Recent content by hatchelhoff

Thanks I understood the concept, and I have no questions.

I have found the deflection of the beam due to bending to be 2.66 mm. I have found the deflection of the beam due to the deformation of the bracket to be 2.74 mm. this gives me a total deflection of 5.4 mm

CivilSigma I do mean the rotation in radans. In need this because the force F causes a deflection in the beam, but it also causes a deformation of the upright bracket which causes an additional dip at the far end. PhanthomJay Merry christmas to you also. Thanks for your link. Chestermiller I...

I can't seem to find any such table. Can you point me int the right direction.

MB = F*Distance = 5KN*410 mm = 2050 NM R1 = R2 = MB/L = 2050NM/0.41M = 5000 N Im not sure how to use these values to find the angular deflection.

Homework Statement The Figure shows a welded steel bracket loaded by a force F = 5 kN. Homework Equations [/B]The Attempt at a Solution I know that the total deflection is the deflection of the beam due to F. And I also need to take into account the angular deflection at the joint...

Thanks pongo38. I have taken into account the deflection at E due to F and due to Ra. this gives me 2FL(^3))/3EI. I then add this to the deflection at F. This gives me a total of 4FL(^3))/3EI.

Homework Statement Homework Equations The answer in the book is (4F(L^3))/3EI The Attempt at a Solution I felt the above question could be simplified into a standard beam problem. I choose the following beam. I used the Yc equation, but when I replace a with L I get an answer of...

I performed one integration and I got the correct answer. I integrated using the limits b and L. My answer was (1/6)*((L^2)*(2L-3B) + B^3). haruspex thanks for you patience and help in getting me to the correct answer. I take it that I only needed to integrate as far as the discontinuity and...

You would have to perform two integrations. the limits of Integration1 would be from the end to the discontinuity. The limits of Integration2 would be from the discontinuity to the other end. you would then add these integrations together.

ok. So i guess I have to integrate between the limits of <x-b> greater than 0 and less that L. but how do i know if x-b is greater than 0 if I don't have a value for b.

I have plotted two charts with different values for b. One chart plots b at 0.1. the other chart plots b at 0.9. I see in both cases that some of the <x-B> values are below the x-axis and some of the <x-b> are above the x-axis.

you also mentioned about integrating from 0 to L. This would give me ((L^2)/6)*(2L-3B)) - ((0^2)/6)*(2(0)-3B)). is this correct ?

I am having difficulty drawing the graph of <x - b> because I don't have a value for b yet. can you give me a further hint please.

for max{x-b, 0}. I think x-b was chosen as the max in the attached example. so does this mean that <x-b> can now be change to (x-b) ?