Formwork for Concrete Structures by R.L.Peurifoy and G.D- By EasyEngineering.net

114 Chapter Five
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For this beam the physical properties are
l= 96 in.
E= 1,600,000 lb per sq in.
I= 20.8 in. 4
Using Beam 4 in Table 5-2 to calculate the deflection from the two
80-lb loads:
∆=Pa [3(l 2 )/4 – (a) 2 ]/6EI
= 80(9)[3(96) 2 /4 – (9) 2 ]/6(1,600,000)(20.8)
= 0.0245 in.
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Using Beam 4 in Table 5-2 to calculate the deflection from the two
120-lb loads:
∆=Pa [3(l 2 )/4 – (a) 2 ]/6EI
= 120(35)[3(96) 2 /4 – (35) 2 ]/6(1,600,000)(20.8)
= 0.1196 in.
By superposition, the total deflection is obtained by adding the
two deflections as follows:
Total deflection = deflection from 80-lb loads + deflection from
120-lb loads
Example 5-11
∆= 0.0245 in. + 0.1196 in.
= 0.144 in.
The beam shown below has both a uniformly distributed load of
150 lb per lin ft and two 200-lb concentrated loads. Use the method of
superposition to calculate the total deflection of the beam.
For this beam the physical properties are
l = 108 in.
E = 1,600,000 lb per sq in.
I = 79.39 in. 4
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