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 ?
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2 Answers
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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.
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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.
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$\begingroup$ 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 $\sqrt 2*F$ $\endgroup$ Oct 21, 2019 at 6:39
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$\begingroup$ @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. $\endgroup$ Oct 21, 2019 at 8:47
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$\begingroup$ @alephzero can you please more elaborate? Why $F_{12}$ and $F_{24}$ must both be zero when they are right angle to each other ? $\endgroup$ Oct 21, 2019 at 9:01
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$\begingroup$ @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. $\endgroup$– kamranOct 21, 2019 at 15:32
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$\begingroup$ @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. $\endgroup$ Oct 21, 2019 at 18:23