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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Textbook Question
Chapter 1.2, Problem 7P

Each of the four vertical links has an 8 × 36-mm uniform rectangular cross section, and each of the four pins has a 16-mm diameter. Determine the maximum value of the average normal stress in the links connecting (a) points B and D, (b) points C and E.

Fig. P1.7

Chapter 1.2, Problem 7P, Each of the four vertical links has an 8  36-mm uniform rectangular cross section, and each of the

(a)

Expert Solution
Check Mark
To determine

The maximum value of average normal stress in the links connecting at point B and D.

Answer to Problem 7P

The maximum value of average normal stress in the links connecting at point B and D is 101.6MPa_.

Explanation of Solution

Given information:

The size of rectangular cross section is 8×36mm.

The diameter (d) of the each pin is 16mm.

Calculation:

Sketch the free body diagram of link ABC as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 1.2, Problem 7P

Here, FBD and FCE are force applied by the link BD and the link CE.

Refer to Figure 1.

Apply the moment equilibrium condition at the point C.

MC=0

(0.4)FBD(0.25+0.4)(20)=00.4FBD13=00.4FBD=13FBD=130.4

FBD=32.5kN(103N1kN)FBD=32.5×103N(Tension)

Refer to Figure 1.

Apply the moment equilibrium condition at the point B.

MB=0

(0.4)FCE(0.25)(20)=00.4FCE5=00.4FCE=5FCE=12.5kN(103N1kN)FCE=12.5×103N(compression)

Calculate the net area of one link for tension as follows:

Anet=b(hdPin) (1)

Here, b is the width of the rectangular cross section, h is the depth of the rectangular cross section, and dPin is the diameter of each pin.

Substitute 8mm for b, 36mm for h, and 16mm for dPin in Equation (1).

Anet=8mm(1m103mm)(36mm(1m103mm)16mm(1m103mm))=0.008(0.0360.016)=0.008×(0.02)=160×106m2

Find the area of network for two parallel links as follows:

Anet=2(160×106m2)=320×106m2

Find the average normal stress (σBD) in the links connecting at point B and D.

σBD=FBDAnet (2)

Here, FBD is force in link BD and Anet is area of network of one link in tension.

Substitute 32.5×103N for FBD and 320×106m2 in Equation (2).

σBD=32.5×103N320×106m2=101,562,500Nm2×(1MPa106Nm2)=101.563MPa101.6Mpa

Thus, the maximum value of average normal stress in the links connecting at point B and D is 101.6MPa_.

(b)

Expert Solution
Check Mark
To determine

The maximum value of average normal stress in the links connecting at point C and E.

Answer to Problem 7P

The maximum value of average normal stress in the links connecting at point C and E is 21.7MPa_.

Explanation of Solution

Calculation:

Calculate the net area of one link for tension as follows:

Anet=bh (3)

Substitute 8mm for b and 36mm for h in Equation (3).

Anet=8mm(1m103mm)(36mm(1m103mm))=0.008×0.036=288×106m2

Find the area of network for two parallel links as follows:

Anet=2(288×106m2)=576×106m2

Find the average normal stress (σCE) in the links connecting at point B and D.

(σCE)=(FCE)Anet (4)

Here, FCE is force in link CE and Anet is area of network of one link in compression.

Substitute 12.5×103N for FCE and 576×106m2 for Anet in Equation (4).

σCE=12.5×103N576×106m2=21,701,388.8Nm2×(1MPa106Nm2)=21.701MPa

Thus, the maximum value of average normal stress in the links connecting at point C and E is 21.7MPa_.

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Chapter 1 Solutions

Mechanics of Materials, 7th Edition

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