## Mechanics of Materials, 7th Edition

$$\sigma_{BD}=101.5\space MPa\space (Tension)$$ $$\sigma_{CE}=21.7 \space MPa \space (Compression)$$
1)Free Body Diagram (FBD) of element $\overline{ABC}$: Equilibrium of Moments $\sum M_{B} = 0$ $F_{CE}*0.4 - 20 * 0.25 = 0$ $\rightarrow F_{CE} = 12.5\space kN$ Equilibrium of Forces in the y-direction $\sum F_{y} = 0$ $20 + 12.5 - F_{BD} = 0$ $\rightarrow F_{BD} = 32.5\space kN$ 2)Internal Forces in Links BD and CE Each of the links consists of two parallel elements so the internal force in each one is halved. $\Diamond$ For links BD: Since the link is in tension, the maximum tensile stress will occur at the pin opening. Therefore, the effective cross sectional area is reduced by the opening width: $A_{BD} = (36-16)*8 = 160\space mm^2$ The max axial stress in BD is then: $\sigma_{BD,max} = \frac{F_{BD}}{A_{BD}} = \frac{(32.5 \div 2)}{160} = 101.5 \space MPa \space(Tension)\space\space\leftarrow ANS1$ $\Diamond$ For links CE: Since the link is in compression, the maximum compressive stress will occur at the pin opening. Therefore, the effective cross sectional area is: $A_{CE} = (36)*8 = 288\space mm^2$ The max axial stress in CE is then: $\sigma_{CE,max} = \frac{F_{CE}}{A_{CE}} = \frac{(12.5 \div 2)}{288} = 21.7 \space MPa \space(Compression)\space\space\leftarrow ANS2$