Statics Shear Force and Bending Moment Diagrams Problem

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Calculating the reactions at points C and E is essential for constructing shear and moment diagrams. The discussion highlights that the reaction at C is zero, which raises concerns about the beam's equilibrium. The reaction at E is determined to be 2.43 kN based on the sum of moments. However, without a reaction at C, the beam cannot be in equilibrium due to a net moment around C. Properly assessing all reactions is crucial for accurate analysis in statics problems.
jordan35
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Figured it out.
 

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For one thing, you haven't calculated the reactions at C and E. Until you do, you won't be able to construct the shear and moment diagrams.

If I am interpreting what you posted correctly, you have all of your reaction at E and none at C. Why?
 
Cx= 0 b/c there are no other forces in the x direction. Ey = 2.43 from sum of moments, and Cy = 0 after plugging into sum of the forces in the y direction. I thought this was strange to have no reactions at C, but don't know what I could have messed up with that part.
 
Sorry, but if you take the reaction at E = 2.43 kN and a zero reaction at C, the beam is NOT in equilibrium because there is a net moment around C.
 
Thread 'Have I solved this structural engineering equation correctly?'

Thread 'Have I solved this structural engineering equation correctly?'

Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...
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