
MECHANICS  EXAMPLE

Support Structure


Example


A mechanical equipment support structure is made from three metal members, each 10 mm by 10 mm. The vertical members, CA and DB, are fixed to the ground. The horizontal member, AB, is pinned to the vertical members at A and B. If the equipment is expected to placed 10 cm from the the left, as shown, what is the maximum load F? Assume all columns are solid steel bars with a stiffness, E, of 200 GPa and a allowable stress of 400 MPa. 




Solution

Basic W8x31 IBeam 

With the slender vertical columns, 10 mm by 10 mm, it would be expect this structure to fail due to column buckling. However, it could also fail due to beam bending or column crushing. These other possible failure modes will be checked after load P is determined for just column buckling.
Both columns, CA and DB, can be considered as fixedpinned columns that have a critical buckling as
P_{cr} = 2.046 π^{2}EI / L^{2}
Moment of inertia for both columns is,
I
= ab^{3 }/ 12 = (0.01)(0.01)^{3} / 12 = 8.333x10^{10} m^{4}
The allowable buckling load for each column is
P_{CA} = 2.046 π^{2} (200x10^{9}) (8.333x10^{10})/0.25^{2}
=
53.85 kN
P_{BD} = 2.046 π^{2} (200x10^{9}) (8.333x10^{10})/0.35^{2}
= 27.47 kN 



Top Beam Section 

Now, the critical buckling loads can be used to find F. First, the P_{CA} will be used to find F, and then P_{BD}. Taking moments about A gives,
(10)F
= P_{BD}(30) = 27.47(30)
P = 82.41 kN
Next, using P_{BD} and taking moments about B gives,
(20)F = P_{AC}(30) = 53.85(30)
F = 80.78 kN
Thus the largest load F before either column will be buckle when F = 80.78 kN. 



Maximum Crushing Load, 40 kN,
in Column CA 

Check column normal stress (crushing stress):
σ = P/A = 80.78/0.01^{2} = 807 MPa
There is a problem. This is higher than the allowable of 400 MPa. It will fail due to crushing before buckling. The maximum column load is,
P = (400 MPa) (0.01 m) (0.01 m) = 40 kN
This means the largest F (moments about point B) will be
20(F) = 40(30)
F = 60 kN 



Maximum Bending Moment 

Check beam bending stress:
The largest bending moment due to a load of F = 60 kN (maximum due to column crushing) will be at the location of the load.
M_{max} = 40(10) = 400 kNm
The bending stress will be
σ = My/I
= (400 kNm) (0.005 m) / 8.333x10^{10} m^{4}
= 2,400,000 MPa






There is another serious problem. This is much higher than the allowable of 400 MPa. It will fail due to beam bending before buckling or column crushing. The maximum moment is,
400 MPa = M (0.005 m) / 8.333x10^{10} m^{4}
M = 66.66 Nm
This moment will be generated a reaction at A of
P_{A} = (66.66 Nm)/ (0.1 m) = 666.66 N
Thus, the maximum load F will be
F (20) = 666.6 (30)
F = 999.9 N = 1.0 kN
Thus, the structure will fail due to beam bending before column buckling or column crushing. It is important to check all failure modes. 



