A rigid steel plate is supported by three posts of high

A rigid steel plate is supported by three posts of high

A rigid steel plate is supported by three posts of high-strength concrete each having an
effective cross-sectional area A=40,000 mm2 and Length L=2 m (see attachment). 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 P allow if the allowable compressive stress in
the concrete is equal to 20 MPa. (Use E=30 GPa for concrete)
Under the influence of P, S would move downwards, creating compressive stresses in
both the right and left columns. Suppose the maximum compressive stress of 20MPa is
realized before S touches the middle concrete bar.
Total area on which P is acting = 2*Cross-sectional area of one column = 2*40,000
= 80,000 mm2
Pmax = Smax*Area = 20*80,000 N = 1.6 MN
Under a maximum compressive stress of 20 MPa, and given Young’s Modulus 30GPa,
Strain created = Stress/Young’s Modulus = 20/30*103 = 6.67*10-4
But strain = Change in length/ Original length
 Change in length = Strain*Original Length = 2*6.67*10-4 = 0.001334m = 1.334 mm.
This indicates that our assumption that the maximum compressive stress of 20MPa is
realized before S touches the middle concrete bar is wrong, since the gap is only 1 mm.
Let us calculate the compressive stress created when S just touches the middle bar.
Here, strain = 0.001/2 = 0.0005
Stress = E*strain = 30*103*0.0005 = 15 MPa, which is created by a force of P =
Stress*Area = 15MPa*80,000*10-6 = 1.2 MN
We want to find out how much additional force would bring the compressive stress up
from 15MPa to 20MPa. Any further load would act on the cross sectional areas of all
there columns because all three are now touching S.
Cross sectional area=3* Cross-sectional area of one column = 3*40,000= 120,000 mm.
Additional P = Additional Stress*Area = 5MPa*120,000*10-6 = 0.6 MN
Therefore, maximum force allowable = Pmax = 1.2 MN + 0.6 MN = 1.8 MN