Chapter 1 - Problems - Page 22: 1.10

(a) $σ_{BD}= 11.09 ksi $ $(tension)$ (b) $σ_{BD}=12.00 ksi $ $( cmpression)$

In the given problem we have to find the normal stress member BD for two different cases of angle $θ$. So, we will consider the free body diagram of the entire system take moments about point A to calculate the force in member BD, and subsequently, calculate the stress from the given cross-sectional area. For Part (a): When θ=0 Calculate the perpendicular distance of all the forces from point A. The distance of 4kpis force. $d_{4}=18 \times sin (30-θ)$ $d_{4}= 18 \times (30)$ $d_{4}= 9 in$ Distance of $F_{BD}$ $d_{BD}= 12 \times cos (30)$ $d_{BD}= 12 \times \frac{\sqrt 3}{2}$ $d_{BD}= 10.39 in $ So, now take a moment of all the forces about point A $∑M_{A}=0$ $4 \times d_{4}- F_{BD}\times d_{BD}=0$ $F_{BD}= \frac{4 \times 9}{10.39}$ $F_{BD}= 3.464 kips (tensions)$ Now, Calculate the area of member BD for tensile force $A=(b-d)t$ $A=(1-\frac{3}{8})(\frac{1}{2)}$ $A= 0.3125 in^{2}$ Calculate tensile stress in member BD using the formula $σ_{BD}=\frac{F_{BD}}{A}$ $σ_{BD}=\frac{3.464 kips}{0.3125}$ $σ_{BD}=11.09 ksi$ For Part (b): When θ=$90 '$ Calculate the perpendicular distance of all the forces from point A. The distance of 4kpis force. $d_{4}=18 \times sin (30-θ)$ $d_{4}= 18 \times sin (30-90)$ $d_{4}= 18 \times \frac{\sqrt 3}{2}$ $d_{4}= 15.58 in$ Distance of $F_{BD}$ $d_{BD}= 12 \times cos (30)$ $d_{BD}= 12 \times \frac{\sqrt 3}{2}$ $d_{BD}= 10.39 in $ So, now take a moment of all the forces about point A $∑M_{A}=0$ $4 \times d_{4}+ F_{BD}\times d_{BD}=0$ $F_{BD}= \frac{4 \times 15.58}{10.39}$ $F_{BD}= -6 kips (compression)$ Now, Calculate the area of member BD for tensile force $A=(b)t$ $A=(1)(\frac{1}{2)}$ $A= 0.5 in^{2}$ Calculate tensile stress in member BD using the formula $σ_{BD}=\frac{F_{BD}}{A}$ $σ_{BD}=\frac{6 kips}{0.5}$ $σ_{BD}=12 ksi$