Mechanics of Materials, 7th Edition

Published by McGraw-Hill Education
ISBN 10: 0073398233
ISBN 13: 978-0-07339-823-5

Chapter 1 - Problems - Page 22: 1.9

Answer

$$A_{BE} = 285.3 \space mm^2$$

Work Step by Step

1) Reactions and internal forces $\bullet$Start by examining the link CD, applying equilibrium, we get: $\sum{M_D} = 0; \space\space 4*0.15 + 4*0.25 - V_C * 0.25 \space \Rightarrow \space V_C = 6.4 \space kN \uparrow$ $\bullet$ Now examine the portion of the structure to the left of the internal hinge at C Applying the equilibrium of moments around point A we can find force in link BE $\sum{M_A} = 0; \space\space 4*0.35 +6.4 * 0.45 - BE * 0.15 = 0$ $BE = 28.53 \space kN$ 2) Average Normal Stress: Normal Stress equation: $\sigma_{AVG} = \frac{N_{INT}}{A} \space\Rightarrow\space A = \frac{N_{INT}}{\sigma_{AVG}}$ $A = \frac{28.53 * 10^3 \space N}{100 * 10^6 \space Pa} = 2.853 *10^{-4}\space m^2 = \boxed{285.3 \space mm^2} \space\space\space\leftarrow ANS$
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