MECHANICS  EXAMPLE



Example

Example Graphic 

Determine the safe uniformly varying load, W_{o }that can be applied to a simply supported span of 18 ft long. The crosssectional area of the beam is 10 in by 6 in and reinforced at the bottom by a 4 in by 0.5 in steel plate. The failure strength of steel and wood should not exceed 18 ksi and 1.2 ksi, respectively. The Young's Modulus of steel can be considered as 20 times of wood ( i.e. E_{s}/E_{w} = 20 ). 



Solution


Finding the location of the neutral axis (NA) is one of the first steps in solving bending stresses in a composite beam. Both the moment of inertia and the distance to the outside edges require knowing the location of the NA. 





The location of NA is generally found from the bottom of the beam (but could be from any reference point). Using the summation of the modulus times Q equation gives, 





0 = ΣEQ = E_{s}Q_{s} + E_{w}Q_{w
} 0 = ( E_{s}/E_{w })( y_{s}A_{s }) + ( E_{w}/E_{w})(y_{w}A_{w})
0 = 20 (0.25  h)(0.5)(4) + (5 + 0.5  h)(6)(10)
0 = 10  40h + 330  60h
h = 3.40 in
Now that the distance h is known, the distance to the center of each material to the NA can be found.
y_{w}= 5.50  3.40 = 2.10 in
y_{s}= 0.25  3.40 = 3.15 in
With the neutral axis (NA) located, the moment of inertia about the NA for each material can be calculated. Using the basic formula for the moment of inertia for rectangular areas and the parallel axis theorem, gives
I_{w}= h^{3} (b/12) + y_{w}^{2}A_{w}
= 10^{3} (6/12) + 2.1^{2} (6)(10)
= 764.6 in^{4}
I_{s} = h^{3 }(b/12 ) + y_{s}^{2}A_{s}
= 0.5^{3 }(4/12) + 3.15^{2} (4)(0.5)
= 19.89 in^{4}






The bending stress in each material areas can be determined by using the bending stress equations for a two material composite beam. There are two equations since each material has a different stress.
The y distance is the maximum distance from the neutral axis for each material section.
y_{smax} = h = 3.40 in
y_{wmax} = 10 + 0.5  h = 10 + 0.5  3.40 = 7.10 in
Substituting values in both the stress equations and using E_{s}/E_{w} = 20 gives
18,000 = M_{s} (3.40) / [764.6/20 + 19.89]
M_{s} = 307,700 lbin = 25,640 lbft
1,200 = M_{w} (7.10) / [764.6 + 20 (19.89)]
M_{w} = 196,500 lbin = 16,380 lbft 





Thus, when the moment reaches 16.38 kipft, the wood will fail in compression. So what load w will generate a moment of 16,380 lbft? To answer this, the moment equation needs to be determined in terms of w.
The reaction at A is
ΣM_{B} = 0 = 18 F_{A} + 6 [0.5 w 18]
F_{A} = 3w (assume all units are in ft and kip)
Next, find the moment equation
ΣM_{cut} = 0 = M_{1} + [0.5 x (wx/18)] (x/3)  3wx
M_{1} = 3wx  x^{3}w/108
Maximum moment will occur at,
dM_{1} / dx = 0 = 3w  x^{2}w/36
x = 10.39
Maximum moment in terms of w is
M_{max} = 20.78 w 





To generate a moment of 16,380 lbft (wood failure), the loading function w must be
16,380 = 20.78 w
w = 788.3 lb/ft 


