I have the following problem
E= 210 MPa d = 40 mm n = 20 y = 1.5 a =1 calculate the maximum allowed load for the system

I am confused on how to solve the problem, I really need advice.

• Please add to the question what you have tried so far to solve this problem and at which point you struggle. Dec 18 '20 at 16:06
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– Wasabi
Dec 18 '20 at 19:21

It is very important that this is a system of 5 elastic bars. Although its not explicitly stated I will assume that they are connected with pins (ie. each bar only carries an axial force).

# forces

To solve this, First you need to calculate the load carried by the rod in compression ($$\sqrt{2}a$$).

In order to so, you need you need to first calculate the forces on the short rods (length a). Due to the symmetry all rods with length $$a$$, carry the same load. That load can be easily calculated as (all the angles are 45 deg) :

$$P_a = \frac{\sqrt{2}}{2} P$$

From the equilibrium of forces on the y direction, on the node at one end of the beam, you have :

$$\sum F_y=0 \rightarrow P_{\sqrt{2}a} + P_a\frac{\sqrt{2}}{2}+ P_a\frac{\sqrt{2}}{2}=0$$ $$P_{\sqrt{2}a} = - 2 P_a\frac{\sqrt{2}}{2}= -\sqrt{2} P_a$$ $$P_{\sqrt{2}a} = - P$$

The minus sign is for compression.

# Buckling

Now you have the load on the beam that is compressed. So now you can calculate the critical buckling load. Given the assumption above about a pin structure you can use for the following structure the formula:

$$P_cr = \left(\frac{n\pi}{l}\right)^2 EI$$:

where:

• n = 1 (mode of buckling)
• $$l = \sqrt{2}a$$ : the length of the rod in compression
• $$E = 200 GPa$$ : the modulus of elasticity
• $$I = \frac{\pi d^4}{64}= \frac{\pi (40mm)^4}{64}$$

A quick note here: If you can't assume that you have a pinned structure you should use one the factor for effective length below

Since you want a safety factor $$n_b$$, what you need:

$$n_b = \frac{P_cr}{P_{\sqrt{2}a}} \rightarrow P_{\sqrt{2}a}= \frac{1}{n_b} \left(\frac{n\pi}{l}\right)^2 EI$$

# plastic yield.

The member which might yield first is the one with the highest load i.e. $$P_{\sqrt{2}a} = - P$$.

As a result the normal stress on the member would be:

$$\sigma_{\sqrt{2}a,y} = \frac{P_{\sqrt{2}a,y} }{A}= \frac{-P}{\pi \frac{d^2}{4}}$$

However, since you have an safety factor for yield:

$$n_y \le =\frac{\sigma_y}{\sigma_{\sqrt{2}a,y} }=\frac{\sigma_y}{\frac{P}{\pi \frac{d^2}{4}} }=\frac{\sigma_y \cdot \pi d^2}{4P }$$

# Bottom line

Both cases above are now in a form, which is a function of P. And you can estimate a maximum allowable load for buckling $$P_b$$, and one for plastic yield $$P_y$$.

You should select the minimum of those two.

$$P_{allowable} = min (P_b, P_y)$$

The 1m "a" bars each carry a tension load of $$F=\sqrt2*P/2 \$$and they have a vertical component along the vertical bar equal to $$F=P/2\$$compression. The horizontal reactions are canceled by the left side reactions.

Edit. [To clarify how we get the P in vertical bars. The two vertical right-hand components cancel each other and leave the vertical bar with P/2 compression. The same thing applies to the bars on the left-hand side, again leaving the vertical bar with another P/2 compression and canceling the right-side horizontal components of reaction P.]

So the vertical bar is under $$F=P/2+P/2= P \$$ compression

we check for two cases, buckling and yield. And we assume the $$E_{steel}= 210GPa.$$

$$I_{cylinder}=1/4 \pi R^4$$

$$P_{cr} = { \pi^2 \, E \, I \over L^2 }$$

$$P_{cr} = { \pi^2 \, 210GPa \, 1/4 \pi R^4\over L^2 }$$

We just plug the R=0.02m and L=1.41m then and multiply by 1/n to get the allowable buckling load.

For allowable compression load, we just figure axial stress.

$$\sigma=P/A{bar} \quad 240MPa=P/ \pi*0.02^2$$

and then multiply by1/1.5 SF.