New answers tagged statics
2
Yes, it's a determinate structure and you can solve for T4 and T5 using statics.
Method of Joints
One approach is to use Method of Joints to progressively solve for the member forces, and it looks like this is the approach you've chosen. With two equations of equilibrium ($\sum F_x = 0$, $\sum F_y = 0$) we can solve for two unknown member forces at any ...
0
At balance you will have the situation:
$ f\times cos(a)\times r = F \times R $
To get the bridge to lift you will need f greater than the balance value.
Note that with the bridge pivot as drawn you can't get the bridge fully up and if you lift it fully up the actuator can't pull it down as it is past "top dead-centre". Top dead-centre will occur when the ...
1
Six seems to be the correct answer:
3 translational DoFs $\left(u_x\right.$, $u_y$, $\left.u_z\right)$ and 3 rotational DoFs $\left(\phi_x\right.$, $\phi_y$, $\left.\phi_z\right)$
You can even have a 3D beam element with 7 degrees of freedom. The 7th is for warping
-3
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
2
Because of symmetry we can cut the truss from the center and assign half of 123 kN to each half.
$$ \Sigma M_a=0 \quad 123*7.5+ 123*15+123*22.5+ 123/2*30- F_{H, \ Horizontal}*10.8=0$$
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$$ F_{H Horizontal}= 7380/10.8 =683kN $$
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$$ F_{HF}= 683sec(\frac{10.8-10.5}{7.5} )=683* sec(0.000698)= 683*1.0000000001=683000000051kN \ copression$$
** Edit**
A ...
1
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 ...
0
One easy way of drawing the shear and moment diagram is to separate the loading, draw the diagrams, and then superpose them.
Let's call the uniformly distributed load W1 and the triangular load W2.
In the diagram, this load and its shear and moment are shown in blue color.
The top part is the W1 loading and its shear is shown as a blue rectangular on the ...
3
When you've done an exercise and got the wrong answer, it's always useful to check to see if your result ever passed the "smell test". That is, does your result make much sense.
Now, we can see a few strange things from a quick glance.
The biggest thing which should call our attention is your moment diagram. It starts at 0 at the support and ends at 128 at ...
1
If your beam cross section is symmetrical about Y axis that same axis is the vertical neutral axis.
If not, say your beam cross section is a T or z or just a random shape the definition of of neutral axis is were the first area moment about it taken on top and bottom are equal.
There are several ways of finding the neutral axis of a random shape.
One is ...
0
You can't set the moment to zero unless the body isn't constrained in its movement. If you could set any moment to zero that didn't move, statics wouldn't work.
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