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.

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    – Wasabi
    Commented Dec 18, 2020 at 19:21

2 Answers 2


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).


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.


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:

enter image description here

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


  • 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

enter image description here

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.

The lower load between the two is the allowable load.


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