Solution Manual Mechanics of Material by Ansel C.Ugural Chapter 7 PDF

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Solution Manual Mechanics of Material by Ansel C.Ugural Chapter 7...

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PROBLEM (7.1) A copper wire of diameter d and modulus of elasticity E is coiled around a spool of radius r . Compute the maximum stress and the bending moment in the wire. Given: d = 1/32 in., r = 6 in., E = 17 x 10 6 psi SOLUTION d

ρ

r C

r = 6 i n. ρ = 6.015625 in. 1 d= = 0.03125 in. 32

Equation (7.4): Ec 17 × 106 (0.015625) = = 44.2 ksi σ max = ρ 6.015625 Mc 32 M = σ max = πd 3 I or πd 3 M = σ max 32 3 3 44.2 × 10 (0.03125) π = = 0.138 lb ⋅ in. 32 ________________________________________________________________________

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PROBLEM (7.2) An aluminum bar of modulus of elasticity E, width b, depth h , and length

L , is acted upon by bending moments M at its ends so that the midpoint deflection is δ (Fig. P7.2). Determine the maximum longitudinal strain and the bending moment in the bar. Given: E = 70 GPa, b = 10 mm, h = 30 mm, L = 2 m, δ = 10 mm Assumption: The deflection curve is nearly flat so that the distance between the ends of the bent bar equals the length of the beam as indicated in the figure.

SOLUTION 3

0.01(0.03) 12 = 22.5 ×10− 9 m 4

I=

O

ρ y 1m

ρ 10 mm

1m y

M

M

x

z

h=30 mm

v max From geometry: ρ 2 − 1 = ( ρ − vmax ) 2 = ( ρ − 0.01) 2 = ρ 2 − 0.02 ρ + 10 −4 or ρ = 50 m Equation (7.2), for y = h 2 : h 0.03 ε max = = = 300 μ 2 ρ 2(50) Equation (7.6): EI 70 × 109 (22.5 × 10 −9 ) M = = = 31.5 N ⋅ m 50 ρ

________________________________________________________________________

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PROBLEM (7.3)

A steel beam having the modulus elasticity E and allowable normal stress

σ all experiences pure bending (Fig. P7.3). Calculate the maximum moment M applied and the corresponding radius of curvature ρ for the two cases: (a) The cross section is a circle of diameter d. (b) The cross section is an equilateral triangle of sides b. Given: d = 12 mm, b = 15 mm, E = 200 GPa, σ all = 160 MPa

that may be

SOLUTION

σ all =

(a)

32M π d3

or D=2c=12 mm

M =

σ allπ d 3

32 (160 ×10 6 )π (0.012) 3 M = 32 = 27.14 N ⋅ m Equation (7.4): 1 σ all 160× 106 = = = 0.133 m −1 9 ρ Ec 200 ×10 (0.006) ρ = 7.52 m Alternatively, Eq. (7.6) yields the same result. 3 1 15(12.99) (b) I = bh 3 = 36 36 −12 = 913× 10 m 4 2h c= h=12.99 mm σ all = Mc I 3 C o o or 60 60 M = σall I (2h 3) B=15 mm

M =

160 ×106 (913 × 10−12 ) = 16.87 N ⋅ m 8.66× 10− 3

Equation (7.9): 1 σ all 160× 106 = = = 0.0924 m− 1 ρ Ec 200 ×10 9(8.66 ×10− 3) or ρ = 10.82 m

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________________________________________________________________________ PROBLEM (7.4) If a circular beam and equilateral triangular beam are to resist the same maximum stress and bending moment (Fig. P7.3), what is the ratio of their cross-sectional areas?

SOLUTION

σ max =

y

32M πd 3

d

z

bh 3 b b 3 3 = ( ) 36 36 2 3 4 = b 96

I=

y

b

h=b 3 2 z

b

σ max =

M (2h 3) 4

3b 96

=

32M b3

Thus, 32M 32M = 3 , b π d3

b = 1.46d

πd 2

1 1.46d 3 At = (1.46d ) 4 2 2 At Ac = 3.69 π = 1.19

Ac =

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________________________________________________________________________ PROBLEM (7.5) A steel rod of moludulus of elasticity E and semicircular cross section with radius r is acted upon by a bending moment Mz (Fig. P7.5). Calculate: (a) The maximum tensile and compressive strains in the rod. (b) The radius of curvature of the rod. Given: Mz = - 1.5 kip ⋅ in. , E = 30 x 106 psi, r = 1 in.

SOLUTION

y x

M

M z=M

C

0.576 in.

r

2 in.

4r

=0.424 in.

r =1 in. (a) I z = 0.11r 4 = 0.11(1) 4 = 0.11 in.4 Mc 1500(0.576) σt = = = 7.855 ksi I 0.11 1500(0.424) Mc σc = − =− = −5.782 ksi I 0.11 Then 7855 σ εt = t = = 262 μ E 30 × 106 5782 σ εc = c = − = − 193 μ 30× 106 E (b )

1

ρ

=

M 1500 , = EI 30 ×10 6(0.11)

ρ = 2200 in. = 183.3 ft

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________________________________________________________________________ PROBLEM (7.6) A brass bar having a modulus of elasticity E and semicircular cross section of radius r (Fig. P7.5) is bent over a cylinder of diameter D. What is the bending moment Mz in the bar? Given: E = 15 x 10 6 psi, r = 1 in., D = 36 ft

SOLUTION

Refer to figure in Solution of Prob. 7.5.

C

18 12 0.424 in. = + ρ = ×216.424 Mz

ρ

σ max =

Ec

ρ

=

15× 106 (0.576) = 39.92 ksi 216.424

Thus, Mz =

σ max I c

=

39.92 ×10 3(0.11) = 7.62 kip ⋅in. 0.576

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________________________________________________________________________ PROBLEM (7.7) Moment M acts about the horizontal (x) axis of a square bar of sides b in the two different positions shown in Fig. P7.7. Determine the maximum stress and the radius of curvature of the bar, in terms of the modulus of elasticity E, b, and M, as required.

SOLUTION

Refer to Table B.6 and Appendix A. M (b 2) 6M = 3 b 4 12 b M 1 M 12 M = = = 4 ρ EI E( b 12) Eb4

σ=

y C

z

b

b

For a triangle: bh 3 bh 3 bh h 2 bh 3 Iz = + Ad 2 = + ( ) = 36 36 2 3 12 Hence, for square:

y

4

b 2

b z

C

b 2

b 2

Therefore,

σ max =

6 2M b3

I z = 2[

b b 1 (b 2)( ) 3] = 12 12 2

Mc M (b 2) 6 2M = = Iz b 4 12 b3 1 M 12 M = = ρ EIz Eb4

σ=

ρ=

4

Eb 12 M

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________________________________________________________________________ PROBLEM (7.8) An aluminum bar of rectangular cross section is acted upon by a bending moment M about the z axis as shown in Fig. P7.8. Calculate: (a) The radius of curvature ρ of the beam. (b) The radius of curvature ρ1 of a transverse cross section of the beam. Given: b = 25 mm, h = 75 mm, M = 5 kN ⋅ m, E = 72 GPa, ν = 0.29

SOLUTION

1 3 1 bh = (25)(75) 3 = 878.91 (103 ) mm4 12 12 1 M 5(10 3) = = = 0.07901 (a) ρ EI 72(109 )(878.91×10−9 ) or ρ = 12.657 m I=

(b) Equation(7.3a)

ρ1 = −

ρ 12.657 =− = − 43.645 m ν 0.29

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________________________________________________________________________ PROBLEM (7.9) An S 8 x 23 rolled-steel I beam (see Table B.7) carries a moment M about the z axis, as shown in Fig. P7.9. Determine: (a) The radius of curvature ρ of the beam. (b) The radius of curvature Given: E = 29 x 10 6 psi,

ρ1 of a transverse cross section of the beam. ν = 0.3, M = 400 kip ⋅ in.

SOLUTION

For S150x26 rolled-steel section: I = 64.4 in.4 (a)

M 400 ×10 = = 2.142(10 −4 ) ρ EI (29 × 10 6 )(64.4) ρ = 4,669 in. = 389 ft 1

3

=

(b) ρ1 = −

ρ 4,669 =− = − 15,563 in.= − 1, 297 ft ν 0.3

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________________________________________________________________________ PROBLEM (7.10) A rectangular tube is made of an aluminum alloy with an allowable stress of σ all and is acted upon by a bending moment M about the z axis (Fig. P7.10). Calculate: (a) The bending moment. (b) The radius of curvature ρ of th beam. Given :

σ all = 120 MPa,

E=70 GPa

SOLUTION

bh 3 1 Iz = − (b − 2t )( h − 2t ) 3 12 12 1 1 = (100)(150)3 − (80)(130)3 12 12 = 13.478(106 ) mm 4

y

t t

M C

z

t

(a)

σ all

Mc , = Iz

h

t b

I 13.478(10 −6 ) M = σ all = 120(10 6) = 21.565 kN ⋅ m 0.075 c 1 M 21,565 (b ) = = = 0.02286 9 ρ EI (70 × 10 )(13.478)(10 − 6 ) ρ = 43.74 m

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________________________________________________________________________ PROBLEM (7.11) An aluminum beam with an hollow circular cross section is acted upon by a bending moment M about the z axis (Fig.P7.11). Calculate: (a) The normal stress at point A. (b) The normal stress at point B. (c) The radius of curvature ρ1 of the beam of a transverse cross section. Given: M = 400 N ⋅ m, D = 50 mm,

d = 30 mm,

E = 70 GPa,

ν = 0.29

SOLUTION

I=

π 1 ( D 4 − d 4 ) = (50 4 − 304 ) = 267.035(103 ) mm4 64 64

(a) y A = 25 mm = 0.025 m. MyA (400)(0.025) = = 37.45 MPa σA = I 267.035(10 −9 ) (b) y B = 15 mm = 0.015 m. MyB (400)(0.015) σB = = = 22.47 MPa I 267.035(10 −9 ) (c)

1

ρ

= =

M EI

z

400 70(10 )(267.035 ×10 −9) 9

= 0.0214

ρ = 46.73 m

y

M C B A d D

Hence,

ρ1 = −

46.73 ρ =− = − 161.14 m ν 0.29

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________________________________________________________________________ PROBLEM (7.12) An aluminum, alloy 6061-T6, beam of hollow circular cross section (Fig. P7.11) has an allowable bending strength σ all . Determine: (a) The largest bending moment M about the z axis that may be applied based on a safety factor n s with respect to yielding of the aluminum. (b) The corresponding radius of curvature ρ of the beam. Given: D = 2 in., d = 1.5 in., n s = 1.9

SOLUTION

σ y = 38 ksi π 4

I = =

64

π

64

E = 10 × 10 9 psi

(from Table B.4)

(D − d4 )

(2 4 − 1.54 ) = 248.51(10 −3 ) in .4

σy

38 = 20 ksi 1.9 Mc M (1) ; 20(10 3 ) = σ all = −3 248.51(10 ) I Solving, M = 4.97 kip ⋅ in.

(a) c = 1 in.

(b )

1

ρ or

=

σ all =

ns

=

M 4.97(103 ) = = 0.002 EI (10 × 10 6 )(248.51× 10− 3)

ρ = 500 in = 41.67 ft

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________________________________________________________________________ PROBLEM (7.13) An S 150 x 26 steel I beam (see Table B.8) is simply supported on a span L and carries a uniform load of intensity w. (a) Determine the maximum value of w. (b) Determine the width of a solid rectangular section of the same 152-mm depth to replace the I shape. (c) Compare the areas of the two sections. Given: L = 10 in., σ all = 120 MPa

SOLUTION

S150x26

w 152 mm 10 m

wL 2

wL 2

As =3270 mm2 S =144x10-6

2 M max wL 8 = S S 6 6 8 Sσ all 8(144 × 10− )(120 × 10 ) w= = = 1.38 kN m 2 L2 (10)

(a) σ all =

(b ) bh2 S= 6

152 mm

144× 10− 6 = b

b (0.152)2 , 6

b = 37.4 mm

(c) As = 3270 mm 2

Ar = 37.4(152) = 5685 m m 2

Thus, Ar As = 1.74

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________________________________________________________________________ PROBLEM (7.14) A simply supported channel AB carries a concentrated load P at the midpoint, as shown in Fig.P7.14. Compute the maximum tensile stress compressive stress Given: L = 10 ft,

σc

σt

and largest

in the member.

b = 14 in.,

h = 6 in.,

t = 0.5 in.,

P = 5 kips

SOLUTION y =∑

2 A1 y + A2 y 2 2 A1 + A2 ∑ Ai 2(6 × 1)(3) + (12 × 1)(5.5) = 2(6 × 1) + (12 × 1) = 4.25 in. Ai y i

=

y 1 in. z

A2

A1

c2 c1

1 in.

C

y

6

1 in. 12

So c1 = 4.25 in. and c2 = 1.75 in. 1 I z = 2[ (1)(6) 3 + (6 ×1)(1.25) 2 ] + 12 1 (12)(1)3 + (12 ×1)(1.25) 2 = 74.5 in.4 12 1 1 PL = (5)(10 × 12) = 300 kip ⋅ in. 4 2 300(4.25) Mc = 17.11 ksi σt = 1 = Iz 74.5 300(1.75) Mc = −7.05 ksi σc = − 2 = − Iz 74.5 Mz = M =

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________________________________________________________________________ PROBLEM (7.15) A simple beam AB of a channel cross section is acted upon by a concentrated load P at midpoint as shown in Fig. P7.14. Determine the maximum permissible load P based upon an allowable normal stress σ all in the beam. Given: L = 2 m,

b = 400 mm,

h = 150 mm,

t = 25 mm,

σ all

= 50 MPa

SOLUTION

y =∑

Ai y i

∑A

=

2 A1 y + A2 y 2 2 A1 + A2

y

A2 A1 2(150 × 25)(75) + (350 × 25)(137.5) = c2 2(150× 25)+ (350× 25) z y 25 C c1 = 108.65 mm 25 So c1 = 108.65 mm and c2 = 41.35 mm 350 mm 1 3 2 I z = 2[ (25)(150) + (25 ×150)(33.65) ] + 12 1 (350)(25)3 + (350 × 25)(28.85)2 = 30.29(10 6 ) mm4 12 1 1 M z = M = PL = p (2) = P 2 2 i

25

Therefore Mc P(0.10192) σ t = 1 ; 50(106 ) = , P = 14.86 ksi Iz 30.29(10−6 ) Mc P (0.0481) σ c = 2 50(106 ) = , P = 31.49 k N Iz 30.29(10− 6) Hence Pall = 14.86 k N

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150 mm

________________________________________________________________________ PROBLEM (7.16) An unsymmetrical W beam is acted upon by a bending moment M z ,as shown in Fig. P7.16. For the ratio of bending stresses of 5/8, at the top and bottom of the beam, determine the width b of the bottom flange.

SOLUTION 150 mm

20 mm

y

...