Mechanics of Materials, 7th Edition
Mechanics of Materials, 7th Edition
7th Edition
ISBN: 9780073398235
Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek
Publisher: McGraw-Hill Education
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Chapter 8, Problem 72RP

Forces are applied at points A and B of the solid cast-iron bracket shown. Knowing that the bracket has a diameter of 0.8 in., determine the principal stresses and the maximum shearing stress at (a) point H, (b) point K.

Fig. P8.72

Chapter 8, Problem 72RP, Forces are applied at points A and B of the solid cast-iron bracket shown. Knowing that the bracket

(a)

Expert Solution
Check Mark
To determine

The principal stresses and the maximum shearing stress at point H.

Answer to Problem 72RP

The maximum principal stress at point H is 31.9ksi_.

The minimum principal stress at H is -6.99ksi_.

The shear stress at point H is 19.42ksi_.

Explanation of Solution

Given information:

The diameter (d) of the bracket is 0.8in.

Calculation:

Sketch the free body diagram of solid cast iron as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 8, Problem 72RP , additional homework tip  1

Refer to Figure 1.

Find the value of P at the section containing point H and K.

P=2,500lb

Find the shear force about y axis as follows:

Vy=600lb

Find the shear force about x axis as follows:

Vx=0

Find the moment about x axis as follows:

Mx=(3.51)(600)=2.5×600=1,500lbin

Find the moment about y axis as follows:

My=0

Find the moment about z axis as follows:

Mz=2.5×600=1,500lbin

Find the value of radius (c) using the relation:

c=d2 (1)

Here, d is the diameter of bracket.

Substitute 0.8in. for d in Equation (1).

c=0.82=0.4in.

Find the area (A) of the circular section using the equation:

A=πc2 (2)

Here, c is the half of the diameter.

Substitute 0.4in. for c in Equation (2).

A=π(0.4)2=0.50265in2

Find the moment of inertia (I) of section using the relation:

I=π4c4 (3)

Substitute 0.4in. for c in Equation (3).

I=π4(0.4)4=20.106×103in4

Find the moment of inertia (J) of section using the relation:

J=2I (4)

Substitute 20.106×103in4 for I in Equation (4).

J=2(20.106×103)=40.106×103in4

Find the value of Q for semicircle using the relation:

Q=23c3 (5)

Substitute 0.4in. for c in Equation (5).

Q=23(0.4)3=42.667×103in3

Determine the normal stress at point H using the relation:

σH=PA+McI (6)

Here, P is the centric force, A is the area of circular cross section, I is the moment of inertia, M is the moment, and c is the centroid distance.

Substitute 2,500lb for P, 0.50265in2 for A, 20.106×103in4 for I, 0.4in. for c, and 1,500lbin for M in Equation (6).

σH=2,5000.50265+1,500×0.420.106×103=4,973.639+29,841.838=24,868.199psi(1ksi103psi)=24.87ksi

Determine the shear stress at point H using the relation:

τH=TcJ (7)

Here, T is the Torque and J is the polar moment of inertia.

Substitute 1,500lb for T, 40.106×103in4 for J, and 0.4in. for c, in Equation (7).

τH=1,500×0.440.106×103=60040.106×103=14,960.355psi(1ksi103psi)=14.92ksi

Sketch the stresses at point H as shown in Figure 2.

Mechanics of Materials, 7th Edition, Chapter 8, Problem 72RP , additional homework tip  2

Find the average (σave) normal stress at point H using the relation:

σave=σH2 (8)

Here, normal stress at point H.

Substitute 24.87ksi for σH in Equation (8).

σave=24.872=12.435ksi

Find the R using the relation:

R=(σH2)2+τH2 (9)

Here, shear stress at point H.

Substitute 24.87ksi for σH and 14.92ksi for τH in Equation (9).

R=(24.872)2+14.922=154.629+222.6064=19.423ksi

Determine the maximum principal stress (σmax) using the relation:

σmax=σave+R (10)

Substitute 12.435ksi for σave and 19.423ksi for R in Equation (10).

σmax=12.435+19.423=31.858=31.9ksi

Thus, the maximum principal stress at point H is 31.9ksi_.

Determine the minimum principal stress (σmin) at point H using the relation:

σmin=σaveR (11)

Substitute 12.435ksi for σave and 19.423ksi for R in Equation (11).

σmin=12.43519.423=6.99ksi

Thus, the minimum principal stress at H is 6.99ksi_.

Determine the maximum shear stress at point H using the relation:

τmax=12(σmaxσmin) (11)

Here, σmax is the principal maximum stress and σmin is the principal minimum stress.

Substitute 6.99ksi for σmin and 31.9ksi for σmax in Equation (11).

τmax=12(31.9(6.99))=19.44ksi

Thus, the shear stress at point H is 19.42ksi_.

(b)

Expert Solution
Check Mark
To determine

The principal stresses and the maximum shearing stress at point K.

Answer to Problem 72RP

The maximum principal stress at point K is 14.21ksi_.

The minimum principal stress at K is 19.18ksi_.

The shear stress at point K is 16.70ksi_.

Explanation of Solution

Calculation:

Determine the normal stress at point K using the relation:

σK=PA (12)

Substitute 2,500lb for P and 0.50265in2 for A, in Equation (12).

σK=2,5000.50265=4,973.639psi(1ksi103psi)=4.974ksi

Determine the shear stress at point K using the relation:

τK=TcJ+VQIt (13)

Substitute 1,500lb for T, 40.106×103in4 for J, 60lb for V, 42.667×103in3 for Q, 20.106×103in4 for I, 0.8in. for t, and 0.4in. for c, in Equation (13).

τH=1,500×0.440.106×103+600×42.667×10320.106×103×0.8=14,960.355+1,591.577=16,551.932psi(1ksi103psi)=16.512ksi

Sketch the stresses at point K as shown in Figure 3.

Mechanics of Materials, 7th Edition, Chapter 8, Problem 72RP , additional homework tip  3

Find the average (σave) normal stress using the relation:

σave=σK2 (14)

Here, normal stress at point H.

Substitute 4.974ksi for σK in Equation (14).

σave=4.9742=2.487ksi

Find the R using the relation:

R=(σH2)2+τH2 (15)

Here, shear stress at point H.

Substitute 4.974ksi for σK and 16.512ksi for τK in Equation (15).

R=(4.9742)2+16.5122=6.185+272.646=16.698ksi

Determine the maximum principal stress (σmax) at point K using the relation:

σmax=σave+R (16)

Substitute 2.487ksi for σave and 16.698ksi for R in Equation (16).

σmax=2.487+16.698=14.21ksi

Thus, the maximum principal stress at point K is 14.21ksi_.

Determine the minimum principal stress (σmin) at point K using the relation:

σmin=σaveR (17)

Substitute 2.487ksi for σave and 16.698ksi for R in Equation (17).

σmin=2.48716.698=19.18ksi

Thus, the minimum principal stress at K is 19.18ksi_.

Determine the maximum shear stress at point K using the relation:

τmax=12(σmaxσmin) (18)

Here, σmax is the principal maximum stress at point K and σmin is the principal minimum stress at point K.

Substitute 19.18ksi for σmin and 14.21ksi for σmax in Equation (18).

τmax=12(14.21(19.18))=16.70ksi

Thus, the shear stress at point K is 16.70ksi_.

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Chapter 8 Solutions

Mechanics of Materials, 7th Edition

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