EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 8220100257063
Author: BEER
Publisher: YUZU
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Chapter 4.5, Problem 55P

4.55 and 4.56 Five metal strips, each 0.5 × 1.5-in. cross section, are bonded together to form the composite beam shown. The modulus of elasticity is 30 × 106 psi for the steel, 15 × 106 psi for the brass, and 10 × 106 psi for the aluminum. Knowing that the beam is bent about a horizontal axis by a couple of moment 12 kip∙in., determine (a) the maximum stress in each of the three metals, (b) the radius of curvature of the composite beam.

Chapter 4.5, Problem 55P, 4.55 and 4.56 Five metal strips, each 0.5  1.5-in. cross section, are bonded together to form the

Fig. P4.55

(a)

Expert Solution
Check Mark
To determine

Find the maximum stress in Aluminum, Brass, and Steel.

Answer to Problem 55P

The maximum stress in Aluminum  is 6.86ksi_.

The maximum stress in Brass is 6.17ksi_.

The maximum stress in Steel is 4.11ksi_.

Explanation of Solution

Given information:

The dimension of each strip is 0.5×1.5in.

The modulus of elasticity of aluminum is Ea=10×106psi.

The modulus of elasticity of brass is Eb=15×106psi.

The modulus of elasticity of steel is Es=30×106psi.

The beam is bent about a horizontal axis by a couple of moment M=12kipin.

Calculation:

Consider aluminum as the reference material.

Calculate the modular ratio (n) as shown below.

n=EbaseEreference (1)

For steel.

Substitute 30×106psi for Es and 10×106psi for Ea in Equation (1).

ns=30×10610×106=3

For brass.

Substitute 15×106psi for Eb and 10×106psi for Ea in Equation (1).

nb=15×10610×106=1.5

For aluminum.

Substitute 10×106psi for Ea and 10×106psi for Ea in Equation (1).

na=10×10610×106=1

Sketch the transformed cross section as shown in Figure 1.

EBK MECHANICS OF MATERIALS, Chapter 4.5, Problem 55P

Refer to Figure 1.

The moment of inertia of the cross sections I1=I5 and I2=I4.

Calculate the moment of inertia for each section as shown below.

For section (1).

I1=I5=na(b1h1312+b1h1d12)

Substitute 1 for na, 1.5in. for b1, 0.5in. for h1, and (1.250.52)in. for d1.

I1=1×(1.5×0.5312+1.5×0.5(1.250.52)2)=0.7656in.4

Hence, I5=0.7656in.4.

For section (2).

I2=I4=nb(b2h2312+b2h2d22)

Substitute 1.5 for nb, 1.5in. for b2, 0.5in. for h2, and (0.750.52)in. for d1.

I2=1.5×(1.5×0.5312+1.5×0.5(0.750.52)2)=0.3047in.4

Hence, I4=0.3047in.4

For section (3).

I3=nsb3h3312

Substitute 3 for ns, 1.5in. for b3, and 0.5in. for h3.

I3=3×1.5×0.5312=0.0469in.4

Calculate the moment of inertia (I) as shown below.

I=I1+I2+I3+I4+I5

Substitute 0.7656in.4 for I1, 0.3047in.4 for I2, 0.0469in.4 for I3, 0.3047in.4 for I4, and 0.7656in.4 for I5.

I=0.7656+0.3047+0.0469+0.3047+0.7656=2.1875in.4

Calculate the maximum stress (σ) as shown below.

|σ|=|MnyI|σ=nMyI (2)

For steel.

Substitute 12kipin. for M, 2.1875in.4 for I, 0.25in. for y, and 3 for n in equation (2).

σ=3×12×0.252.1875=4.11ksi

Hence, maximum stress in steel is 4.11ksi_.

For brass.

Substitute 12kipin. for M, 2.1875in.4 for I, 0.75in. for y, and 1.5 for n in equation (2).

σ=1.5×12×0.752.1875=6.17ksi

Hence, maximum stress in brass is 6.17ksi_.

For aluminum.

Substitute 12kipin. for M, 2.1875in.4 for I, 1.25in. for y, and 1 for n in equation (2).

σ=1×12×1.252.1875=6.86ksi

Therefore, maximum stress in aluminum is 6.86ksi_.

(b)

Expert Solution
Check Mark
To determine

The radius of curvature of the composite beam.

Answer to Problem 55P

The radius of curvature of the composite beam is 151.9ft_.

Explanation of Solution

Given information:

The dimension of each strip is 0.5×1.5in.

The modulus of elasticity of aluminum is Ea=10×106psi.

The modulus of elasticity of brass is Eb=15×106psi.

The modulus of elasticity of steel is Es=30×106psi.

The beam is bent about a horizontal axis by a couple of moment M=12kipin.

Calculation:

Refer to part (a).

The moment of inertia of the beam is I=2.1875in.4.

Calculate the radius of curvature (ρ) as shown below.

1ρ=MEaI

Substitute 12kipin. for M, 2.1875in.4 for I, and 10×106psi for Ea.

1ρ=12kipin.×1,000lb1kip10×106psi×2.1875in.4ρ=15.4857×104=1,823in.×1ft12in.=151.9ft

Therefore, the radius of curvature of the composite beam is 151.9ft_.

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Chapter 4 Solutions

EBK MECHANICS OF MATERIALS

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