491.912]66.482MPa τ =++−=
( c ) Direction cosines are:
Equation (1.40) is therefore