# Problem: Determine the maximum value of the average normal stress

Hi guys.

Normal stress is difined as: σ= P/A. And the Maximum normal stress σ = P/ (b*h/cos(θ)) ??

The right answers should be σAB= 97,7 MPa and σBC = -66,5 MPa. But how do I calculate it?

What I have tried:
Force AB: 40* sin(60deg) = 34,641 kN = 34,641 * 10^3 N

σAB= 34,641 * 10^3 N/((45-20)*12)/sin(60deg)) = 100 MPa

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PhanthomJay
Homework Helper
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Hi guys.

Normal stress is difined as: σ= P/A. And the Maximum normal stress σ = P/ (b*h/cos(θ)) ??

The right answers should be σAB= 97,7 MPa and σBC = -66,5 MPa. But how do I calculate it?

What I have tried:
Force AB: 40* sin(60deg) = 34,641 kN = 34,641 * 10^3 N

σAB= 34,641 * 10^3 N/((45-20)*12)/sin(60deg)) = 100 MPa
I think you are misreading the figure regarding the cross sectional area of the link.....its just b*h, where b is (45-20) = 25 mm, and h is 12 mm. Thus, A = 300 mm^2.
But beyond that, you are not calculating the force correctly. You might want to draw a free body diagram of the joint at B and use Newton's 1st law in the x direction (sum of forces in x direction = 0) and y directions (sum of forces in y direction = 0), to solve for the force components in each link. Then the member force can be found from the sq rt of the sum of the squares.

Thank you very much, that explanation helped me to could compute it :)