Problem: Determine the maximum value of the average normal stress

[zartiox]

Hi guys.

Please look at the uploaded picture.
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

 

[PhanthomJay]

Hi guys.

Please look at the uploaded picture.
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.

 

[zartiox]

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