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

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

The largest load which can be applied is:

P=67.62 kN

Step-by-step 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\\(4)/(5)F_(BC) - P=0\\F_(BC)=(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((3)/(5))=0\\N_(a-a)=0.75P`

∑ F(y)= 0 (Upwards is positive)

`F_(BC)sin\theta- V_(a-a)= 0\\((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)= ((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=(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=(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