The figure below shows the frame divided into simple free bodies. External loads

Question

The figure below shows the frame divided into simple free bodies. External loads, support reactions and unknown end forces are applied to each free body. Note that there are two unknown end forces where free bodies are connected in the complete structure by pin. There are two unknown end forces and a moment where free bodies are rigidly connected in the complete structure. For equilibrium of connections unknown end forces and moments are applied to the connected free bodies in opposite directions. Consider equilibrium of each free body and calculate unknown end forces.

Correct answers provided:

Hc = 1

Vc = 1

He = 1

Ve = 1

Me = -6

Hf = 0

Vf = 2

Explanation / Answer

First consider ABC

Let the direction of Vc be X axis and Hc be Y axis

As the body is in equilibrium,

Equationg forces in X direction, we get,

Vc - 1 = 0 ( since both are in the opposite direction)

Vc = 1

In the Y direction,

Hc + 1 - 2 = 0

Hc = 1

Now consider CDE,

Here X direction is along Hc

Y direction along Ve

Equationg forces in X direction, we get,

Hc - He = 0

He = 1

Equationg forces in Y direction, we get,

Ve - Vc = 0

Ve = 1

Now equate moments about the point E,

moment due to Vc = 1*8 = 8 in the anti clockwise direction

(moment due to Hc = 0 since the line of action passes through the point considered)

moment due to moment at D = 2 in the clockwise direction

Me + 8 - 2 = 0

Me = -6

Now consider EFG

Let the direction of He be X axis and 3 be Y axis

Equating forces along X direction, we get He - Hf -1 = 0

Hf = 0

Equating forces along Y direction,

3 -Ve - Vf = 0

Vf = 2

The same can be confirmed by substituting in the equations for the figure FJK.

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