Materials Science and Engineering
Materials Science and Engineering
9th Edition
ISBN: 9781118324578
Author: Jr. William D. Callister
Publisher: WILEY
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Chapter 4.11, Problem 42QP
To determine

(a)

To explain:

The reason whether the surface energy is greater than or less than the grain boundary energy for the same material.

(b)

To determine

To explain:

The reason for grain boundary energy of a small-angle grain boundary is less than for a high-angle one.

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250 mm 400 mm A B C E F 250 mm PROBLEM 1.52 Each of the two vertical links CF connecting the two horizontal members AD and EG has a 10 × 40-mm uniform rectangular cross section and is made of a steel with an ultimate strength in tension of 400 MPa, while each of the pins at C and F has a 20-mm diameter and are made of a steel with an ultimate strength in shear of 150 MPa. Determine the overall factor of safety for the links CF and the pins connecting them to the horizontal members. 24 kN
50 mm 12 mm B O C OA 300 mm 450 mm E PROBLEM 1.51 Each of the steel links AB and CD is connected to a support and to member BCE by 25-mm-diameter steel pins acting in single shear. Knowing that the ultimate shearing stress is 210 MPa for the steel used in the pins and that the ultimate normal stress is 490 MPa for the steel used in the links, determine the allowable load P if an overall factor of safety of 3.0 is desired. (Note that the links are not reinforced around the pin holes.)
P 4.4-22 Determine the values of the node voltages V1, V2, and v3 for the circuit shown in Figure P 4.4-22. 202 ww 4ia 202 w + + ±12 V V₁ ΖΩ V2 ΖΩ V3 11 A + 하
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