# 2D- Truss Member Analysis

I am a graduate student trying to sharpen the direct stiffness method skills. However, I am stuck at the beginning.

The problem is given in the image.

Can someone help me to write down all the nodal and member forces with equations ? Because the specific value are not given, I will give a general overview of the method. When applying the direct stiffness method for trusses, the following steps must be used:

Find the local stiffness matrix for each individual element as following:

$$\begin{bmatrix}AE/L & 0 & -AE/L & 0 \\0 & 0 & 0 & 0\\-AE/L & 0 & AE/L & 0 \\0 & 0 &0 &0\end{bmatrix}$$

You also need a beta matrix in order to convert the local coordinate system into global coordinates.

$$\beta = \begin{bmatrix} cos\theta & sin\theta & 0 & 0 \\-sin\theta & cos\theta & 0 & 0\\0 & 0 & cos\theta & sin\theta \\0 & 0 &-sin\theta & cos\theta\end{bmatrix}$$

Then, $$[\beta]^T[k_l][\beta] = [k_g]$$

Complete the same process for all matrices and then assemble global matrices to create the system matrix [K]. Apply boundary conditions for the problem.

Then solve for the nodal displacements by using the equation, $$[K]^{-1}[F] = [U]$$

Where F is the forces at the joints and U is the nodal displacements.

Expand [U] into elemental displacements $$[U_{1,2,3}]$$

Multiply each by their own original beta matrices in order to produce $$[\delta_{1,2,3}]$$

Multiply each original stiffness matrices by their delta values like so: $$[k_l][\delta_{1,2,3}] = [F_{e 1,2,3,}]$$

This will produce the elemental forces.

Let's call the length of square truss members L then the diagonal angle is 45 degrees (let me know if the diagram is not to scale).

$$\Sigma M_1= -LF_{4,2}+ LF_{4,3}=0 \quad F_{4,2}=F_{4,3} \ pointing\ left \\ F_{3}= -F \ pointing \ right \quad member_{ 3,4}= \ in\ tension\quad R_3= F \ pionting\ left$$

$$\Sigma F_{v1}=0 \quad F_{v1}= F\ pointing\ down \ R_{v1}=F\ pointing\ up$$

$$\Sigma F_{h1}=0 \quad F_{h1}= F\ pointing \ left \quad R_{h1}=F\ pointing\ right \\ member \ 1,4 \ in \ compression \\ F_{1,4}= \sqrt{2}*F$$

Members 1,2 and 2,4 are zero stressed.

• I cannot make sense of your answer. You take a moment sum at $M1$ but how do you multiply $F_{4,2}$ with $L$ it is parallel to the $F$ direction. Besides, how do you come up with members that are zero stressed ? Also, the solution says $F_{1,4}$ must be $\sqrt 2*F$ Oct 21, 2019 at 6:39
• @Yirmidokuz If point 2 is a pin joint, members $F_{12}$ and $F_{24}$ must both have zero force because the two members are at right angles to each other. I don't know how Kamran got the wrong value for $F_{14}$, but you can find $F_{14}$ and $F_{34}$ from equilibrium at point 4. Oct 21, 2019 at 8:47
• @alephzero can you please more elaborate? Why $F_{12}$ and $F_{24}$ must both be zero when they are right angle to each other ? Oct 21, 2019 at 9:01
• @Yirmidokuz, you are right. I corrected my arithmetic error on F14. I got it by just vector addition of forces at node 4. I multiplied F42 by L because we assumed the length=L and moment of a force about a point is F*distance. Whythe other to members are zero? because they are just a hinge mechanism. angle 421 can change offering no resistance. Oct 21, 2019 at 15:32
• @Yirmidokuz, take a look at node #2. If member F42 had carried a force, nothing would be able to resist it since the only other member is F12 which is connected perpendicularly to F42. Write down the equations of this node to realize. Oct 21, 2019 at 18:23