# Strength of materials question

## Homework Statement

A rigid steel plate is supported by three posts of high-strength concrete each having an effective cross-sectional area A =40000 mm2 and length L = 2 m. Before the load P is applied, the middle post is shorter than the others by an amount s = 1.0 mm.

Determine the maximum allowable load Pallow if the allowable compressive stress in the concrete is σallow = 18 MPa. (Use E = 30 GPa for concrete.)

## The Attempt at a Solution

Working aswell as diagram are shown in the pdf attached. Just looking for feedback on how I completed the question as it just seemed a little to easy for an assignment question?? cheers

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PhanthomJay
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## Homework Statement

A rigid steel plate is supported by three posts of high-strength concrete each having an effective cross-sectional area A =40000 mm2 and length L = 2 m. Before the load P is applied, the middle post is shorter than the others by an amount s = 1.0 mm.

Determine the maximum allowable load Pallow if the allowable compressive stress in the concrete is σallow = 18 MPa. (Use E = 30 GPa for concrete.)

## The Attempt at a Solution

Working aswell as diagram are shown in the pdf attached. Just looking for feedback on how I completed the question as it just seemed a little to easy for an assignment question?? cheers
Yes, not that easy. The middle column doesn't take nearly as much load as the outer ones, which have taken up most of the load before the middle column sees any load at all. The load is equally shared amongst the three only after the first 1 mm of deflection in the outer ones.

To start with, I do not feel that the assumption of a "rigid" steel plate is realistic. In structural analysis, everything deforms under load, and that is the basis of inderterminate structures like this present one.

However, since no physical information is available for the steel plate, we will have to live with the assumption of the plate being rigid.

The loading of the three columns progresses in two stages, with respective load capacities of F1 and F2.

F1 represents the total force required to deform the two outer columns to a deformation of 1 mm, as you have correctly calculated.
F2 represents the additional load shared by all three columns until one or more columns bust under σallow = 18 MPa. It is obvious that F2 is equally shared by all three columns because of the "rigid" assumption, and therefore the additional stress capacity of
σ = 18 MPa - 15 MPa = 3 Mpa.

If you do the detailed calculations, you will find that F1+F2 = 1.56 Mpa.

Post your calculations if you need more details.

Hi mathmate, following your logic (which seems very reasonable), I get F1 = 1200kN and F2 = 120kN, therefore F1 + F2 = 1320kN

Thank you ;)

Sorry, I have slipped the wrong unit. I meant 1.56 MN, or in your notation, 1560 kN.

The way I looked at it, when the two outer columns were supporting the load,the total load that creaed the 1 mm shortening is:
0.04 m2 * 2 columns * 15 Mpa = 1200 kN.

After the middle column is engaged, the outer columns have a remaining capacity of 3 Mpa before failure, so the additional load would be:
0.04 m2 * 3 columns * 3 Mpa = 360 kN.

Let me know if I slipped somewhere in my calculations, which tends to happen if I need to use the calculator.