Offset Cantilever Beam - Horizontal & Vertical Deflection

[MEstudent890]

Homework Statement

The offset cantilever beam, ABCD, is fabricated from structural steel. Es = 30,000,000 lbs/in², I = 60 in⁴. A load of 2000 lbs is placed at point D. Neglect the weight of the beam. Determine the horizontal and vertical deflection at D with respect to A in inches.

Homework Equations

δ=Phl²/2EI

The Attempt at a Solution

First attempt at the problem was to move the load up the distance of the vertical member BC in order to create a moment and then use the point load equation of a cantilever beam. Professor told me this was incorrect and I needed to add a dummy load, Q, placed at point D to serve as a horizontal component of the load. Then integrate each section of the beam over P or Q.

Very confused. Any direction would be helpful

 

[KataKoniK]

.Thank you for your post. I am a scientist and I would be happy to assist you with this problem.

Firstly, let's clarify the given information. The offset cantilever beam ABCD is made of structural steel, with a Young's modulus (Es) of 30,000,000 lbs/in2 and a moment of inertia (I) of 60 in4. A load of 2000 lbs is placed at point D, and we are neglecting the weight of the beam itself. The task is to determine the horizontal and vertical deflection at point D with respect to point A, in inches.

As you correctly mentioned, the deflection equation for a point load on a cantilever beam is δ=Phl2/2EI, where P is the load, h is the distance from the load to the fixed end of the beam, l is the length of the beam, E is the Young's modulus, and I is the moment of inertia.

However, in this case, we have a combination of both vertical and horizontal loads at point D. This means that we need to consider both the vertical and horizontal components of the load separately.

To do this, we can use the concept of superposition. This means that we can consider the vertical and horizontal components of the load separately, and then add them together to get the total deflection at point D.

For the vertical component, we can use the same equation as before, with P being the vertical load of 2000 lbs and h being the distance from point D to the fixed end of the beam (which is also the length of the beam). So the deflection due to the vertical load alone would be δv=2000(l2/2EI).

For the horizontal component, we can use the same equation, but this time with P being the dummy load Q, and h being the distance from point D to the fixed end of the beam (which is also the length of the beam). So the deflection due to the horizontal load alone would be δh=Q(l2/2EI).

Now, to find the total deflection at point D, we can simply add the two deflections together: δtotal=δv+δh.

I hope this helps to clarify the problem for you. If you have any further questions, please don't hesitate to ask. Good luck with your work!