Solving Displacement of Prismatic Bar AD with Loads P1,P2,P3

[Dyls]

I'm a little lost with this question:

A prismatic bar AD is subjected to loads P1 = 3k P2 = 3k P3 = 11k. Cross-sectional Area A = 1.40 sq in. Modulus of E = 17000 ksi. Find displacement of the free end of bar. What should P3 be displacement at end is to be reduced by half of original value?

So the bar has three sections. AB section has a cantilever beam attached to a wall on the left, the right side has a force P1 going to the left. BC has no forces in it. CD has P2 force on the left going to the right. Point D is the end of the beam and there is a force coming straight out of the beam to the right. Each section is 20 in long.

So I first find external forces. then I get lost...

 

[Nate810]

To find the displacement of the free end of the bar, we need to use a beam bending equation. The equation is as follows: Displacement = (P1*L1^3 + P2*L2^3 + P3*L3^3)/(3*E*A)Where P1, P2, and P3 are the forces applied at each end of the beam (in kips), L1, L2, and L3 are the lengths of each segment (in inches), E is the modulus of elasticity (in ksi) and A is the cross-sectional area (in sq. in.). Using the given data, this equation simplifies to: Displacement = (3*20^3 + 3*20^3 + 11*20^3)/(3*17000*1.4) = 0.0066 inchesTo reduce the displacement at the end by half, P3 must be equal to 8.5 kips.

 

[wais]

To solve this problem, we need to use the principles of statics and mechanics of materials. First, we can draw a free body diagram of the bar, indicating all the external forces acting on it. This will help us determine the reactions at the supports and the internal forces within the bar.

Next, we can use equilibrium equations to find the reactions at the supports. Since the bar is in static equilibrium, the sum of all the forces and moments acting on it must be equal to zero. This will give us the values for P1, P2, and P3.

Once we have the external forces, we can use the equations of mechanics of materials to find the internal forces within the bar. This will allow us to determine the stress and strain in the bar.

To find the displacement of the free end of the bar, we can use the equation for axial deformation of a prismatic bar: δ = PL/AE, where δ is the displacement, P is the applied load, L is the length of the bar, A is the cross-sectional area, and E is the modulus of elasticity.

To reduce the displacement at the end of the bar by half, we can use the principle of superposition. This means that we can add a new load, P3', that is equal to half of the original P3, in the opposite direction. Then, we can solve for the new displacement at the end of the bar using the same equation.

In summary, to solve for the displacement of the free end of the prismatic bar, we need to:

1. Draw a free body diagram and determine the external forces.
2. Use equilibrium equations to find the reactions at the supports.
3. Use mechanics of materials equations to find the internal forces and stress/strain within the bar.
4. Use the equation for axial deformation to find the displacement of the free end.
5. Use the principle of superposition to determine the new displacement if P3 is reduced by half.