Given information:

The uniform cross section is (50×150) mm.

Calculation:

Sketch the free body diagram of ABC as shown in Figure 1.

Here, W is the weight of the loading.

Find the weight of the loading as follows:

W=mg (1)

Here, m is the mass of the loading and g is the acceleration due to gravity.

Substitute 80 Mg for m and 9.81 m/s2 in Equation (1).

W=80×9.81=784.8 kN

Taking moment about A,

FCD×15−W×28=015FCD−28W=0FCD=2815W (2)

Substitute 784.8 kN for W in Equation (2).

FCD=2815(784.8 kN)=1,464.96=1,465 kN

Find the area of the cross section using the relation:

A=bd (3)

Here, b is the width of the cross section and d is the depth of the cross section.

Substitute 50 mm for b and 150 mm for d in Equation (3).

A=50×150=7,500 mm2(1 m103 mm)2=7.5×10−3 m2

Determine the normal stress in the central portion of the link as follows:

σCD=FCDA (4)

Here, A is the area of the cross section and FCD is force in member CD.

Substitute 7.5×10−3 m2 for A and 1,465 kN for FCD in Equation (4).

σCD=1,4657.5×10−3=195.3×106 Pa=195.3×106 Pa×1 Mpa106 Pa=195.3 Mpa

Thus, the normal stress in the centre portion of the link is 195.3 MPa_.