# 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
Commented May 5, 2020 at 14:28
• Thank you I appreciate your help Commented May 5, 2020 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. Commented May 7, 2020 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
Commented May 5, 2020 at 18:58
• Ok i look at it. Commented May 5, 2020 at 19:00
• Just need to double it since you have the tension coming from both sides of the truss.
– Wasabi
Commented May 5, 2020 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. Commented May 7, 2020 at 23:59
• @Manu G, off the top of my head no. I am sure if you Google it you find a lot. Commented May 8, 2020 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
Commented May 5, 2020 at 18:22
• Gotcha, seems like I messed up a little there. Commented May 5, 2020 at 18:25
• both of you are a little up mistaken the answer is actually compressed Commented May 5, 2020 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
Commented May 5, 2020 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. Commented May 6, 2020 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. Commented May 7, 2020 at 20:19