Answer:
The value of Modulus of elasticity E = 85.33 × [tex]10^{6}[/tex] [tex]\frac{lbm}{in^{2} }[/tex]
Beam deflection is = 0.15 in
Explanation:
Given data
width = 5 in
Length = 60 in
Mass of the person = 125 lb
Load = 125 × 32 = 4000[tex]\frac{ft lbm}{s^{2} }[/tex]
We know that moment of inertia is given as
[tex]I = \frac{bt^{3} }{12}[/tex]
[tex]I = \frac{5 (1.5^{3} )}{12}[/tex]
I = 1.40625 [tex]in^{4}[/tex]
Deflection = 0.15 in
We know that deflection of the beam in this case is given as
Δ = [tex]\frac{PL^{3} }{48EI}[/tex]
[tex]0.15 = \frac{4000(60)^{3} }{48 E (1.40625)}[/tex]
E = 85.33 × [tex]10^{6}[/tex] [tex]\frac{lbm}{in^{2} }[/tex]
This is the value of Modulus of elasticity.
Beam deflection is = 0.15 in
