Solutions Chapter 4

C

PROBLEM 4.1

Knowing that the couple shown acts in a vertical plane, determine the stress at (a) point A, (b) point B.

SOLUTION

For rectangle: 1 3

12 I= bh For cross sectional area:

3 3 3 4 1 2 3 1 1 1 (2)(1.5) (2)(5.5) (2)(1.5) 28.854 in 12 12 12 I = I +I + I = + + = (a) y_A = 2.75 in. (25)(2.75) 28.854 A A My I σ = − = − _σA = −2.38 ksi  (b) y_B =0.75 in. (25)(0.75) 28.854 B B My I σ = − = − _σB = −0.650 ksi 

PROBLEM 4.2

Knowing that the couple shown acts in a vertical plane, determine the stress at (a) point A, (b) point B.

SOLUTION For rectangle: 1 3 12 I = bh Outside rectangle: 3 1 1 (80)(120) 12 I = 6 4 6 4 1 11.52 10 mm 11.52 10 m I = × = × − Cutout: 2 3 1 (40)(80) 12 I = 6 4 6 4 2 1.70667 10 mm 1.70667 10 m I = × = × − Section: 6 4 1 2 9.81333 10 m I = I −I = × − (a) y_A =40 mm =0.040 m 3 6 6 (15 10 )(0.040) 61.6 10 Pa 9.81333 10 A A My I σ = − = − × ₋ = − × × 61.6 MPa A σ = −  (b) y_B = −60 mm = −0.060 m 3 6 6 (15 10 )( 0.060) 91.7 10 Pa 9.81333 10 B B My I σ = − = − × − ₋ = × × 91.7 MPa B σ = 

PROBLEM 4.3

Using an allowable stress of 16 ksi, determine the largest couple that can be applied to each pipe.

SOLUTION (a)

(

4 4

)

4 4 3 4 3 (0.6 0.5 ) 52.7 10 in 4 4 0.6 in. (16)(52.7 10 ) : 0.6 o i I r r c Mc I M I c π π σ σ − − = − = − = × = × = = = 1.405 kip in M = ⋅  (b) 4 4 3 4 3 (0.7 0.5 ) 139.49 10 in 4 0.7 in. (16)(139.49 10 ) : 0.7 I c Mc I M I c π σ σ − − = − = × = × = = = 3.19 kip in M = ⋅ 

PROBLEM 4.4

A nylon spacing bar has the cross section shown. Knowing that the allowable stress for the grade of nylon used is 24 MPa, determine the largest couple Mz that can be applied to the bar.

SOLUTION 3 4 rect circle 3 4 6 4 6 6 6 3 1 12 4 1 (100)(80) (25) 3.9599 10 mm 12 4 3.9599 10 m 80 40 mm 0.040 m 2 (24 10 )(3.9599 10 ) : 2.38 10 N m 0.040 z I I I bh r c Mc I M I c π π σ σ − − = − = − = − = × = × = = = × × = = = = × ⋅ 2.38 kN m z M _{= ⋅ }

PROBLEM 4.5

A beam of the cross section shown is extruded from an aluminum alloy for which _σY =250 MPa and _σU =450 MPa. Using a factor of safety of 3.00, determine the largest couple that can be applied to the beam when it is bent about the z-axis.

SOLUTION Allowable stress. 6 450 150 MPa . . 3 150 10 Pa U F S σ = = = = ×

Moment of inertia about z-axis.

3 3 4 1 3 3 4 2 3 4 3 1 1 (16)(80) 682.67 10 mm 12 1 (16)(32) 43.69 10 mm 12 682.67 10 mm = = × = = × = = × I I I I 6 4 6 4 1 2 3 6 6 3 1.40902 10 mm 1.40902 10 m 1 with (80) 40 mm 0.040 m 2 (1.40902 10 )(150 10 ) 5.28 10 N m 0.040 I I I I Mc c I I M c σ σ − − = + + = × = × = = = = × × = = = × ⋅ M =5.28 kN m⋅ 

PROBLEM 4.6

Solve Prob. 4.5, assuming that the beam is bent about the y-axis.

PROBLEM 4.5 A beam of the cross section shown is extruded from an

aluminum alloy for which _σY =250 MPa and _σU =450 MPa. Using a factor of safety of 3.00, determine the largest couple that can be applied to the beam when it is bent about the z-axis.

SOLUTION Allowable stress: 6 450 150 MPa . . 3.00 150 10 Pa σ = = = = × U F S

Moment of inertia about y-axis.

3 2 3 4 1 3 3 4 2 3 4 3 1 1 (80)(16) (80)(16)(16) 354.987 10 mm 12 1 _{(32)(16) 10.923 10 mm} 12 354.987 10 mm = + = × = = × = = × I I I I 3 4 9 4 1 2 3 9 6 3 720.9 10 mm 720.9 10 m 1 with (48) 24 mm 0.024 m 2 (720.9 10 )(150 10 ) 4.51 10 N m 0.024 I I I I Mc c I I M c σ σ − − = + + = × = × = = = = × × = = = × ⋅ M =4.51 kN m⋅ 

PROBLEM 4.7

Two W4 13× rolled sections are welded together as shown. Knowing that for the steel alloy used _σY =36 ksi and _σU =58 ksi and using a factor of safety of 3.0, determine the largest couple that can be applied when the assembly is bent about the z axis.

SOLUTION

Properties of W4 × 13 rolled section. (See Appendix C.) 2 4 Area 3.83 in Depth 4.16 in. 11.3 in x I = = =

For one rolled section, moment of inertia about axis a-a is

2 _{11.3 (3.83)(2.08)}2 _{27.87 in}4

a x

I ₌I ₊Ad _{= + =}

For both sections, _{2 55.74 in}4

depth 4.16 in. z a I I c = = = = all all all 58 19.333 ksi . . 3.0 (19.333) (55.74) 4.16 σ σ σ σ = = = = = = U Mc F S I I M c Mall=259 kip in⋅ 

PROBLEM 4.8

Two W4 13× rolled sections are welded together as shown. Knowing that for the steel alloy used _σY =36 ksi and _σU =58 ksi and using a factor of safety of 3.0, determine the largest couple that can be applied when the assembly is bent about the z axis.

SOLUTION

Properties of W4 × 13 rolled section. (See Appendix C.) 2 4 Area 3.83 in Width 4.060 in. 3.86 in y I = = =

For one rolled section, moment of inertia about axis b-b is

2 _{3.86 (3.83)(2.030)}2 _{19.643 in}4

b y

I ₌I ₊Ad _{= + =}

For both sections, _{2 39.286 in}4

width 4.060 in. z b I I c = = = = all all all 58 19.333 ksi . . 3.0 (19.333) (39.286) 4.060 σ σ σ σ = = = = = = U Mc F S I I M c Mall=187.1 kip in⋅ 

PROBLEM 4.9

Two vertical forces are applied to a beam of the cross section shown. Determine the maximum tensile and compressive stresses in portion BC of the beam.

SOLUTION 2 2 2 1 1 2 2 2 4 (4)(25) (25) 981.7 mm 10.610 mm 2 2 3 3 25 (50)(25) 1250 mm 12.5 mm 2 2 r A r y h A bh y π π π π = = = = = = = = = = − = − = − 1 1 2 2 1 2 (981.7)(10.610) (1250)( 12.5) 2.334 mm 981.7 1250 A y A y y A A + + − = = = − + + 1 2 4 2 4 2 6 4 1 1 1 1 1 1 1 2 3 2 3 4 1 1 1 1 3 3 3 4 2 2 2 (25) (981.7)(10.610) 42.886 10 mm 8 8 10.610 ( 2.334) 12.944 mm 42.866 10 (981.7)(12.944) 207.35 10 mm 1 1 (50)(25) 65.104 10 mm 12 12 12.5 x I I A y r A y d y y I I A d I bh d y y π π = − = − = − = × = − = − − = = + = × + = × = = = × = − = − 2 3 2 3 4 2 2 2 2 3 4 9 4 1 2 top bot ( 2.334) 10.166 mm 65.104 10 (1250)(10.166) 194.288 10 mm 401.16 10 mm 401.16 10 m 25 2.334 27.334 mm 0.027334 m 25 2.334 22.666 mm 0.022666 m I I A d I I I y y − − − = = + = × + = × = + = × = × = + = = = − + = − = − 3 3 0 : (4 10 )(300 10 ) 1200 N m M −Pa = M = Pa = × × − = ⋅ top 6 top 9 (1200)(0.027334) 81.76 10 Pa 401.16 10 My I σ = − = − ₋ = − × × σtop = −81.8 MPa 6 bot bot 9 (1200)( 0.022666) 67.80 10 Pa 401.16 10 My I σ = − = − − ₋ = × × σbot =67.8 MPa

PROBLEM 4.10

Two vertical forces are applied to a beam of the cross section shown. Determine the maximum tensile and compressive stresses in portion BC of the beam.

SOLUTION 2 , mm A y₀, mm 3 0, mm Ay  600 30 _{18 10×} 3  600 30 _{18 10×} 3  300 5 _{1.5 10×} 3 1500 _{37.5 10×} 3 3 0 37.5 10 25mm 1500 Y = × =

Neutral axis lies 25 mm above the base.

3 2 3 4 4 1 2 1 3 2 3 4 3 3 4 9 4 1 2 3 1 (10)(60) (600)(5) 195 10 mm 195 mm 12 1 (30)(10) (300)(20) 122.5 10 mm 12 512.5 10 mm 512.5 10 m I I I I I I I I − = + = × = = = + = × = + + = × = × top bot 3 3 3 35 mm 0.035 m 25 mm 0.025 m 150 mm 0.150 m 10 10 N (10 10 )(0.150) 1.5 10 N m = = = − = − = = = × = = × = × ⋅ y y a P M Pa 3 top 6 top 9 (1.5 10 )(0.035) 102.4 10 Pa 512.5 10 My I σ = − = − × ₋ = − ×

× σtop = −102.4 MPa (compression) 

3 6 bot bot 9 (1.5 10 )( 0.025) _{73.2 10 Pa} 512.5 10 M y I σ = − = − × − ₋ = ×

PROBLEM 4.11

Two vertical forces are applied to a beam of the cross section shown. Determine the maximum tensile and compressive stresses in portion BC of the beam. SOLUTION A y₀ A y₀  8 7.5 60  6 4 24  4 0.5 2 Σ 18 86 86 4.778 in. 18 = = o Y Neutral axis lies 4.778 in. above the base.

3 2 3 2 4 1 1 1 1 1 3 2 3 2 4 2 2 2 2 2 3 2 3 2 4 3 3 3 3 3 4 1 2 3 top b 1 1 (8)(1) (8)(2.772) 59.94 in 12 12 1 1 (1)(6) (6)(0.778) 21.63 in 12 12 1 1 (4)(1) (4)(4.278) 73.54 in 12 12 59.94 21.63 73.57 155.16 in 3.222 in. I b h A d I b h A d I b h A d I I I I y y = + = + = = + = + = = + = + = = + + = + + = = ot = −4.778 in. 0 (25)(20) 500 kip in M Pa M Pa − = = = = ⋅ top top (500)(3.222) 155.16 My I

σ = − = − σtop = −10.38 ksi (compression) 

bot bot (500)( 4.778) 155.16 My I

PROBLEM 4.12

Knowing that a beam of the cross section shown is bent about a horizontal axis and that the bending moment is 6 kN ⋅ m, determine the total force acting on the top flange.

SOLUTION

The stress distribution over the entire cross section is given by the bending stress formula:

x

M y I σ = −

where y is a coordinate with its origin on the neutral axis and I is the moment of inertia of the entire cross sectional area. The force on the shaded portion is calculated from this stress distribution. Over an area element dA, the force is

x My dF dA dA I σ = = −

The total force on the shaded area is then

* *

My M M

F dF dA ydA y A

I I I

=



= −



= −



= −

where _y* _{is the centroidal coordinate of the shaded portion and A}* _{is its area.} 1 2 54 18 36 mm 54 36 54 36 mm d d = − = = + − =

PROBLEM 4.12 (Continued)

Moment of inertia of entire cross section:

3 2 3 2 6 4 1 1 1 1 1 3 2 3 2 6 4 2 2 2 2 2 6 4 6 4 1 2 1 1 (216)(36) (216)(36)(36) 10.9175 10 mm 12 12 1 1 (72)(108) (72)(108)(36) 17.6360 10 mm 12 12 28.5535 10 mm 28.5535 10 m− = + = + = × = + = + = × = + = × = × I b h A d I b h A d I I I

For the shaded area,

* 2 * * * 3 3 6 3 * * 3 6 6 3 (216)(36) 7776 mm 36 mm 279.936 10 mm 279.936 10 m (6 10 )(279.936 10 ) 28.5535 10 58.8 10 N A y A y MA y F I − − − = = = = × = × × × = − = × = × F =58.8 kN 

PROBLEM 4.13

Knowing that a beam of the cross section shown is bent about a horizontal axis and that the bending moment is 6 kN ⋅ m, determine the total force acting on the shaded portion of the web.

SOLUTION

The stress distribution over the entire cross section is given by the bending stress formula:

x

M y I σ = −

where y is a coordinate with its origin on the neutral axis and I is the moment of inertia of the entire cross sectional area. The force on the shaded portion is calculated from this stress distribution. Over an area element dA, the force is

x My dF dA dA I σ = = −

The total force on the shaded area is then

* *

M y M M

F dF dA ydA y A

I I I

=



= −



= −



= −

where _y* _{is the centroidal coordinate of the shaded portion and A}* _{is its area.} 1 2 54 18 36 mm 54 36 54 36 mm d d = − = = + − =

PROBLEM 4.13 (Continued)

Moment of inertia of entire cross section:

3 2 3 2 6 4 1 1 1 1 1 3 2 3 2 6 4 2 2 2 2 2 6 4 6 4 1 2 1 1 (216)(36) (216)(36)(36) 10.9175 10 mm 12 12 1 1 (72)(108) (72)(108)(36) 17.6360 10 mm 12 12 28.5535 10 mm 28.5535 10 m I b h A d I b h A d I I I − = + = + = × = + = + = × = + = × = ×

For the shaded area,

* 2 * * * 3 3 6 * * 3 6 6 3 (72)(90) 6480 mm 45 mm 291.6 10 mm 291.6 10 m (6 10 )(291.6 10 ) 28.5535 10 61.3 10 N A y A y MA y F I − − − = = = = × = × × × = = × = × F =61.3 kN 

PROBLEM 4.14

Knowing that a beam of the cross section shown is bent about a horizontal axis and that the bending moment is 50 kip ⋅ in., determine the total force acting (a) on the top flange, (b) on the shaded portion of the web.

SOLUTION

The stress distribution over the entire cross-section is given by the bending stress formula: _x My I σ = −

where y is a coordinate with its origin on the neutral axis and I is the moment of inertia of the entire cross sectional area. The force on the shaded portion is calculated from this stress distribution. Over an area element dA, the force is

x My dF dA dA I σ = = −

The total force on the shaded area is then * *

My M M

F dF dA ydA y A

I I I

=



= −



= −



= −

where _y* _{is the centroidal coordinate of the shaded portion and A}* _{is its area.} Calculate the moment of inertia.

3 3 4

1 1

(6 in.)(7 in.) (4 in.)(4 in.) 150.17 in

12 12 50 kip in I M = − = = ⋅

(a) Top flange: A*=(6 in.)(1.5 in.) =9 in2 y*= 2 in. 0.75 in.+ = 2.75 in.

2 4 50 kip in (9 in )(2.75 in.) 8.24 kips 150.17 in F = ⋅ = F=8.24 kips 

(b) Half web: A*=(2 in.)(2 in.) = 4 in2 y*=1 in.

2 4 50 kip in (4 in )(1 in.) 1.332 kips 150.17 in F = ⋅ = F=1.332 kips 

PROBLEM 4.15

The beam shown is made of a nylon for which the allowable stress is 24 MPa in tension and 30 MPa in compression. Determine the largest couple M that can be applied to the beam.

SOLUTION 2 , mm A y0, mm Ay0, mm3  600 22.5 _{13.5 10×} 3  300 7.5 _{2.25 10×} 3 Σ 900 _{15.75 10}3 × 3 0 15.5 10

17.5 mm The neutral axis lies 17.5 mm above the bottom. 900 Y = × = top bot 3 2 3 2 3 4 1 1 1 1 1 3 2 3 2 3 4 2 2 2 2 2 3 4 9 4 1 2 30 17.5 12.5 mm 0.0125 m 17.5 mm 0.0175 m 1 1 (40)(15) (600)(5) 26.25 10 mm 12 12 1 1 (20)(15) (300)(10) 35.625 10 mm 12 12 61.875 10 mm 61.875 10 m y y I b h A d I b h A d I I I − = − = = = − = − = + = + = × = + = + = × = + = × = × | | My M I I y σ σ = =

Top: (tension side) (24 10 )(61.875 10 )6 9 118.8 N m 0.0125

M ₌ × × − _{= ⋅}

Bottom: (compression) (30 10 )(61.875 10 )6 9 106.1 N m 0.0175

M ₌ × × − _{= ⋅}

PROBLEM 4.16

Solve Prob. 4.15, assuming that d=40 mm.

PROBLEM 4.15 The beam shown is made of a nylon for which the allowable stress is 24 MPa in tension and 30 MPa in compression. Determine the largest couple M that can be applied to the beam.

SOLUTION 2 , mm A y0, mm Ay0, mm3  600 32.5 _{19.5 10}3 ×  500 12.5 _{6.25 10}3 × Σ 1100 _{25.75 10×} 3 3

0 25.75 10 23.41 mm The neutral axis lies 23.41 mm above the bottom. 1100 Y ₌ × ₌ top bot 3 2 3 2 3 4 1 1 1 1 1 2 2 3 2 3 4 2 2 2 2 2 3 4 1 2 40 23.41 16.59 mm 0.01659 m 23.41 mm 0.02341 m 1 1 (40)(15) (600)(9.09) 60.827 10 mm 12 12 1 1 (20)(25) (500)(10.91) 85.556 10 mm 12 12 146.383 10 mm 146.3 y y I b h A d I b h A d I I I = − = = = − = − = + = + = × = + = + = × = + = × = _{83 10 m×} −9 4 | | My M I I y σ σ = =

Top: (tension side) (24 10 )(146.383 10 )6 9 212 N m 0.01659

M ₌ × × − _{= ⋅}

Bottom: (compression) (30 10 )(146.383 10 )6 9 187.6 N m 0.02341

M ₌ × × − _{= ⋅}

PROBLEM 4.17

Knowing that for the extruded beam shown the allowable stress is 12 ksi in tension and 16 ksi in compression, determine the largest couple M that can be applied. SOLUTION A y₀ Ay₀  2.25 1.25 2.8125  2.25 0.25 0.5625 4.50 3.375 3.375 0.75 in. 4.50 Y _{= =}

The neutral axis lies 0.75 in. above bottom.

top bot 3 2 3 2 4 1 1 1 1 1 2 2 3 2 4 2 2 2 2 2 4 1 2 2.0 0.75 1.25 in., 0.75 in. 1 1 (1.5)(1.5) (2.25)(0.5) 0.984375 in 12 12 1 1 (4.5)(0.5) (2.25)(0.5) 0.609375 in 12 12 1.59375 in y y I b h A d I b h A d I I I My I M I y σ σ = − = = − = + = + = = + = + = = + = = =

Top: (compression) (16)(1.59375) 20.4 kip in 1.25

M = = ⋅

Bottom: (tension) (12)(1.59375) 25.5 kip in 0.75

M = = ⋅

PROBLEM 4.18

Knowing that for the casting shown the allowable stress is 5 ksi in tension and 18 ksi in compression, determine the largest couple M

that can be applied.

SOLUTION

Locate the neutral axis and compute the moment of inertia.

3 2 1 for rectangle 12 ( ) i i i i i i i i i A y Y I b h A d y Y I A d I  = _ = = − =  + Part _{A, in}2 _{, in.} i y _{, in}3 i i A y d_i, in. 2_{, in}4 i i A d I, in4 1 1.5 1.25 1.875 0.3333 0.1667 0.03125 2 1.0 0.75 0.75 0.1667 0.0277 0.02083 3 0.5 0.25 0.125 0.6667 0.2222 0.01042 Σ 3.0 2.75 0.4166 0.0625 2 4 2.75 0.9167 in. ( ) 0.4166 0.0625 0.479 in 3.0 Ay Y I I Ad A  = _ = = =  + = + =

Allowable bending moment.

or Mc _M I I c σ σ = = Tension at A: _{σA ≤}5 ksi 0.583 in. A c ₌ (5)(0.479) _{4.11 kip in} 0.583 M _{≤ = ⋅}

Compression at B: _{σB ≤}18 ksi c_{B =}0.917 in.

(18)(0.479)

9.40 kip in 0.917

M _{≤ = ⋅}

PROBLEM 4.19

Knowing that for the extruded beam shown the allowable stress is 120 MPa in tension and 150 MPa in compression, determine the largest couple M that can be applied.

SOLUTION

 rectangle  circular cutout

3 2 1 2 3 2 2 3 2 1 2 1 2 (150)(250) 37.5 10 mm (50) 7.85398 10 mm 29.64602 10 mm 0 mm 50 mm A A A A A y y Ay Y A π = = × = − = − × = + = × = = − Σ = Σ 3 3 3 (37.5 10 )(0) ( 7.85393 10 )( 50) 29.64602 10 13.2463 mm Y ₌ × + − × − × = 2 1 2 3 3 2 4 3 2 6 6 6 4 6 4 ( ) 1 (150)(250) (37.5 10 )(13.2463) 12 (50) (7.85398 10 )(50 13.2463) 4 201.892 10 36.3254 10 165.567 10 mm 165.567 10 m X I I Ad I I π ′ − = Σ + = −   =_ + × _     −_ + × + _   = × − × = × = ×

Top: (tension side) c₌125 13.2463 111.7537 mm 0.11175 m_{− = =}

6 6 3 (165.567 10 )(120 10 ) 0.11175 177.79 10 N m σ σ = = = × − × = × ⋅ Mc _M I I c

Bottom: (compression side) 125 13.2463 138.2463 mm 0.13825 m c= + = = 6 6 3 (165.567 10 )(150 10 ) 0.13825 179.64 10 N m σ σ = = = × − × = × ⋅ Mc I M I c

PROBLEM 4.20

Knowing that for the extruded beam shown the allowable stress is 120 MPa in tension and 150 MPa in compression, determine the largest couple M that can be applied.

SOLUTION 2 , mm A y0, mm Ay0, mm3 d, mm  2160 27 58320 3  1080 36 38880 3 Σ 3240 97200 97200

30 mm The neutral axis lies 30 mm above the bottom. 3240 Y = = top bot 3 2 3 2 3 4 1 1 1 1 1 2 2 3 2 3 4 2 2 2 2 2 3 4 9 4 1 2 54 30 24 mm 0.024 m 30 mm 0.030 m 1 1 (40)(54) (40)(54)(3) 544.32 10 mm 12 12 1 1 1 (40)(54) (40)(54)(6) 213.84 10 mm 36 36 2 758.16 10 mm 758.16 10 m y y I b h A d I b h A d I I I − = − = = = − = − = + = + = × = + = + = × = + = × = × | |_{σ =} M y |M|₌ σI I y

Top: (tension side) (120 10 )(758.16 10 )6 9 _{3.7908 10 N m}3 0.024

M ₌ × × − _{= × ⋅}

Bottom: (compression) (150 10 )(758.16 10 )6 9 _{3.7908 10 N m}3 0.030

M ₌ × × − _{= × ⋅}

PROBLEM 4.21

A steel band saw blade, that was originally straight, passes over 8-in.-diameter pulleys when mounted on a band saw. Determine the maximum stress in the blade, knowing that it is 0.018 in. thick and 0.625 in. wide. Use _{E=29 10 psi.×} 6

SOLUTION

Band blade thickness: t₌0.018 in.

Radius of pulley: 1 4.000 in.

2 r₌ d₌ Radius of curvature of centerline of blade:

1 4.009 in. 2 1 0.009 in. 2 r t c t ρ= + = = = Maximum strain: 0.009 0.002245 4.009 m c ε ρ = = = Maximum stress: _{(29 10 )(0.002245)}6 m E m σ = ε = × 3 65.1 10 psi m σ = × _{σm =}65.1 ksi 

PROBLEM 4.22

Straight rods of 0.30-in. diameter and 200-ft length are sometimes used to clear underground conduits of obstructions or to thread wires through a new conduit. The rods are made of high-strength steel and, for storage and transportation, are wrapped on spools of 5-ft diameter. Assuming that the yield strength is not exceeded, determine (a) the maximum stress in a rod, when the rod, which is initially straight, is wrapped on a spool, (b) the corresponding bending moment in the rod. Use E ₌ 29 10 psi_× 6 .

SOLUTION

Radius of cross section: 1 1(0.30) 0.15 in.

2 2 r ₌ d _{= =} Moment of inertia: 4 _(0.15)4 _{397.61 10 in}6 4 4 4 I ₌ πr ₌ π _{= ×} − 1 5 ft 60 in. 30 in. 2 0.15 in. D D c r ρ = = = = = = (a) 6 3 max (29 10 )(0.15) 145 10 psi 30 Ec σ ρ × = = = × σmax =145 ksi  (b) (29 10 )(397.61 10 )6 6 30 EI M ρ − × × = = M ₌384 lb in_⋅ 

PROBLEM 4.23

A 900-mm strip of steel is bent into a full circle by two couples applied as shown. Determine (a) the maximum thickness t of the strip if the allowable stress of the steel is 420 MPa, (b) the corresponding moment M of the couples. Use E ₌200 GPa.

SOLUTION

When the rod is bent into a full circle, the circumference is 900 mm. Since the circumference is equal to 2π times ρ, the radius of curvature, we get

900 mm 143.24 mm 0.14324 m 2 ρ π = = = Stress: E Ec or c E ρσ σ ε ρ = = =

For _{σ =} 420 MPa and E =200 GPa,

6 3 9 (0.14324)(420 10 ) 0.3008 10 m 200 10 c = × = × − ×

(a) Maximum thickness: t =2c =0.6016 10 m× −3

0.602 mm

t = 

Moment of inertia for a rectangular section.

3 3 3 3 15 4 (8 10 )(0.6016 10 ) 145.16 10 m 12 12 bt I = = × − × − = × − (b) Bending moment: M EI ρ = 9 15 (200 10 )(145.16 10 ) 0.203 N m 0.14324 M = × × − = ⋅ M = 0.203 N m⋅ 

PROBLEM 4.24

A 60 N ⋅ m couple is applied to the steel bar shown. (a) Assuming that the couple is applied about the z axis as shown, determine the maximum stress and the radius of curvature of the bar. (b) Solve part a, assuming that the couple is applied about the y axis. Use E=200 GPa.

SOLUTION

(a) Bending about z-axis.

3 3 3 4 9 4 1 1 (12)(20) 8 10 mm 8 10 m 12 12 20 10 mm 0.010 m 2 − = = = × = × = = = I bh c 6 9 (60)(0.010) 75.0 10 Pa 8 10 Mc I σ = = ₋ = × × σ =75.0 MPa  3 1 9 9 1 60 37.5 10 m (200 10 )(8 10 ) M EI ρ = = _{× ×} − = × − − ρ=26.7 m 

(b) Bending about y-axis.

3 3 3 4 9 4 6 9 1 1 (20)(12) 2.88 10 mm 2.88 10 m 12 12 12 6 mm 0.006 m 2 (60)(0.006) 125.0 10 Pa 2.88 10 I bh c Mc I σ − − = = = × = × = = = = = = × × σ =125.0 MPa  3 1 9 9 1 60 104.17 10 m (200 10 )(2.88 10 ) M EI ρ = = _{× ×} − = × − − ρ=9.60 m 

PROBLEM 4.25

A couple of magnitude M is applied to a square bar of side a. For each of the orientations shown, determine the maximum stress and the curvature of the bar.

SOLUTION 4 3 3 1 1 12 12 12 2 a I bh aa a c = = = = max ₄2 12 a M Mc I a σ = = max ₃ 6M a σ =  4 1 12 M M EI _Ea ρ = = 4 1 12M Ea ρ = 

For one triangle, the moment of inertia about its base is

( )

3 4 3 1 4 2 1 4 1 2 1 1 2 12 12 2 24 24 12 a a I bh a a I I a I I I   = =   =   = = = + = max 4 3 / 2 6 2 /12 2 a Mc Ma M c I a a σ = = = = max 3 8.49M a σ =   1 ₄ 12 ρ = = M M EI _Ea  4 1 12M Ea ρ = 

PROBLEM 4.26

A portion of a square bar is removed by milling so that its cross section is as shown. The bar is then bent about its horizontal axis by a couple M. Considering the case where h = 0.9h0, express the maximum stress in the bar in the form σm =kσ0, where σ0 is the maximum stress that would have occurred if the original square bar had been bent by the same couple M, and determine the value of k.

SOLUTION 1 2 3 3 0 4 3 3 3 4 0 0 2 3 4 0 0 4 2 1 1 (4) (2) (2 2 ) ( ) 12 3 1 4 4 4 3 3 3 3 3 4 _{(4 3 )} 3 σ = +     = _{ } + _{ } −     = + − = − = = = = − − I I I h h h h h h h h h h h h h c h Mc Mh M I _{h h h} h h h

For the original square, _{0 0}

0 _{2 3} 0 0 0 0 , . 3 3 (4 3 ) h h c h M M h h h h σ = = = = − 3 3 0 0 2 2 0 0 0 0 0 0 0.950 (4 3 ) (4 (3)(0.9) )(0.9 ) 0.950 σ σ σ σ = = = − − = h h h h h h h h 0.950 k= 

PROBLEM 4.27

In Prob. 4.26, determine (a) the value of h for which the maximum stress

m

σ is as small as possible, (b) the corresponding value of k.

PROBLEM 4.26 A portion of a square bar is removed by milling so that its cross section is as shown. The bar is then bent about its horizontal axis by a couple M. Considering the case where h = 0.9h0, express the maximum stress in the bar in the form σm =kσ0,where σ0 is the maximum stress that would have occurred if the original square bar had been bent by the same couple M, and determine the value of k.

SOLUTION 1 2 3 3 0 4 3 3 3 4 0 0 2 3 0 4 2 1 1 (4) (2) (2 2 ) 12 3 1 4 4 4 3 3 3 3 4 3 I I I hh h h h h h h h h h h I c h h h h c = +     = _ _ +  _{ } − = − − = − = = − I c is maximum at 2 3 0 4 _0. 3 d h h h dh  ₋ ₌     2 0 8 3 0 3h h− h = 0 8 9 h= h   2 3 3 0 0 0 0 ₃ 0 4 8 8 256 729 3 9 9 729 256 I _{h h h h} Mc M c σ I h     = _ _ −_ _ = = = 

For the original square, 0 3

0 0 0 0 1 3 I h h c h h c = = = 0 0 2 0 0 3 Mc M I h σ = = 0 729 1 729 0.949 256 3 768 σ σ = ⋅ = = k=0.949 

PROBLEM 4.28

A couple M will be applied to a beam of rectangular cross section that is to be sawed from a log of circular cross section. Determine the ratio d/b, for which (a) the maximum stress σm will be as small as possible,

(b) the radius of curvature of the beam will be maximum.

SOLUTION

Let D be the diameter of the log.

2 2 2 2 2 2 3 2 1 1 1 12 2 6 D b d d D b I I bd c d bd c = + = − = = =

(a) σm is the minimum when I

c is maximum. 2 2 2 3 2 2 1 1 1 ( ) 6 6 6 1 3 1 0 6 6 3 I b D b D b b c d I D b b D db c = − = −   = − = =     2 1 2 2 3 3 d = D − D = D d 2 b =  (b) EI M ρ =

ρ is maximum when I is maximum, 1 3

12bd is maximum, or 2 6 b d is maximum. 2 2 6 (D −d d) is maximum. 2 5 7 6D d −8d = 0 3 2 d = D 2 3 2 1 4 2 b = D − D = D d 3 b = 

PROBLEM 4.29

For the aluminum bar and loading of Sample Prob. 4.1, determine (a) the radius of curvature ρ′ of a transverse cross section, (b) the angle between the sides of the bar that were originally vertical. Use E = 10.6 × 106 _{psi and v = 0.33.}

SOLUTION

From Sample Prob. 4.1, _{I=12.97 in}4 _{M =103.8 kip in⋅}

3 6 1 6 1 103.8 10 755 10 in (10.6 10 )(12.97) M EI ρ − − × = = = × × (a) 1 _v1 _{(0.33)(755 10 ) 249 10 in}6 6 1 ρ′= ρ = × − = × − − 4010 in. ρ′ = _ρ′ =334 ft 

(b) length of arc 3.25 810 10 rad6

radius 4010 θ ρ − = = = = × ′ b 0.0464 θ = ° 

PROBLEM 4.30

For the bar and loading of Example 4.01, determine (a) the radius of curvature ,ρ (b) the radius of curvature ρ′ of a transverse cross section, (c) the angle between the sides of the bar that were originally vertical. Use

6 29 10 psi

= ×

E and v=0.29.

SOLUTION

From Example 4.01, _{M =30 kip in, ⋅ I =1.042 in}4

(a) 3 6 1 6 1 (30 10 ) 993 10 in (29 10 )(1.042) M EI ρ − − × = = = × × ρ=1007 in.  (b) _ε1 _vε vc _v c ρ ρ = = = ′ 6 1 6 1 1 1 (0.29)(993 10 )in. 288 10 in. v ρ′= ρ = × − − = × − − ρ′ =3470 in. 

(c) length of arc 0.8 _{230 10 rad}6

radius 3470 b θ ρ − = = = = × ′ θ =0.01320° 

PROBLEM 4.31

A W200 31.3× rolled-steel beam is subjected to a couple M of moment 45 kN ⋅ m. Knowing that E = 200 GPa and v = 0.29, determine (a) the radius of curvature ,ρ (b) the radius of curvature ρ′ of a transverse cross section.

SOLUTION

For W 200 31.3× rolled steel section,

6 4 6 4 31.4 10 mm 31.4 10 m− = × = × I (a) 3 3 1 9 6 1 45 10 _{7.17 10 m} (200 10 ) (31.4 10 ) M EI ρ − − − × = = = × × × ρ=139.6 m  (b) 1 v1 (0.29) (7.17 10 ) 2.07 10 m3 3 1 ρ′= ρ = × − = × − − ρ′ =481 m 

PROBLEM 4.32

It was assumed in Sec. 4.3 that the normal stresses σ_y in a member in pure bending are negligible. For an initially straight elastic member of rectangular cross section, (a) derive an approximate expression for σ_y as a function of y, (b) show that (_σy)_max = −( /2 )(c _{ρ σx})_max and, thus, that σ_y can be neglected in all practical situations. (Hint: Consider the free-body diagram of the portion of beam located below the surface of ordinate y and assume that the distribution of the stress σ_x is still linear.)

SOLUTION

Denote the width of the beam by b and the length by L. θ

ρ = L Using the free body diagram above, with cos 1

2 θ _≈ 0 : 2 sin 0 2 2 1 sin 2 y y y _c x y y y y _c x _c x _c x F bL bdy dy dy dy L L θ σ σ θ θ σ σ σ σ ρ − − − − Σ = + = = − ≈ − = −









But, _x ( _x)_max y c σ = −σ

(a) ( )max ( )max 2

2 y y x x y _c c y ydy c c σ σ σ ρ − ρ ₋ =



= ( )max₍ 2 2₎ 2 x y σ _c y c σ ρ = − 

The maximum value _σy occurs at y= . 0 (b) 2 max max max ( ) ( ) ( ) 2 2 x x y c c c σ σ σ ρ ρ = − = − 

PROBLEM 4.33

A bar having the cross section shown has been formed by securely bonding brass and aluminum stock. Using the data given below, determine the largest permissible bending moment when the composite bar is bent about a horizontal axis.

Aluminum Brass

Modulus of elasticity 70 GPa 105 GPa

Allowable stress 100 MPa 160 MPa

SOLUTION

Use aluminum as the reference material.

For aluminum, n=1.0

For brass, n E E= _b/ _a =105 / 70 1.5= Values of n are shown on the figure.

For the transformed section,

3 3 3 4 1 1 1 1 3 3 3 4 2 2 2 2 3 4 3 1 3 4 9 4 1 2 3 1.0 (8)(32) 21.8453 10 mm 12 12 1.5_{(32)(32) 131.072 10 mm} 12 12 21.8453 10 mm 174.7626 10 mm 174.7626 10 m n I b h n I b h I I I I I I − = = = × = = = × = = × = + + = × = × | |_σ = n M y M = σI I ny Aluminum: 6 6 9 3 1.0, | | 16 mm 0.016 m, 100 10 Pa (100 10 )(174.7626 10 ) 1.0923 10 N m (1.0)(0.016) σ − = = = = × × × = = × ⋅ n y M Brass: 6 6 9 3 1.5, | | 16 mm 0.016 m, 160 10 Pa (160 10 )(174.7626 10 ) 1.1651 10 N m (1.5)(0.016) σ − = = = = × × × = = × ⋅ n y M

PROBLEM 4.34

A bar having the cross section shown has been formed by securely bonding brass and aluminum stock. Using the data given below, determine the largest permissible bending moment when the composite bar is bent about a horizontal axis.

Aluminum Brass

Modulus of elasticity 70 GPa 105 GPa

Allowable stress 100 MPa 160 MPa

SOLUTION

Use aluminum as the reference material. For aluminum, n=1.0

For brass, n E E= _b/ _a =105/70 1.5= Values of n are shown on the sketch. For the transformed section,

(

)

3 3 3 4 1 1 1 1 3 3 3 3 3 4 2 2 2 2 2 3 4 3 1 3 4 9 4 1 2 3 1.5 (8)(32) 32.768 10 mm 12 12 1.0 (32)(32 16 ) 76.459 10 mm 12 12 32.768 10 mm 141.995 10 mm 141.995 10 m n I b h n I b H h I I I I I I − = = = × = − = − = × = = × = + + = × = × | |_σ = nMy M = σI I ny Aluminum: 6 6 9 1.0, | | 16 mm 0.016 m, 100 10 Pa (100 10 )(141.995 10 ) 887.47 N m (1.0)(0.016) σ − = = = = × × × = = ⋅ n y M Brass: 6 6 9 1.5, | | 16 mm 0.016 m, 160 10 Pa (160 10 )(141.995 10 ) 946.63 N m (1.5)(0.016) σ − = = = = × × × = = ⋅ n y M

PROBLEM 4.35

For the composite bar indicated, determine the largest permissible bending moment when the bar is bent about a vertical axis.

PROBLEM 4.35 Bar of Prob. 4.33.

Aluminum Brass

Modulus of elasticity 70 GPa 105 GPa

Allowable stress 100 MPa 160 MPa

SOLUTION

Use aluminum as the reference material.

For aluminum, n=1.0

For brass, n E E= _b/ _a =105/70 1.5= Values of n are shown on the figure.

For the transformed section,

3 2 3 2 3 4 1 1 1 1 1 1 1 3 3 3 4 2 2 2 2 3 4 3 1 3 4 9 4 1 2 3 1.0 (32)(8) (1.0)[(32)(8)](20) 103.7653 10 mm 12 12 1.5_{(32)(32) 131.072 10 mm} 12 12 103.7653 10 mm 338.58 10 mm 338.58 10 m n I h b n A d n I h b I I I I I I − = + = + = × = = = × = = × = + + = × = × | |_σ = n My M = σI I ny Aluminum: 6 6 9 3 1.0, | | 24 mm 0.024 m, 100 10 Pa (100 10 )(338.58 10 ) 1.411 10 N m (1.0)(0.024) σ − = = = = × × × = = × ⋅ n y M Brass: 6 6 9 3 1.5, | | 16 mm 0.016 m, 160 10 Pa (160 10 )(338.58 10 ) 2.257 10 N m (1.5)(0.016) σ − = = = = × × × = = × ⋅ n y M

PROBLEM 4.36

For the composite bar indicated, determine the largest permissible bending moment when the bar is bent about a vertical axis.

PROBLEM 4.36 Bar of Prob. 4.34.

Aluminum Brass

Modulus of elasticity 70 GPa 105 GPa

Allowable stress 100 MPa 160 MPa

SOLUTION

Use aluminum as the reference material. For aluminum, n=1.0

For brass, n E E= _b/ _a =105/70 1.5= Values of n are shown on the sketch. For the transformed section,

(

3 3

)

3 3 3 4 1 1 1 1 1 3 3 3 4 2 2 2 2 3 4 3 2 3 4 9 4 1 2 3 1.5 (32)(48 32 ) 311.296 10 mm 12 12 1.0 (8)(32) 21.8453 10 mm 12 12 21.8453 10 mm 354.99 10 mm 354.99 10 m n I h B b n I h b I I I I I I − = − = − = × = = = × = = × = + + = × = × | |_σ = nMy M = σI I ny Aluminum: 6 6 9 3 1.0, | | 16 mm 0.016 m, 100 10 Pa (100 10 )(354.99 10 ) 2.2187 10 N m (1.0)(0.016) σ − = = = = × × × = = × ⋅ n y M Brass: 6 6 9 3 1.5 | | 24 mm 0.024 m 160 10 Pa (160 10 )(354.99 10 ) 1.57773 10 N m (1.5)(0.024) σ − = = = = × × × = = × ⋅ n y M Choose the smaller value.

3 1.57773 10 N m

PROBLEM 4.37

Wooden beams and steel plates are securely bolted together to form the composite member shown. Using the data given below, determine the largest permissible bending moment when the member is bent about a horizontal axis.

Wood Steel

Modulus of elasticity: 2 10 psi× 6 _{29 10 psi×} 6

Allowable stress: 2000 psi 22 ksi

SOLUTION

Use wood as the reference material. 1.0 in wood / 29/2 14.5 in steel s w n n E E = = = =

For the transformed section,

3 3 4 1 1 1 1 3 3 4 2 2 2 2 4 3 1 4 1 2 3 1.0 (3)(10) 250 in 12 12 14.5 1 (10) 604.17 in 12 12 2 250 in 1104.2 in n I b h n I b h I I I I I I = = =   = = _{ } =   = = = + + = nMy I M I ny σ σ = ∴ = Wood: 3 1.0, 5 in., 2000 psi (2000)(1104.2) 441.7 10 lb in (1.0)(5) n y M σ = = = = = × ⋅ Steel: 3 3 3

14.5, 5 in., 22 ksi 22 10 psi

(22 10 )(1104.2) 335.1 10 lb in (14.5)(5) n y M σ = = = = × × = = × ⋅

PROBLEM 4.38

Wooden beams and steel plates are securely bolted together to form the composite member shown. Using the data given below, determine the largest permissible bending moment when the member is bent about a horizontal axis.

Wood Steel

Modulus of elasticity: 2 10 psi× 6 _{29 10 psi×} 6

Allowable stress: 2000 psi 22 ksi

SOLUTION

Use wood as the reference material. 1.0 in wood / 29/2 14.5 in steel s w n n E E = = = =

For the transformed section,

3 2 1 1 1 1 1 1 1 3 2 4 2 3 4 2 2 2 2 4 3 1 4 1 2 3 12 14.5 1 1 (5) (14.5)(5) (5.25) 999.36 in 12 2 2 1.0 (6)(10) 500 in 12 12 999.36 in 2498.7 in n I b h n A d n I b h I I I I I I = +     = _{ } + _{ } =     = = = = = = + + = nMy I M I ny σ σ = ∴ = Wood: 3 1.0, 5 in., 2000 psi (2000)(2499) 999.5 10 lb in (1.0)(5) n y M σ = = = = = × ⋅ Steel: 3 3 3

14.5, 5.5 in., 22 ksi 22 10 psi

(22 10 )(2499) 689.3 10 lb in (14.5)(5.5) n y M σ = = = = × × = = × ⋅

PROBLEM 4.39

A steel bar and an aluminum bar are bonded together to form the composite beam shown. The modulus of elasticity for aluminum is 70 GPa and for steel is 200 GPa. Knowing that the beam is bent about a horizontal axis by a couple of moment M =1500 N m,⋅ determine the maximum stress in (a) the aluminum, (b) the steel.

SOLUTION

Use aluminum as the reference material.

For aluminum, 1 For steel, _s/ _a 200/70 2.8571 n n E E = = = = Transformed section: 0 0 0 109714 37.65 mm 2914.3 Y = = d= y −Y 3 2 3 2 3 4 1 1 1 1 1 1 1 3 2 3 2 3 4 2 2 2 2 2 2 2 3 4 9 4 1 2 2.8571 (30)(20) (1714.3)(12.35) 318.61 10 mm 12 12 1 _{(30)(40) (1200)(17.65) 533.83 10 mm} 12 12 852.44 10 mm 852.44 10 m 1500 N m n I b h n A d n I b h n A d I I I M − = + = + = × = + = + = × = + = × = × = ⋅ Stress: n My I σ = − (a) Aluminum: n=1, y= −37.65 mm= −0.03765 m 6 9 (1)(1500)( 0.03765) 66.2 10 Pa 852.44 10 a σ = − − ₋ = × × σa =66.2 MPa  (b) Steel: n=2.8571, y=60 37.65 22.35 mm 0.02235 m− = = 6 9 (2.8571)(1500) (0.02235) 112.4 10 Pa 852.44 10 s n My I σ = − = − ₋ = − × × 112.4 MPa s σ = − 

Part A, mm2 nA, mm2 y_o, mm nAy_o, mm3 d, mm

1 600 1714.3 50 85714 12.35

2 1200 1200 20 24000 17.65

PROBLEM 4.40

A steel bar and an aluminum bar are bonded together to form the composite beam shown. The modulus of elasticity for aluminum is 70 GPa and for steel is 200 GPa. Knowing that the beam is bent about a horizontal axis by a couple of moment M =1500 N m,⋅ determine the maximum stress in (a) the aluminum, (b) the steel.

SOLUTION

Use aluminum as the reference material.

For aluminum, 1 For steel, _s/ _a 200/70 2.8571 n n E E = = = = Transformed section: Part A, mm2 _{nA, mm}2 _{, mm} o y _{, mm}3 o nAy d, mm 1 600 600 50 30000 25.53 2 1200 3428.5 20 68570 4.47 Σ 4028.5 98570 0 98570 24.47 mm 4028.5 Y = = d = y0 −Y0 3 2 3 2 3 4 1 1 1 1 1 1 1 3 2 3 2 3 4 2 2 2 2 2 2 2 3 4 9 4 1 2 1 (30)(20) (600)(25.53) 411.07 10 mm 12 12 2.8571 (30)(40) (3428.5)(4.47) 525.64 10 mm 12 12 936.71 10 mm 936.71 10 m 1500 N m n I b h n A d n I b h n A d I I I M − = + = + = × = + = + = × = + = × = × = ⋅ Stress: n M y I σ= − (a) Aluminum: n=1, y=60 24.47 35.53 mm 0.03553 m− = = 6 9 (1)(1500)(0.03553) 56.9 10 Pa 936.71 10 a σ = − ₋ = − × × σa = −56.9 MPa  (b) Steel: n=2.8571, y= −24.47 mm= −0.02447 m 6 9 (2.8571)(1500) ( 0.02447) 111.9 10 Pa 936.71 10 s σ = − −₋ = × × σs =111.9 MPa 

PROBLEM 4.41

The 6 12-in.× timber beam has been strengthened by bolting to it the steel reinforcement shown. The modulus of elasticity for wood is _{1.8 10 psi×} 6 _{and for} steel, _{29 10 psi.×} 6 _{Knowing that the beam is bent about a horizontal axis by a} couple of moment M =450 kip in.,⋅ determine the maximum stress in (a) the wood, (b) the steel.

SOLUTION

Use wood as the reference material.

For wood, 1 For steel, _s/ _w 29 /1.8 16.1111 n n E E = = = =

Transformed section:  = wood  = steel

421.931 112.278 3.758 in. o Y = =

The neutral axis lies 3.758 in. above the wood-steel interface.

3 2 3 2 4 1 1 1 1 1 1 1 3 2 3 2 4 2 2 2 2 2 2 2 4 1 2 1 (6)(12) (72)(6 3.758) 1225.91 in 12 12 16.1111_{(5) (0.5) (40.278)(3.578 0.25) 647.87 in} 12 12 1873.77 in 450 kip in n I b h n A d n I b h n A d I I I nMy M I σ = + = + − = = + = + + = = + = = ⋅ = − (a) Wood: n=1, y=12 3.758 8.242 in− = (1) (450) (8.242) 1.979 ksi 1873.77 w σ = − = − _σw= −1.979 ksi  (b) Steel: n=16.1111, y= −3.758 0.5− = −4.258 in (16.1111) (450) ( 4.258) 16.48 ksi 1873.77 s σ = − − = _σs =16.48 ksi  2 , in A _{nA, in}2 o y _{, in}3 o nAy  72 72 6 432  2.5 40.278 −0.25 −10.069 112.278 421.931

PROBLEM 4.42

The 6 12-in.× timber beam has been strengthened by bolting to it the steel reinforcement shown. The modulus of elasticity for wood is _{1.8 10 psi×} 6 and for steel, _{29 10 psi.×} 6 _{Knowing that the beam is bent about a} horizontal axis by a couple of moment M =450 kip in.,⋅ determine the maximum stress in (a) the wood, (b) the steel.

SOLUTION

Use wood as the reference material.

6 6 For wood, 1 29 10 For steel, 16.1111 1.8 10 s w n E n E = × = = = ×

For C8 11.5× channel section,

2 4

3.38 in , _w 0.220 in., 0.571 in., _y 1.32 in

A= t = x= I =

For the composite section, the centroid of the channel (part 1) lies 0.571 in. above the bottom of the section. The centroid of the wood (part 2) lies 0.220+6.00 =6.22 in. above the bottom.

Transformed section: 3 0 2 478.93 in 3.787 in. 126.456 in Y = = d = y0 −Y0

The neutral axis lies 3.787 in. above the bottom of the section.

2 2 4 1 1 1 1 1 1 3 2 3 2 4 2 2 2 2 2 2 2 4 1 2 (16.1111)(1.32) (54.456)(3.216) 584.49 in 1 (6)(12) (72)(2.433) 1290.20 in 12 12 1874.69 in 450 kip in I n I n A d n I b h n A d I I I n My M I σ = + = + = = + = + = = + = = ⋅ = −

(a) Wood: n=1, y=12 0.220 3.787 8.433 in.+ − =

(1)(450)(8.433) 2.02 ksi 1874.69 w σ = − = − _σw = −2.02 ksi  (b) Steel: n₌16.1111, y_{= −}3.787 in. (16.1111) (450) ( 3.787) 14.65 ksi 1874.67 s σ = − − = _σs =14.65 ksi 

Part A, in2 nA, in2 y, in. nAy, in3 d, in.

1 3.38 54.456 0.571 31.091 3.216

2 72 72 6.22 447.84 2.433

PROBLEM 4.43

For the composite beam indicated, determine the radius of curvature caused by the couple of moment 1500 N ⋅ m.

Beam of Prob. 4.39.

SOLUTION

See solution to Prob. 4.39 for the calculation of I.

9 4 9 1 9 9 852.44 10 m 70 10 Pa 1 1500 0.02513 m (70 10 )(852.44 10 ) a I E M EI ρ − − − = × = × = = = × × 39.8 m ρ= 

PROBLEM 4.44

For the composite bar indicated, determine the radius of curvature caused by the couple of moment 1500 N ⋅ m.

Beam of Prob. 4.40.

SOLUTION

See solution to Prob. 4.40 for calculation of I.

9 4 9 1 9 9 936.71 10 m 70 10 Pa 1 1500 0.02288 m (70 10 )(936.71 10 ) a I E M EI ρ − − − = × = × = = = × × 43.7 m ρ= 

PROBLEM 4.45

For the composite beam indicated, determine the radius of curvature caused by the couple of moment 450 kip ⋅ in.

Beam of Prob. 4.41.

SOLUTION

See solution to Prob. 4.41 for calculation of I.

4 6 3 3 6 1 6 1873.77 in 1.8 10 psi 450 kip in 450 10 lb in 1 450 10 133.42 10 in (1.8 10 )(1873.77) w I E M M EI ρ − − = = × = ⋅ = × ⋅ × = = = × × 7495 in. 625 ft ρ= = 

PROBLEM 4.46

For the composite beam indicated, determine the radius of curvature caused by the couple of moment 450 kip ⋅ in.

Beam of Prob. 4.42.

SOLUTION

See solution to Prob. 4.42 for calculation of I.

4 6 3 3 6 1 6 1874.69 in 1.8 10 psi 450 kip in 450 10 lb in 1 450 10 133.36 10 in (1.8 10 )(1874.69) w I E M M EI ρ − − = = × = ⋅ = × ⋅ × = = = × × 7499 in. 625 ft ρ= = 

PROBLEM 4.47

The reinforced concrete beam shown is subjected to a positive bending moment of 175 kN ⋅ m. Knowing that the modulus of elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete.

SOLUTION 2 2 3 2 3 2 200 GPa 8.0 25 GPa 4 (4) (25) 1.9635 10 mm 4 4 15.708 10 mm s c s s E n E A d nA π π = = =   = ⋅ = _{ } = ×   = ×

Locate the neutral axis:

3 2 3 6 300 (15.708 10 )(480 ) 0 2 150 15.708 10 7.5398 10 0 x x x x x − × − = + × − × = Solve for x: 3 3 2 6 15.708 10 (15.708 10 ) (4)(150)(7.5398 10 ) (2)(150) 177.87 mm, 480 302.13 mm x x x − × + × + × = = − = 3 3 2 3 3 2 9 4 3 4 1 (300) (15.708 10 )(480 ) 3 1 (300)(177.87) (15.708 10 )(302.13) 3 1.9966 10 mm 1.9966 10 m I x x nMy I σ − = + × − = + × = × = × = − (a) Steel: y= −302.45 mm= −0.30245 m 3 6 3 (8.0)(175 10 )( 0.30245) 212 10 Pa 1.9966 10 σ = − × −₋ = × × σ =212 MPa  (b) Concrete: y=177.87 mm 0.17787 m= 3 6 3 (1.0)(175 10 )(0.17787) 15.59 10 Pa 1.9966 10 σ = − × ₋ = − × × σ = −15.59 MPa 

PROBLEM 4.48

Solve Prob. 4.47, assuming that the 300-mm width is increased to 350 mm.

PROBLEM 4.47 The reinforced concrete beam shown is subjected to a positive bending moment of 175 kN ⋅ m. Knowing that the modulus of elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete.

SOLUTION 2 2 3 2 3 2 200 GPa 8.0 25 GPa 4 (4) (25) 4 4 1.9635 10 mm 15.708 10 mm s c s s E n E A d nA π π = = =   = =  _{ } = × = ×

Locate the neutral axis:

3 2 3 6 350 (15.708 10 )(480 ) 0 2 175 15.708 10 7.5398 10 0 x x x x x − × − = + × − × = Solve for x: 3 3 2 6 15.708 10 (15.708 10 ) (4)(175)(7.5398 10 ) (2)(175) 167.48 mm, 480 312.52 mm x x x − × + × + × = = − = 3 3 2 3 3 2 9 4 3 4 1 (350) (15.708 10 )(480 ) 3 1 (350)(167.48) (15.708 10 )(312.52) 3 2.0823 10 mm 2.0823 10 m I x x nMy I σ − = + × − = + × = × = × = − (a) Steel: y= −312.52 mm= −0.31252 m 3 6 3 (8.0)(175 10 )( 0.31252) 210 10 Pa 2.0823 10 σ = − × −₋ = × × σ =210 MPa  (b) Concrete: y=167.48 mm 0.16748 m= 3 6 3 (1.0)(175 10 )(0.16748) 14.08 10 Pa 2.0823 10 σ = − × ₋ = − × × σ = −14.08 MPa 

PROBLEM 4.49

A concrete slab is reinforced by 16-mm-diameter steel rods placed on 180-mm centers as shown. The modulus of elasticity is 20 GPa for concrete and 200 GPa for steel. Using an allowable stress of 9 MPa for the concrete and of 120 MPa for the steel, determine the largest bending moment in a portion of slab 1 m wide.

SOLUTION 200 GPa 10 20 GPa s c E n E = = =

Consider a section 180-mm wide with one steel rod.

2 2 2 3 2 (16) 201.06 mm 4 4 2.0106 10 mm s s A d nA π π = = = = ×

Locate the neutral axis:

3 2 3 3 180 (100 )(2.0106 10 ) 0 2 90 2.0106 10 201.06 10 0 x x x x x − − × = + × − × = Solving for x: 3 3 2 3 2.0106 10 (2.0106 10 ) (4)(90)(201.06 10 ) (2)(90) 37.397 mm 100 62.603 mm x x x − × + × + × = = − = 3 3 2 3 3 2 6 4 6 4 1 (180) (2.0106 10 )(100 ) 3 1 (180)(37.397) (2.0106 10 )(62.603) 3 11.018 10 mm 11.018 10 m I x x nMy I M I ny σ σ − = + × − = + × = × = × = − ∴ = Concrete: _{n=1, y=37.397 mm 0.037397 m,= σ = ×9 10 Pa}6 6 6 3 (9 10 )(11.018 10 ) 2.6516 10 N m (1.0)(0.037397) M = × × − = × ⋅

PROBLEM 4.49 (Continued) Steel: _{n=10, y=62.603 mm 0.062603 m,= σ =120 10 Pa×} 6 6 6 3 (120 10 )(11.018 10 ) 2.1120 10 N m (10)(0.062603) M = × × − = × ⋅

Choose the smaller value. _{M = 2.1120 10 N m×} 3 _⋅

The above is the allowable positive moment for a 180-mm wide section. For a 1-m = 1000-mm width, mutiply by 1000 5.556

180 =

3 3

(5.556)(2.1120 10 ) 11.73 10 N m

PROBLEM 4.50

A concrete slab is reinforced by 16-mm-diameter steel rods placed on 180-mm centers as shown. The modulus of elasticity is 20 GPa for concrete and 200 GPa for steel. Using an allowable stress of 9 MPa for the concrete and of 120 MPa for the steel, determine the largest allowable positive bending moment in a portion of slab 1 m wide. Solve Prob. 4.49, assuming that the spacing of the 16-mm-diameter rods is increased to 225 mm on centers.

SOLUTION 200 GPa 10 20 GPa s a E n E = = =

Consider a section 225-mm wide with one steel rod.

2 2 2 3 2 (16) 201.06 mm 4 4 2.0106 10 mm s s A d nA π π = = = = ×

Locate the neutral axis:

3 2 3 225 (100 )(2.0106 10 ) 0 2 112.5 2.0106 201.06 10 0 x x x x x − − × = + − × = Solving for x: 3 3 2 3 3 3 2 3 3 2 6 4 6 4 2.0106 10 (2.0106 10 ) (4)(112.5)(201.06 10 ) (2)(112.5) 34.273 mm 100 65.727 1 (225) 2.0106 10 (100 ) 3 1 (225)(34.273) (2.0106 10 )(65.727) 3 11.705 10 mm 11.705 10 m x x x I x x − − × + × + × = = − = = + × − = + × = × = × | | nMy M I I ny σ σ = ∴ = Concrete: 6 6 6 3 1, 34.273 mm 0.034273 m, 9 10 Pa (9 10 )(11.705 10 ) 3.0738 10 N m (1)(0.034273) n y M σ − = = = = × × × = = × ⋅

PROBLEM 4.50 (Continued) Steel: 6 6 6 3 10, 65.727 mm 0.065727 m, 120 10 Pa (120 10 )(11.705 10 ) 2.1370 10 N m (10)(0.065727) n y M σ − = = = = × × × = = × ⋅

Choose the smaller value. M =2.1370 10 N m× 3 ⋅

The above is the allowable positive moment for a 225-mm-wide section. For a 1-m = 1000-mm section, multiply by 1000 4.4444

225 =

3 3

(4.4444)(2.1370 10 ) 9.50 10 N m

PROBLEM 4.51

A concrete beam is reinforced by three steel rods placed as shown. The modulus of elasticity is 3 10 psi× 6 for the concrete and 29 10 psi× 6 for the steel. Using an allowable stress of 1350 psi for the concrete and 20 ksi for the steel, determine the largest allowable positive bending moment in the beam.

SOLUTION 6 6 2 2 2 2 29 10 9.67 3 10 7 3 (3) 1.8040 in 17.438 in 4 4 8 s c s s E n E A π d π nA × = = = ×    = =   _{  } = =

Locate the neutral axis:

2 8 (17.438)(14 ) 0 2 4 17.438 244.14 0 x x x x x − − = + − = Solve for x. 2 17.438 17.438 (4)(4)(244.14) 5.6326 in. (2)(4) x=− + + = 14− =x 8.3674 in. 3 2 3 2 4 1 1 8 (14 ) (8)(5.6326) (17.438)(8.3674) 1697.45 in 3 s 3 I = x +nA −x = + = nMy I M I ny σ σ = ∴ =

Concrete: n=1.0, y =5.6326 in., _σ =1350 psi

3

(1350)(1697.45)

406.835 10 lb in 407 kip in (1.0)(5.6326)

M = = × ⋅ = ⋅

Steel: _{n=9.67, y =8.3674 in., σ =20 10 psi×} 3

3

(20 10 )(1697.45)

419.72 lb in 420 kip in (9.67)(8.3674)

M = × = ⋅ = ⋅

PROBLEM 4.52

Knowing that the bending moment in the reinforced concrete beam is +100 kip ⋅ ft and that the modulus of elasticity is _{3.625 10 psi×} 6 _{for the}

concrete and 29 10 psi× 6 for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete.

SOLUTION 6 6 2 2 2 29 10 8.0 3.625 10 (4) (1) 3.1416 in 25.133 in 4 s c s s E n E A π nA × = = = ×   = _{ } = =   

Locate the neutral axis.

(24)(4)( 2) (12 ) (25.133)(17.5 4 ) 0 2 x x+ + x  _{ }− − −x =   2 2 96x+192+6x −339.3+25.133x = 0 or 6x +121.133x−147.3= 0 Solve for x. 2 121.133 (121.133) (4)(6)(147.3) 1.150 in. (2)(6) x ₌ − + + ₌ 3 17.5 4 12.350 in. d _{= − − =}x 3 2 3 2 4 1 1 1 1 1 3 3 4 2 2 2 2 4 3 3 3 4 1 2 3 1 1 (24)(4) (24)(4)(3.150) 1080.6 in 12 12 1 1 (12)(1.150) 6.1 in 3 3 (25.133)(12.350) 3833.3 in 4920 in I b h A d I b x I nA d I I I I = + = + = = = = = = = = + + = nMy I

σ = − where 100 kip ftM = ⋅ =1200 kip in.⋅

(a) Steel: n ₌8.0 y _{= −}12.350 in.

(8.0)(1200)( 12.350) 4920 s σ = − − _σs = 24.1 ksi  (b) Concrete: n =1.0, y = +4 1.150 =5.150 in. (1.0)(1200)(5.150) c σ = − _σc = −1.256 ksi 

PROBLEM 4.53

The design of a reinforced concrete beam is said to be balanced if the maximum stresses in the steel and concrete are equal, respectively, to the allowable stresses σ_s and _σc. Show that to achieve a balanced design the distance x from the top of the beam to the neutral axis must be

1 σ σ = + s c c s d x E E

where E_c and E_s are the moduli of elasticity of concrete and steel, respectively, and d is the distance from the top of the beam to the reinforcing steel.

SOLUTION ( ) ( ) 1 1 1 1 s c s c s c s c s c c s s c nM d x Mx I I n d x d n n x x E d x n E d x E E σ σ σ σ σ σ σ σ σ σ − = = − = = − = + = + = + 

PROBLEM 4.54

For the concrete beam shown, the modulus of elasticity is 3.5 10× 6psi for the concrete and 6

29 10× psi for the steel. Knowing that b= in. and 8 d =22 in., and using an allowable stress of 1800 psi for the concrete and 20 ksi for the steel, determine (a) the required area As

of the steel reinforcement if the beam is to be balanced, (b) the largest allowable bending moment. (See Prob. 4.53 for definition of a balanced beam.)

The design of a reinforced concrete beam is said to be balanced if the maximum stresses in the steel and concrete are equal, respectively, to the allowable stresses σ_s and _σc.

SOLUTION 6 6 3 29 10 8.2857 3.5 10 ( ) ( ) 1 1 20 10 1 1 2.3410 8.2857 1800 0.42717 (0.42717)(22) 9.398 in. 22 9.398 12.602 in. s c s c s c s c E n E nM d x Mx I I n d x d n n x x d x n x d d x σ σ σ σ σ σ × = = = × − = = − = = − × = + = + ⋅ = = = = − = − =

Locate neutral axis: ( ) 0

2 s x bx −nA d x− = (a) 2 (8)(9.398)2 _{3.3835 in}2 2 ( ) (2)(8.2857)(12.602) s bx A n d x = = = − 2 3.38 in s A =  3 2 3 2 4 1 _{( )} 1_{(8)(9.398) (8.2857)(3.3835)(12.602) 6665.6 in} 3 s 3 I bx nA d x n My I M I ny σ σ = + − = + = = =

(b) Concrete: 1.0 9.398 in. 1800 psi (1800)(6665.6) 1.277 10 lb in6

(1.0)(9.398) c n= y= _σ = M = = × ⋅ Steel: 3 3 6 8.2857 | | 12.602 in. 20 10 psi (20 10 )(6665.6) 1.277 10 lb in (8.2857)(12.602) s n y M σ = = = × × = = × ⋅

PROBLEM 4.55

Five metal strips, each 40 mm wide, are bonded together to form the composite beam shown. The modulus of elasticity is 210 GPa for the steel, 105 GPa for the brass, and 70 GPa for the aluminum. Knowing that the beam is bent about a horizontal axis by a couple of moment 1800 N ⋅ m, determine (a) the maximum stress in each of the three metals, (b) the radius of curvature of the composite beam.

SOLUTION

Use aluminum as the reference material. 1 in aluminum. / 210 / 70 3 in steel. / 105 / 70 1.5 in brass. s a b a n n E E n E E = = = = = = =

Due to symmetry of both the material arrangement and the geometry, the neutral axis passes through the center of the steel portion.

For the transformed section,

3 2 3 2 3 4 1 1 1 1 1 1 1 3 2 3 2 3 4 2 2 2 2 2 2 2 3 3 3 4 3 3 3 3 3 4 3 4 4 2 5 1 1 (40)(10) (40)(10)(25) 253.33 10 mm 12 12 1.5 (40)(10) (1.5)(40)(10)(15) 140 10 mm 12 12 3.0 (40)(20) 80 10 mm 12 12 140 10 mm 253.33 10 mm n I b h n A d n I b h n A d n I b h I I I I = + = + = × = + = + = × = = = × = = × = = × 3 4 9 4 866.66 10 mm 866.66 10 m I =



I = × = × − (a) nMy I σ = − where M =1800 N m⋅ Aluminum: n =1.0 y = −30 mm =0.030 m 6 9 (1.0)(1800)(0.030) 62.3 10 Pa 866.66 10 a σ = ₋ = × × σa =62.3 MPa  Brass: n=1.5 y= −20 mm= −0.020 m 6 9 (1.5)(1800)(0.020) 62.3 10 Pa 866.66 10 b σ = ₋ = × × σb=62.3 MPa  Steel: n=3.0 y= −10 mm= −0.010 m 6 9 (3.0)(1800)(0.010) 62.3 10 Pa 866.66 10 s σ = ₋ = × × σs =62.3 MPa (b) Radius of curvature. 1 ₉ 1800 ₉ 0.02967 m 1 (70 10 )(866.66 10 ) a M E I ρ = = _{× ×} − = − ρ=33.7 m 