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Total Force In Portions Of A Bar Under Axial Loading


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Strings (SiPjAjk) = S7P3A32     Base Sequence = 12735     String Sequence = 12735 - 3 - 32

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Total Force In Portions Of A Bar Under Axial Loading

Figure 125.1 illustrates an axially loaded uniform bar that is rigidly fixed at its ends A and B. Axial loading is at points B and C.

(a) Determine the total force in portions AB, BC and CD of the bar. Assume consistent deformation and superposition.
(b) Interprete the result of problem (a) if the bar is not rigidly held at its ends A and D, but instead, prestressed by wedging it between rigid walls under an initial compression of 10,000 lb.
(c) What happens if the initial compression in problem (b) is less than 8000 lb?

The strings: S7P3A32 (Force - Push).

The math:
Pj Problem of Interest is of type force (push). Tension and compression are at play. On the whole more compression than tension, hence compression (push) is chosen.

Total Force In Portions Of A Bar Under Axial Loading

Consistent deformation implies, load is distributed in portions of the bar in inverse proportion to their respective lengths.
Superposition implies that total force in each portion is the algebraic sum of the forces imposed by the individual loads.
So, for load 9,000 lb:
(7/9)9000 = 7000 is carried in tension by part AB
(2/9)9000 = 2000 is carried in compression by part BD

For load 18,000 lb:
(4/9)18000 = 8000 is carried in compression by part AC
(5/9)18000 = 10000 is carried in tension by part CD

Denoting compression by negative sign and tension by positive sign, we have actual stresses in parts as follows:
In part AB = 7000 - 8000 = -1000 lb.
In part BC = -2000 - 8000 = -10,000 lb.
In part CD = -2000 + 10,000 = +8000 lb.

(b) The results of problem (a) will be interpreted as the changes in force, the parts of the bar will undergo. So, final force:
In part AB = 7000 - 8000 -10000 = -11,000 lb (compression).
In part BC = -2000 - 8000 - 10000 = -20,000 lb (compression).
In part CD = -2000 + 10000 - 10000 = -2000 lb (compression).

If the initial compression is less than 8,000 lb, the bar will break contact with the wall at D (zero tension).
So, there will be no force in part CD
So, force in part AB = 9000 lb (compression).
So, force in part BC = 18,000 lb (compression).
So in essence, the forces in AB and BC becomes statistically deterministic.

Figure 123.1 illustrates Pascal's Principle which states that pressure applied to an enclosed
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