I know this is a long one but I really need help on this badly. If you are not w
ID: 1710908 • Letter: I
Question
I know this is a long one but I really need help on this badly. If you are not willing to work out the entire problem in detail please do not attempt to answer it. HELP!!!
The pin connected planar truss shown below is one of several used to support a sign and is to be constructed of three steel two-force members to support a constant vertical (weight) load of 1.00 kip at joint B and a variable horizontal (wind) load of 0.300 kip at joint B. The location of joints A and B are fixed, but the vertical position of joint C can be changed by varying the lengths of members AC and BC. Joint A is constrained in both the horizontal and vertical directions. Joint C is only constrained in the horizontal direction. For the steel truss members, the allowable stress in tension () is 20.0 ksi and the allowable stress in compression (Oe) is 12.0 ksi. The weight density of the steel is 0.284 pounds/inch3 You are to design a truss of minimum weight with the length of member AC varying from 18 to 50 inches and each member of the truss must have a cross-sectional area sufficient to meet the strength requirements above. Consider the ends of the bars to be reinforced to eliminate stress concentrations but do not consider the weight of the reinforcement. Do not consider the possibility of any members buckling What value of L (i.e. the length of member AC to the closest 0.1 in) provides the minimum weight truss and what is the weight of the truss? You may use Excel, MATLAB, or Mathcad. The program is to be written such that inputs for (G), (Oe), Pi, and P2 can be changed by only changing one cell for each input. Your submission should include, in report form, the necessary Free Body Diagrams, each of the equations used (written using the Microsof equation tool included in Word or equivalent), a printout of the program, and a graph of the truss weight (y-axis) vs. the length of member AC (x-axis) on a separate page. Additionally, the program is to be submitted via e- 30.00" 16.00" L1 P2-+/-0.30 kip PI = 1.00 kip NOT TO SCALEExplanation / Answer
First of all we have to understand that since the joint A and B both are constrained in both horizontal and vertical direction therefore joint C can not move because it is also constrained in the horizontal direction. Now we have to Find out the value of reactions for that we have to apply the equilibrium conditions. But before that We will resolve the joint B. (I am assuming the direction of wind toward negative direction and if the length of AC is 16 then BC become horizontal )
Now we can calculate the value of TAC and TBC.
Now if we apply the equilibrium condition on complete truss then AY = 1.00 Kip (because only these two are vertical forces).
Now resolve joint A.
Will give us the value of Ax , TAC
Hence we can found the C from equilibrium condition i.e. C - Ax = 0.3 Kip
At last we resolve the joint C and find the remaining forces.
Once we have all the values we will check whether all the values are under permissible limit or not.
If forces in any member exceeds the allowable limit then truss get fail.
Similarly we will design it for other value of L1 like 17,18,19,20 and find the value of length when truss will not fail for minimum length.
Because minimum length will have minimum weight.
We know the weight per unit length so once we have the value of length we can find the weight of truss.