just part a and b A 40 lb. rigid structure consists of a 60 inch uniform beam re

Question

just part a and b

A 40 lb. rigid structure consists of a 60 inch uniform beam resting on simple support" with a 12 inch square uniform block attached to the right hand end. Both the beam and the block weigh 20 pounds The structure is enclosed in a rocket fairing. During launch the package is subjected to a vertical acceleration of 4.8 g's. Calculate the reaction forces at the supports. Calculate the internal bending internal bending moment at a point B Calculate the bending moment under static conditions

Explanation / Answer

>> As, the whole structure is accelerating upwards with an acceleration of 4.8g

So, Effectively, g' = g + 4.8g = 5.8*g

>> Now, Considering all forces in our structure:

1. Rxn at A, Ra , vertically upwards

2. Rxn at B,. Rb, vertically upwards, at 30 in from A

3. W1 = Weight of Beam = 20*5.8g lb-ft/s2 , at 60/2 = 30 in from A, vertically downwards

4. W2 = Weight of Block = 20*5.8g lb-ft/s2 , at 60 + 12/2 = 66 in from A, vertically downwards

>> Now, Considering Equilibrium,

=> Net Vertical Force = 0

=> Ra + Rb = W1 + W2 = 2*20*5.8*32.2 = 7470.4 lbf ....(1)...

>> Also, Moment about A, Ma = 0

=> Ma = (30/12)*Rb - (30/12)*W1 - (66/12)*W2 = 0

=> 2.5*Rb = 2.5*3735.2 + 5.5*3735.2

=> Rb = 11952.64 lbf = 371.2 lb

and, Ra = - 4482.4 lbf = - 139.0205 lb

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