Statics Question

I have shared the question statement, its diagram and my solution but my solution is wrong and I have no idea why is it wrong, please help me out.

• I never thought in any of my wild dreams to help DemonLordKing to solve his static homework but here you go : Instead of setting the maximum values, try to find the forces in any member in function of P, then fill the maximum allowable stresses in the most critic member. If i calculate everything correctly then it should be something around 5.19 Nov 17 '18 at 14:16
• Thanks a lot, and its not my homework, im solving questions on my own to practice and learn the concepts, and I understood that we have to take the members in function of P but I need to understand why my solution is wrong? Nov 18 '18 at 10:40
• Suppose the maximum stress is $8$ kN in DC, then it implies the stress of say $\sigma$ > $8$kN in the CE if recall correctly or some other member, that's why your solution is wrong, the real stress in DC is not $8$ kN or $6$kN. Your assumptions are wrong. Nov 18 '18 at 11:44

you should think first about the whole structure, and then look into each joint. Apply the sum of the forces on the horizontal, the sum of the momentum and vertical forces must all be equal to zero to the structure as an object, similar to what you have been doing all the course.

I have found a mistake on your solution since you didn't consider the first gender support in $$E$$. Correct that and try to express, in the equations of ballance, expressions for the reactions relative to $$P$$ from the equations of equilibrium of the whole structure. This will result in a problem which will be in function of $$P$$. In this step $$R_a$$ and $$R_e$$ will be, if my calculations are correct, $$3P$$ and $$2P$$ in modulo, respectively.

Then, and only then, go to the joints and do their ballance of equilibrium. As it is all in function of $$P$$ you then separate those suffering compression from tension. Among those suffering compression, which is taking the higher force? Equal it to $$6$$kN. Among those suffering tension, which is taking the higher force? Equal it to $$8$$kN. The smallest of the two is the correct value of $$P$$, since the highest will cause a violation in the other restriction.

• The reactions will actually be $P$ and $-2P$, but otherwise good answer.
– Wasabi
Nov 18 '18 at 19:27

Using method of joints we can calculate the maximum compression according to the values given being 6kN. As that strain is greater than 8kN we should assume 6kN to be achieved first.

Force P should not exceed the maximum compression value of 6kN at the vertical component of the truss CD.

$$6 \sin 60=3 \sqrt {3}=5.196kN$$

There is no way that tension can even come close to 8kN in the truss without exceeding the 6kN compression.

• No, you can't assume that. You must calculate the tension in terms of P that each link can sustain. You can make deductions about pairs of links from symmetry conditions that reduce the number of calculations. For instance, the reaction at A must be P, and the reaction at E must be -2P. Then we can use bilateral symmetry to reduce the number of links from 7 to 4, and simplify the vertex formula to trivial equations if taken in a sensible order. Jan 7 '19 at 22:30
• I checked it with computer software designed for this problem and it is accurate. Jan 7 '19 at 23:25
• But by changing the angles a little, it wouldn't be. You can have a quite similar looking truss that would be limited by tension. The fact that you guessed correctly isn't the point. Your reasoning is faulty. You must do the calcs and apply reason. That is what they are trying to teach here. Jan 7 '19 at 23:33
• 6sin90 is the max vertical force allowable which means that F_p would be equal but opposite and therefore 6kN whilst the horizontal members would be zero length. Jan 7 '19 at 23:38