So it can react moments through a couple of horizontal forces - one at each wheel? If so, this is quite a different diagram than the one you showed in your OP.
If one takes a free body diagram of the horizontal member and applies a vertical load it will reveal itself unstable. Summing moments about the right end (hinge point) must equal zero so to satisfy the condition of the hinge - i.e. zero moment. For example, say there is a 2 kip vertical...
Returning to your original question:
1. I agree with your follow-up post regarding the applied loads being the components of the resultant load for the grapple placed exactly centered on the frame.
2. I suggest you consider the idea the actual placement of the grapple on the frame may be...
So then it’s a matter of determining the moments of inertia about the principal axes per DIK’e post and resolving the moment to the components about the principal axes, no?
Am I missing something? You state you have Ix and Iy. The stress at a given point for a moment about the x-axis is then Fb=(M Y)/Ix. Fb is the stress, M is the moment, Y is the perpendicular distance from the X-axis to the point under consideration.
I agree with LAZAR90's analysis. This setup is similar to a ladder on the wall of a house. No vertical load at the top of the rafter.
I checked by converting the calculated and given forces to the components parralell to and perpendicular to the longitudinal axis of the rafter. the axial...
The analysis of a pole such as you describe is often viewed as analogous to a cantilever sheet pile wall. There is a discussion about it in "foundation Analysis and Design" by Bowles. It is in the section covering cantilever sheet pile design.
My original post (since delted) is wrong: rb1957's formula is correct. The length is 2L (assuming the vertical member is at the midspan of the horizontal member.
can you provide some more info about this table 9source, what is L1xL2 referring to, end conditions, etc.?
I deleted my post. I don't believe it was correct. I mistook L as the full length of the horizontal member, but he sketch indicates it is not.
That said, a question I do have are the support conditions are? It appears it is simply supported at the ends? However; there are values reported...
Clarifying question: Your system consists of a cantilever beam. the unsupported end of the beam has beams framing into it (spandrel beams perpendicular to the cantilever beam & supporting wall loads). Is this correct? If so, are the spandrel beams in torsion for your load case you are looking at?
Assuming the beam yields, then you have a hinge (i.e. rotation) and an unstable system. This likely explains why the program reports large shear.
Imagine a beam simply supported at one end and apply a load at the other end. What do you have for a shear if you run the numbers by hand? The...
I am completely confused. Your last post talks about the wall loads at the end of the cantilever, but your third post from the start of this string states your are not concerned about gravity loads. Which is it?
Looking at the sketch you submitted for the case of a beam supported at both ends, it appears the moments at the ends are due to the fixity of the joints at the columns and the lateral deflection of the frame. The moments then result in equal and opposite shears at the beam ends.
For a...
Please clarify the situation you have:
1. is this a steel duct with stiffener "rings" spaced every so often (such as you might see at an industrial plant or power plant for exhaust gas/intake air)?
2. can you add a sketch?
"But the weld groups for the straps aren't even in the same plane as the bead that goes around the end cantilevered tube?"
The plates/straps and the top & bottom welds are in the same plane. Calculate the moment of inertia using the plate cross section and the top/bottom welds. You will then...
you will need to calculate the moment of inertia for the entire weld group using the parallel axis theorem. You will then us MC/I to calculate the weld stress (c= dist. from neutral axis to top (or bottom) weld).
If you are considering it as a truss, then CJ and EH are zero force members. Removing these from the analytical model then reveals CE and EI are zero force members. With those now removed, DI is also found to be a zero force member. For load applied to the top chords, then you have bending ad...
One approach is to calculate the maximum stress in the plate (assuming the assembly is uniformly loaded for this discussion). You can then determine the moment in the steel at this point (Moment x section modulus). Next you can calculate the uniform load transferred to the steel beam (Moment x...
chrislaope
I interpret the AWWA the same as the company that uses r=1.0. The bolt closest to the end of the plate has a force of 0.5T but the plate has a force of 1.0T; therefore, the r value is 1.0. It is a question of bolt force vs. plate force at the section considered.
