Solutions manual for mechanics of materials si 9th edition by hibbeler ibsn 9789810694364

SOLUTIONS MANUAL for Mechanics of Materials SI 9th Edition by Hibbeler IBSN 9789810694364 Download: https://downloadlink.org/p/solutions-manual-for-mechanics-of-materialssi-9th-edition-by-hibbeler-ibsn-9789810694364/

2-1

The center portion of th e rubber balloon has a diameter of d = 100 mm. If the air pressure within it cau ses the balloon's diameter to become d = 125 mm, determine the av erage normal strain in the rubber.

Given:

d0 := 100mm

d := 125mm

Solution: ε :=

π d − π d0 π d0

ε = 0.2500

mm mm

Ans

Ans: CE

10 5

0.00250 mm mm,

BD

0.00107 mm mm

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2–2. A thin strip of rubber has an unstretched length of 15 If it is aroundaround a pipe having outer diameter 375in.mm.

If stretched it is stretched a pipeanhaving an outer of 5 in., determine the average normal strain in normal the strip.strain diameter of 125 mm, determine the average in the strip.

L0 = 15 in.

L0 = 375 mm L == p(125 mm) p(5 in.)

L – L0 125p – 375 e5 L - L = 0.0472 5p - 15 05 = = = 0.0472 in.>in.

15 mm/mmL 0 375 L0

Ans.

Ans: P = 0.0472 mm>mm 10 6

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2–3. The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD.

D

E

4m P A

B

3m

C

2m

2m

¢LBD ¢ LCE = 7 3 3 (10) = 4.286 mm 7 ¢ LCE 10 = = 0.00250 mm>mm = L 4000

¢LBD = PCE

PBD =

Ans.

4.286 ¢ LBD = 0.00107 mm>mm = 4000 L

Ans.

Ans: PCE = 0.00250 mm>mm, PBD = 0.00107 mm>mm 10 7

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*2–4. The force applied at the handle of the rigid lever causes the lever to rotate clockwise about the pin B through an angle of 2°. Determine the average normal strain developed in each wire. The wires are unstretched when the lever is in the horizontal position.

G 200 mm

H

dA = 200(0.03491) = 6.9813 mm dC = 300(0.03491) = 10.4720 mm dD = 500(0.03491) = 17.4533 mm Average Normal Strain: The unstretched length of wires AH, CG, and DF are LAH = 200 mm, LCG = 300 mm, and LDF = 300 mm. We obtain dA 6.9813 = = 0.0349 mm>mm (Pavg)AH = LAH 200 dC 10.4720 = = 0.0349 mm>mm (Pavg)CG = L 300 CG dD = 17.4533 = 0.0582 mm>mm (Pavg)DF = LDF 300

Ans. Ans.

Ans.

10 8

E C

200 mm

2° b p rad = 0.03491 rad.

180 Since u is small, the displacements of points A, C, and D can be approximated by

200 mm 300 mm

300 mm B

A

Geometry: The lever arm rotates through an angle of u = a

F

D

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2–5. The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2 mm, determine the normal strain developed in each wire.

C

30 30

A

P

B

œ = 23002 + 22 - 2(300)(2) cos 150° = 301.734 mm LAC

PAC = PAB =

LœAC - LAC 301.734 - 300 = = 0.00578 mm>mm 300 LAC

Ans.

Ans: PAC = PAB = 0.00578 mm>mm 10 9

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2–6. The rubber band of unstretched length 2r0 is forced down the frustum of the cone. Determine the average normal strain in the band as a function of z.

r0

z

h

2r0

Geometry: Using similar triangles shown in Fig. a, h¿ h¿ +h = ; 2r0 r0

h¿ = h

Subsequently, using the result of h¿ r r = 0; z+h h

r =

r0 (z + h) h

Average Normal Strain: The length of the rubber band as a function of z is 2pr0 (z + h). With L0 = 2r0, we have L = 2pr = h

Pavg

L - L0 = L0

2pr0 (z + h) h 2r0 =

2r0

p (z + h) - 1 h

Ans.

=

Ans: Pavg = 11 0

p (z + h) - 1 h

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2–7. The pin-connected rigid rods AB and BC are inclined at u = 30° when they are unloaded. When the force P is applied u becomes 30.2°. Determine the average normal strain developed in wire AC.

P B

u

u 600 mm

A

C

Geometry: Referring to Fig. a, the unstretched and stretched lengths of wire AD are LAC = 2(600 sin 30°) = 600 mm LAC ¿ = 2(600 sin 30.2°) = 603.6239 mm Average Normal Strain: (Pavg)AC =

LAC ¿ - LAC 603.6239 - 600 = 6.04(10 - 3) mm>mm = LAC 600

Ans.

Ans: (Pavg)AC = 6.04(10 - 3) mm>mm 11 1

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*2–8. Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes it to rotate by u = 0.3°, determine the normal strain in the cable.

Originally the cable is unstretched.

u D

P 300 mm

B 300 mm A

C

400 mm

AB = 24002 + 3002 = 500 mm AB¿ = 24002 + 3002 - 2(400)(300) cos 90.3° = 501.255 mm PAB =

AB¿ -AB 501.255 - 500 = AB 500

= 0.00251 mm>mm

Ans.

11 2

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2–9. Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes a normal strain in the cable of 0.0035 mm>mm, determine the displacement of point D. Originally the cable is unstretched.

u D

P 300 mm

B 300 mm A

C

400 mm

AB = 23002 + 4002 = 500 mm AB¿ = AB + eABAB = 500 + 0.0035(500) = 501.75 mm 501.752 = 3002 + 4002 - 2(300)(400) cos a a = 90.4185° u = 90.4185° - 90° = 0.4185° = ¢ D = 600(u) = 600(

p (0.4185) rad 180°

p )(0.4185) = 4.38 mm 180°

Ans.

Ans: ¢ D = 4.38 mm 11 3

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2–10. The corners of the square plate are given the displacements indicated. Determine the shear strain along the edges of the plate at A and B.

A

16 mm B

D 3 mm

3 mm 16 mm 16 mm

C

16 mm

At A: 9.7 u¿ b = 43.561° = tan - 1 a 10.2 2 u¿ = 1.52056 rad (gA)nt =

p - 1.52056 2

= 0.0502 rad

Ans.

At B: f¿ 10.2 b = 46.439° = tan - 1 a 2 9.7 f¿ = 1.62104 rad (gB)nt =

p - 1.62104 2

= - 0.0502 rad

Ans.

Ans: (gA)nt = 0.0502 rad, (gB)nt = - 0.0502 rad 11 4

x

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2–11. The corners B and D of the square plate are given the displacements indicated. Determine the average normal strains along side AB and diagonal DB.

y A

16 mm B

D 3 mm

3 mm 16 mm 16 mm

Referring to Fig. a, LAB = 2162 + 162 = 2512 mm LAB¿ = 2162 + 132 = 2425 mm LBD = 16 + 16 = 32 mm LB¿ D¿ = 13 + 13 = 26 mm Thus,

A eavg B AB =

LAB¿ - LAB 2425 - 2512 = - 0.0889 mm>mm = LAB 2512

Ans.

A eavg B BD =

26 - 32 LB¿ D¿ - LBD = = - 0.1875 mm>mm LBD 32

Ans.

11 5

C

16 mm

x

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2–12

11 6

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2–13

.

Ans: PDB = PAB cos 2 u + PCB sin 2 u 11 7

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2–14. The force P applied at joint D of the square frame causes the frame to sway and form the dashed rhombus.

Determine the average normal strain developed in wire AC. Assume the three rods are rigid.

D

P

200 mm

E

200 mm C

3 400 mm

A

B

Geometry: Referring to Fig. a, the stretched length of LAC ¿ of wire AC¿ can be determined using the cosine law.

LAC ¿ = 24002 + 4002 - 2(400)(400) cos 93° = 580.30 mm The unstretched length of wire AC is LAC = 24002 + 4002 = 565.69 mm Average Normal Strain: (Pavg)AC =

LAC ¿ - LAC 580.30 - 565.69 = 0.0258 mm>mm = LAC 565.69

Ans.

Ans: (Pavg)AC = 0.0258 mm>mm 11 8

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2–15. The force P applied at joint D of the square frame causes the frame to sway and form the dashed rhombus.

Determine the average normal strain developed in wire AE. Assume the three rods are rigid.

D

P

200 mm

E

200 mm C

3 400 mm

A

B

Geometry: Referring to Fig. a, the stretched length of LAE¿ of wire AE can be determined using the cosine law.

LAE¿ = 24002 + 2002 - 2(400)(200) cos 93° = 456.48 mm The unstretched length of wire AE is LAE = 24002 + 2002 = 447.21 mm Average Normal Strain: (Pavg)AE =

LAE ¿ - LAE 456.48 - 447.21 = = 0.0207 mm>mm 447.21 LAE

Ans.

Ans: (Pavg)AE = 0.0207 mm>mm 11 9

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*2–16. The triangular plate ABC is deformed into the shape shown by the dashed lines. If at A, eAB = 0.0075, PAC = 0.01 and gxy = 0.005 rad, determine the average normal strain along edge BC.

y C 300 mm gxy

A

Average Normal Strain: The stretched length of sides AB and AC are LAC¿ = (1 + ey)LAC = (1 + 0.01)(300) = 303 mm LAB¿ = (1 + ex)LAB = (1 + 0.0075)(400) = 403 mm Also, u =

p 180° - 0.005 = 1.5658 rad a b = 89.7135° 2 p rad

The unstretched length of edge BC is LBC = 23002 + 4002 = 500 mm and the stretched length of this edge is LB¿ C¿ = 23032 + 4032 - 2(303)(403) cos 89.7135° = 502.9880 mm We obtain, PBC =

LB¿ C¿ - LBC 502.9880 - 500 = = 5.98(10 - 3) mm>mm 500 LBC

12 0

Ans.

400 mm

B

x

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2–17. The plate is deformed uniformly into the shape shown by the dashed lines. If at A, gxy = 0.0075 rad., while PAB = PAF = 0, determine the average shear strain at point G with respect to the x¿ and y¿ axes.

y¿

y

F

E x¿

600 mm

C

D gxy

300 mm

180° b = 0.4297°. Thus, p rad c = 90° - 0.4297° = 89.5703° b = 90° + 0.4297° = 90.4297°

Geometry: Here, gxy = 0.0075 rad a

A

G 300 mm

600 mm

B

Subsequently, applying the cosine law to triangles AGF¿ and GBC¿, Fig. a, LGF¿ = 26002 + 3002 - 2(600)(300) cos 89.5703° = 668.8049 mm LGC¿ = 26002 + 3002 - 2(600)(300) cos 90.4297° = 672.8298 mm Then, applying the sine law to the same triangles, sin f sin 89.5703° = ; 668.8049 600 sin 90.4297° sin a ; = 672.8298 300

f = 63.7791°

a = 26.4787°

Thus, u = 180° - f - a = 180° - 63.7791° - 26.4787° = 89.7422° a

p rad b = 1.5663 rad 180°

Shear Strain: (gG)x¿ y¿ =

p p - 1.5663 = 4.50(10 - 3) rad - u = 2 2

Ans.

Ans: (gG)x¿ y¿ = 4.50(10 - 3) rad 12 1

x

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2–18. The piece of plastic is originally rectangular.

Determine the shear strain gxy at corners A and B if the plastic distorts as shown by the dashed lines.

y

5 mm

2 mm 2 mm

B

4 mm

C 300 mm 2 mm D 400 mm

A

x

3 mm

Geometry: For small angles, a = c =

2 = 0.00662252 rad 302

b = u =

2 = 0.00496278 rad 403

Shear Strain: (gB)xy = a + b = 0.0116 rad = 11.6 A 10 - 3 B rad

Ans.

(gA)xy = u + c = 0.0116 rad = 11.6 A 10 - 3 B rad

Ans.

Ans: (gB)xy = 11.6(10 - 3) rad, (gA)xy = 11.6(10 - 3) rad 12 2

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2–19. The piece of plastic is originally rectangular.

Determine the shear strain gxy at corners D and C if the plastic distorts as shown by the dashed lines.

y 5 mm 2 mm 2 mm

B

4 mm

C 300 mm 2 mm D 400 mm

A

x

3 mm

Geometry: For small angles, 2 = 0.00496278 rad 403 2 b = u = = 0.00662252 rad 302 Shear Strain: (gC)xy = a + b a = c =

= 0.0116 rad = 11.6 A 10 - 3 B rad

Ans.

(gD)xy = u + c = 0.0116 rad = 11.6 A 10 - 3 B rad

Ans.

Ans: (gC)xy = 11.6(10 - 3) rad, (gD)xy = 11.6(10 - 3) rad 12 3

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*2–20. The piece of plastic is originally rectangular.

Determine the average normal strain that occurs along the diagonals AC and DB.

y

5 mm

2 mm 2 mm

B

4 mm

C 300 mm 2 mm D 400 mm

A 3 mm

Geometry: AC = DB = 24002 + 3002 = 500 mm DB¿ = 24052 + 3042 = 506.4 mm A¿ C¿ = 24012 + 3002 = 500.8 mm Average Normal Strain: A¿ C¿ -AC AC 500.8 - 500 = PAC = 500 = 0.00160 mm>mm = 1.60 A 10 - 3 B mm>mm PDB =

Ans.

DB ¿ -DB DB 506.4 - 500 = 500

= 0.0128 mm>mm = 12.8 A 10 - 3 B mm>mm

Ans.

12 4

x

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2–21. The rectangular plate is deformed into the shape of a parallelogram shown by the dashed lines. Determine the average shear strain gxy at corners A and B.

y

5 mm D

C

300 mm 5 mm A

B

x

400 mm

Geometry: Referring to Fig. a and using small angle analysis, 5 u = = 0.01667 rad 300 5 f = = 0.0125 rad 400 Shear Strain: Referring to Fig. a, (gA)xy = u + f = 0.01667 + 0.0125 = 0.0292 rad

Ans.

(gB)xy = u + f = 0.01667 + 0.0125 = 0.0292 rad

Ans.

Ans: (gA)xy = 0.0292 rad, (gB)xy = 0.0292 rad 12 5

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2–22. The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5 mm. Determine the shear strain, gxy, at A.

y

45

800 mm 45

x¿

A

45

A¿ 5 mm

800 mm

x

L = 28002 + 52 - 2(800)(5) cos 135° = 803.54 mm sin u sin 135° ; = u = 44.75° = 0.7810 rad 800 803.54 p p - 2u = - 2(0.7810) gxy = 2 2 = 0.00880 rad

Ans.

Ans: gxy = 0.00880 rad 12 6

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2–23. The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5 mm. Determine the average normal strain Px along the x axis.

y

45

800 mm 45 A 45

x¿

A¿ 5 mm

800 mm

x

L = 28002 + 52 - 2(800)(5) cos 135° = 803.54 mm 803.54 - 800 = 0.00443 mm>mm Px = 800

Ans.

Ans: Px = 0.00443 mm>mm 12 7

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*2–24. The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5 mm. Determine the average normal strain Px¿ along the x¿ axis.

y

45

800 mm 45

x¿

A

45

800 mm

x

L = 800 cos 45° = 565.69 mm 5 Px¿ = = 0.00884 mm>mm 565.69

Ans.

12 8

A¿ 5 mm

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y

2–25. The square rubber block is subjected to a shear strain of gxy = 40(10 - 6)x + 20(10 - 6)y, where x and y are in mm. This deformation is in the shape shown by the dashed lines, where all the lines parallel to the y axis remain vertical after the deformation. Determine the normal strain along edge BC. Shear Strain: Along edge DC, y = 400 mm. Thus, (gxy)DC = 40(10 - 6)x + 0.008. dy = tan (gxy)DC = tan 340(10 - 6)x + 0.0084. Then, Here, dx dy =

L0

C

A

B

400 mm

tan [40(10 - 6)x + 0.008]dx

L0

300 mm

dc = -

D

1 e ln cos c 40(10 - 6)x + 0.008 d f ` 40(10 - 6) 0

x

300 mm

= 4.2003 mm Along edge AB, y = 0. Thus, (gxy)AB = 40(10 - 6)x. Here,

dy dx

= tan (gxy)AB =

-6

tan [40(10 )x]. Then, dB

300 mm

tan [40(10 - 6)x]dx

dy = L0

L0

dB = -

300 mm 1 -6 e ln cos c 40(10 )x d f ` 40(10 - 6) 0

= 1.8000 mm Average Normal Strain: The stretched length of edge BC is LB¿ C¿ = 400 + 4.2003 - 1.8000 = 402.4003 mm We obtain, (Pavg)BC =

LB¿ C¿ - LBC 402.4003 - 400 = = 6.00(10 - 3) mm>mm 400 LBC

Ans.

Ans: (Pavg)BC = 6.00(10 - 3) mm>mm 12 9

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2– 26.

Ans: 1Pavg 2 CA = - 5.59110-3 2 mm > mm

13 0

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2–27. The square plate ABCD is deformed into the shape shown by the dashed lines. If DC has a normal strain Px = 0.004, DA has a normal strain Py = 0.005 and at D, gxy = 0.02 rad, determine the shear strain at point E with respect to the x¿ and y¿ axes.

y y¿

x¿ 600 mm A¿

B¿ B

A

600 mm

D

E

C C¿

x

Average Normal Strain: The stretched length of sides DC and BC are LDC¿ = (1 + Px)LDC = (1 + 0.004)(600) = 602.4 mm LB¿ C¿ = (1 + Py)LBC = (1 + 0.005)(600) = 603 mm Also, a =

p 180° - 0.02 = 1.5508 rad a b = 88.854° 2 p rad

f =

p 180° + 0.02 = 1.5908 rad a b = 91.146° 2 p rad

Thus, the length of C¿A¿ and DB¿ can be determined using the cosine law with reference to Fig. a.

LC¿ A¿ = 2602.42 + 6032 - 2(602.4)(603) cos 88.854° = 843.7807 mm LDB¿ = 2602.42 + 6032 - 2(602.4)(603) cos 91.146° = 860.8273 mm Thus, LE¿ A¿ =

LC¿A¿ = 421.8903 mm 2

LE¿ B¿ =

LDB¿ = 430.4137 mm 2

Using this result and applying the cosine law to the triangle A¿ E¿ B¿ , Fig. a, 602.42 = 421.89032 + 430.41372 - 2(421.8903)(430.4137) cos u u = 89.9429° a

p rad b = 1.5698 rad 180°

Shear Strain: (gE)x¿ y¿ =

p p - 1.5698 = 0.996(10 - 3) rad - u = 2 2

Ans.

Ans: (gE)x¿ y¿ = 0.996(10 - 3) rad 13 1

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*2–28.

The wire is subjected to a normal strain that is

(x/L)e–(x/L)

P

2

defined by P = (x>L)e - (x>L) . If the wire has an initial length L, determine the increase in its length.

x x L

L

¢L =

1 - (x>L)2 xe dx L L0 2

= -L c

=

2

L e - (x>L) L d = 31 - (1>e)4 2 2 0

L 3e - 14 2e

Ans.

13 2

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2–29.

Ans: 1Pavg 2AC = 0.0168 mm >mm, 1gA 2 xy = 0.0116 rad 13 3

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2–30. The rectangular plate is deformed into the shape shown by the dashed lines. Determine the average normal strain along diagonal BD, and the average shear strain at corner B.

y 6 mm 400 mm 2 mm

2 mm

6 mm C

D

300 mm 2 mm A

Geometry: The unstretched length of diagonal BD is

x

B 400 mm

3 mm

LBD = 23002 + 4002 = 500 mm Referring to Fig. a, the stretched length of diagonal BD is LB¿ D¿ = 2(300 + 2 - 2)2 + (400 + 3 - 2)2 = 500.8004 mm Referring to Fig. a and using small angle analysis, 2 = 0.004963 rad 403 3 a = = 0.009868 rad 300 + 6 - 2 f =

Average Normal Strain: Applying Eq. 2, (Pavg)BD =

LB¿ D¿ - LBD 500.8004 - 500 = = 1.60(10 - 3) mm>mm Ans. 500 LBD

Shear Strain: Referring to Fig. a, (gB)xy = f + a = 0.004963 + 0.009868 = 0.0148 rad

Ans.

Ans: (Pavg)BD = 1.60(10 - 3) mm>mm, (gB)xy = 0.0148 rad 13 4

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2–31. The nonuniform loading causes a normal strain in the shaft that can be expressed as Px = kx2, where k is a constant. Determine the displacement of the end B. Also, what is the average normal strain in the rod?

L A

B x

d(¢x) = Px = kx2 dx L

(¢x)B =

(Px)avg

2

L0

kx =

(¢x)B = = L

kL 3

3

Ans.

kL3 3 kL2 = L 3

Ans.

Ans: 3

(¢x)B =

13 5

kL kL2 , (Px)avg = 3 3

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*2–32 The rubber block is fixed along edge AB, and edge CD is moved so that the vertical displacement of any point in the block is given by v(x) = (v0>b3)x3. Determine the shear strain gxy at points (b>2, a>2) and (b, a).

y

v (x) A

v0

D

a

B

Shear Strain: From Fig. a,

b

dv = tan gxy dx 3v0 2 x = tan gxy b3 gxy = tan - 1 a

3v0 2 x b b3

Thus, at point (b>2, a>2), gxy = tan - 1 c

3v0 b 2 a b d b3 2

3 v0 = tan - 1 c a b d 4 b and at point (b, a), gxy = tan - 1 c

3v0 b3

= tan - 1 c 3 a

C

Ans.

(b2) d

v0 bd b

Ans.

13 6

x

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2–33. The fiber AB has a length L and orientation u. If its ends A and B undergo very small displacements uA and vB, respectively, determine the normal strain in the fiber when it is in position A¿ B¿ .

y B¿ vB B L A

uA A¿

u x

Geometry: 2 LA ¿ B ¿ = 2(L cos u - u A ) + (L sin u + vB)2 2

2

= 2L2 + uA + vB + 2L(vB sin u - uA cos u) Average Normal Strain: LA¿ B¿ - L PAB = L 2

=

A

1 +

2

uA + v B 2(vB sin u - uA cos u) + - 1 L L2

Neglecting higher terms u2A and v2B PAB = B 1 +

2(vB sin u - uA cos u) 21 R - 1 L

Using the binomial theorem: 2uA cos u 1 2v sin u PAB = 1 + ¢ B ≤ + ... - 1 2 L L =

vB sin u uA cos u L L

Ans.

Ans. PAB = 13 7

vB sin u uA cos u L L

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2–34. If the normal strain is defined in reference to the final length, that is, P¿n = lim ¿ a p:p

¢s¿ -¢s b ¢s¿

instead of in reference to the original length, Eq. 2–2 , show that the difference in these strains is represented as a second-order term, namely, Pn - Pn¿ = Pn Pn¿ .

PB =

¢ S ¿ -¢S ¢S

PB - PœA =

¢S ¿ -¢S ¢S

-

¢ S ¿ -¢S ¢S¿

¢ S ¿ 2 -¢S¢S ¿ -¢S¿ ¢S +¢S2 ¢S¢S¿ ¢ S ¿ 2 +¢S2 - 2¢S¿ ¢S = ¢S¢S¿ =

=

(¢ S ¿ -¢S)2 ¢S¢S¿

= ¢

¢ S¿ -¢S ¢ S¿ -¢S ≤¢ ≤ ¢S¿ ¢S

= PA PœB (Q.E.D)

13 8

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