Answer:
(The diagram of the question is given in Attachment 1)
The largest load which can be applied is:
P=67.62 kN
Explanation:
All the forces are shown in the diagram in Attachment 2.
∑ F(y)= 0 (Upwards is positive)
[tex]F_{BC}sin\theta-P=0\\\frac{4}{5}F_{BC} - P=0\\F_{BC}=\frac{5}{4}P\\\\F_{BC}=1.25P[/tex]
∑ F(x)= 0 (Towards Right is positive)
[tex]N_{a-a} - F_{BC}cos\theta=0\\N_{a-a}-1.25P(\frac{3}{5})=0\\N_{a-a}=0.75P[/tex]
∑ F(y)= 0 (Upwards is positive)
[tex]F_{BC}sin\theta- V_{a-a}= 0\\(\frac{4}{5})1.25P-V_{a-a}=0\\V_{a-a}=P[/tex]
The cross sectional area of a-a:
[tex]A_{a-a}= \frac{(0.026)(0.026)}{3/5}\\A_{a-a}= 1.127\cdot10^{-3}[/tex]
σ = N(a-a)/A(a-a)
Substitute values:
[tex]160\cdot10^6=\frac{0.75P}{1.127\cdot10^{-3}}\\P=240.42\cdot10^3 N\\P=240.42 kN[/tex]
Т= V(a-a)/A(a-a)
Substitute values:
[tex]60\cdot10^6=\frac{P}{1.127\cdot10^{-3}}\\P=67.62\cdot10^{3}N\\P=67.62kN[/tex]
To satisfy both the condition, we have to choose the lower value of P.
P=67.62 kN
