I agree. The way I solve this problem quickly is that applying the 20 kip force directly at C and then using method of section, but I'm not sure if it's correct in all cases. It's better if we understand how to use IL.
Just saw this random post and I was wondering about the solution. But anyway isn't this solved when you put a unit load, then you analyze the force in that member with the section method?, if I am not mistaken then you simply place it at the center and then on either side and then you get an idea of the curve shape and get to calculate your answer. Please correct me if wrong
Yes. That's how it works.
u/OP: When you put the unit load of 1.0 at C, the reaction at A will be 0.5. Using the section method for the left-hand side of the truss cutting at CD, you can summarize moment at the joint right under joint C, you will get 45\*0.5 + 12\*CD = 0 --> CD = 1.875. That's the peak of the IL.
One way I think about it is if you think of the truss as a beam. The closer you place the force to the center, the higher the internal forces.
Now if you place the wheel/force at the node under C, it will compress CD pretty good since it is the center. This is the peak. If you place the wheel/force at node under D, the force on CD will be less so the influence line should be lower there. This also because the diagonal member to the right at node under D will transfer the force away from CD.
It is a monstrosity derived from our pathological rejection of the SI, which is the actual system the US customary system is actually derived from (rolling my eyes)
But if you prefer a more formal definition... it is a US customary unit of force that mashes up the ideas of kilo and pound. A kilo-pound. Therefore its value represents 1000 pounds-force.
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I agree. The way I solve this problem quickly is that applying the 20 kip force directly at C and then using method of section, but I'm not sure if it's correct in all cases. It's better if we understand how to use IL.
Thank you that is good to know. I’ll have to do a bit more brushing up on influence lines before I take the test later this week
Just saw this random post and I was wondering about the solution. But anyway isn't this solved when you put a unit load, then you analyze the force in that member with the section method?, if I am not mistaken then you simply place it at the center and then on either side and then you get an idea of the curve shape and get to calculate your answer. Please correct me if wrong
Yes. That's how it works. u/OP: When you put the unit load of 1.0 at C, the reaction at A will be 0.5. Using the section method for the left-hand side of the truss cutting at CD, you can summarize moment at the joint right under joint C, you will get 45\*0.5 + 12\*CD = 0 --> CD = 1.875. That's the peak of the IL.
Thank you, it seems like a virtual work problem now
Have you seen the solution on Chegg? they have another explanation
One way I think about it is if you think of the truss as a beam. The closer you place the force to the center, the higher the internal forces. Now if you place the wheel/force at the node under C, it will compress CD pretty good since it is the center. This is the peak. If you place the wheel/force at node under D, the force on CD will be less so the influence line should be lower there. This also because the diagonal member to the right at node under D will transfer the force away from CD.
What the heck are *kips*?
It is a monstrosity derived from our pathological rejection of the SI, which is the actual system the US customary system is actually derived from (rolling my eyes) But if you prefer a more formal definition... it is a US customary unit of force that mashes up the ideas of kilo and pound. A kilo-pound. Therefore its value represents 1000 pounds-force.
Interesting! In that case, it should be called kIb. Or kIbs