# I can't even see how can i solve this truss question

I can't even see how can I solve this truss question I'm an engineering student, our professor assigned us this truss problem. I have found an answer saying that it is a zero member.

I have tried but I couldn't even see where should I start from.

i am working on this problem but i couldn't find any value once i reach a joint with 2 unknowns

like this joint for example [B] []3

i have even used the momentum equation but i can't go anymore

• I can't give a full answer right now, but I can tell you HI isn't a zero-force member. It would be if the top chord (FH and its mirror) were horizontal, but they're not.
– Wasabi
May 5 '20 at 14:28
• Thank you I appreciate your help May 5 '20 at 15:24
• Seriously it's much easier to put the fraction values instead of the cosine or sine values. I don't understand the need to calculate the angles. On the side note, check out my answer. May 7 '20 at 20:26

Because of symmetry we can cut the truss from the center and assign half of 123 kN to each half.

$$\Sigma M_a=0 \quad 123*7.5+ 123*15+123*22.5+ 123/2*30- F_{H, \ Horizontal}*10.8=0$$

.

$$F_{H Horizontal}= 7380/10.8 =683kN$$

.

$$F_{HF}= 683sec(\frac{10.8-10.5}{7.5} )=683* sec(0.000698)= 683*1.0000000001=683000000051kN \ copression$$

** Edit**

A quick inspection of the very small cord angle at FH shows the force in IH is very small tension and is negligible.

$$F_{IH}= 683*0.3/7.5=27.3kN 27.3*2 = 54.6kN \quad \text{for both sides}$$

• The question is asking for the force in HI, not HF. Also, I made a simple model on Ftool and got 683.9kN for HF.
– Wasabi
May 5 '20 at 18:58
• Ok i look at it. May 5 '20 at 19:00
• Just need to double it since you have the tension coming from both sides of the truss.
– Wasabi
May 5 '20 at 20:03
• @kamran Can you point me to some materials about the symmetry thing (giving half of 123 kN to each side) that was done in the beginning. May 7 '20 at 23:59
• @Manu G, off the top of my head no. I am sure if you Google it you find a lot. May 8 '20 at 0:09

Step 1: Solve for the reactions at the joints.

Step 2: Take the section across FI, and then solve for the three unknowns.

Step 3: Finally take the FBD at joint H and then solve for the values.

The answer I got seems to put HI in tension at 54.67 kN.

EDIT 1: Mistake in support reaction (@Wasabi thanks for pointing it out)

EDIT 2: Didn't put the external forces when I took the section across FI

P.S. @Wasabi thanks for pointing out the problem. Finally fixed the issue.

• I'd suggest checking your work. Using a program, I go a different result (IH = 54.7kN in tension). One thing that caught my eye is your support reaction doesn't seem right: symmetry demands that the supports have equal vertical reactions, therefore it should be simple to obtain reactions of $7\times123/2 = 430.5\text{ kN}$.
– Wasabi
May 5 '20 at 18:22
• Gotcha, seems like I messed up a little there. May 5 '20 at 18:25
• both of you are a little up mistaken the answer is actually compressed May 5 '20 at 18:50
• @YyyUuu: I can guarantee to you that it isn't in compression. If your book says it is, it's wrong.
– Wasabi
May 5 '20 at 20:43
• @Wasabi Took a little time to get around the interface. But found it much better than the other software that I find when I just Google for Truss Solvers. It's a really good software. Kudos to making it and allowing it free access. May 6 '20 at 20:27

https://mathalino.com/reviewer/engineering-mechanics/problem-430-parker-truss-method-sections

i have the found the answer here

it is a similar question and it is solved in details

good luck

also take a look on the simple model made by wasabi it will help you for sure

he used ftool for it

• Welcome to Engineering! Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference. May 7 '20 at 20:19