Question

Determine the largest load P that can be a applied to the frame without causing either the average normal stress or the average shear stress at section a-a to exceed σ = 160 MPa and τ = 60 MPa , respectively. Member CB has a square cross section of 26 mm on each side.

Answer 1

Answer:

(The diagram of the question is given in Attachment 1)

The largest load which can be applied is:

P=67.62 kN

Explanation:

Make a Free body Diagram:

All the forces are shown in the diagram in Attachment 2.

Analyze the equilibrium of Joint C in Figure (a):

∑ F(y)= 0 (Upwards is positive)

$F_{BC}sin\theta-P=0\\\frac{4}{5}F_{BC} - P=0\\F_{BC}=\frac{5}{4}P\\\\F_{BC}=1.25P$

Substitute F(BC) in Figure (b):

∑ F(x)= 0 (Towards Right is positive)

$N_{a-a} - F_{BC}cos\theta=0\\N_{a-a}-1.25P(\frac{3}{5})=0\\N_{a-a}=0.75P$

∑ F(y)= 0 (Upwards is positive)

$F_{BC}sin\theta- V_{a-a}= 0\\(\frac{4}{5})1.25P-V_{a-a}=0\\V_{a-a}=P$

Find Cross Sectional Area:

The cross sectional area of a-a:

$A_{a-a}= \frac{(0.026)(0.026)}{3/5}\\A_{a-a}= 1.127\cdot10^{-3}$

Find P from Normal Stress Equation:

σ = N(a-a)/A(a-a)

Substitute values:

$160\cdot10^6=\frac{0.75P}{1.127\cdot10^{-3}}\\P=240.42\cdot10^3 N\\P=240.42 kN$

Find P from Shear Stress Equation:

Т= V(a-a)/A(a-a)

Substitute values:

$60\cdot10^6=\frac{P}{1.127\cdot10^{-3}}\\P=67.62\cdot10^{3}N\\P=67.62kN$

Results:

To satisfy both the condition, we have to choose the lower value of P.

P=67.62 kN