Respuesta :
Answer:3142 Pound force
Explanation:
Given
Length(L)=16 in.
diameter=0.5 in.
Modulus of elasticity (E)[tex]=6.4\times 10^6 psi.[/tex]
Allowable stress [tex]\left ( \sigma _{allowable}\right )=17000 psi[/tex]
Max elongation of the bar =0.04 in.
Also we Know
Elongation is given
[tex]\Delta L=\frac{PL}{AE}[/tex]
where
P=Force applied
L=length of bar
A=Cross-section
E=Modulus of Elasticity
[tex]A=\frac{\pi d^2}{4}=\frac{\pi}{16} in.^2[/tex]
[tex]0.04=\frac{P\times 16\times 16}{\pi \times 6.4\times 10^6}[/tex]
[tex]P=3142 pound\ force\approx 13.976 kN[/tex]
corresponding stress is [tex]\frac{3142}{\frac{\pi }{16}}=16,000 psi[/tex]
Which is less than Allowable stress
thus allowable value of Force P is 3142 Pound force or 13.976 kN
The allowable value of the forces P is; 3142 lb.f
How to find the force in a bar?
We are given;
Length; L = 16 in.
Diameter; d = 0.5 in.
Modulus of elasticity; E = 6.4 × 10⁶ psi
Allowable stress; σ_allow = 17000 psi
Max elongation of the bar; ΔL =0.04 in.
Formula for elongation is;
ΔL = PL/AE
A is area = πd²/4 = π × 0.5²/4 = π/16 in²
Making P the subject gives;
P = (ΔL * A * E)/L
P = (0.04 * (π/16) * 6.4 × 10⁶)/16
P = 3142 lb.f
Read more about force in stress at; https://brainly.com/question/14468674
