Question

A beam has a rectangular cross section that is 5 inches wide and 1.5 inches tall. The supports are 60 inches apart and with a 125 lb person standing in the middle of the beam and it deflects .15 in.

Find the modulus of elasticity


Find the beam deflection

Answer 1

The value of Modulus of elasticity E = 85.33 ×
`10^(6)`
`(lbm)/(in^(2) )`

Beam deflection is = 0.15 in

Step-by-step explanation:

Given data

width = 5 in

Length = 60 in

Mass of the person = 125 lb

Load = 125 × 32 = 4000
`(ft lbm)/(s^(2) )`

We know that moment of inertia is given as

`I = (bt^(3) )/(12)`

`I = (5 (1.5^(3) ))/(12)`

I = 1.40625
`in^(4)`

Deflection = 0.15 in

We know that deflection of the beam in this case is given as

Δ =
`(PL^(3) )/(48EI)`

`0.15 = (4000(60)^(3) )/(48 E (1.40625))`

E = 85.33 ×
`10^(6)`
`(lbm)/(in^(2) )`

This is the value of Modulus of elasticity.

Beam deflection is = 0.15 in