Answer:
The elastic modulus of the steel is 139062.5 N/in^2
Explanation:
Elastic modulus = stress ÷ strain
Load = 89,000 N
Area of square cross section of the steel bar = (0.8 in)^2 = 0.64 in^2
Stress = load/area = 89,000/0.64 = 139.0625 N/in^2
Length of steel bar = 4 in
Extension = 4×10^-3 in
Strain = extension/length = 4×10^-3/4 = 1×10^-3
Elastic modulus = 139.0625 N/in^2 ÷ 1×10^-3 = 139062.5 N/in^2
Answer:
they are used for electrical currents so that they can flow along the appropriate wires in the circuit
Explanation:
Answer:
2.8
Explanation:
The ideal mechanical advantage of the pulley IMA = D'/D where D' = diameter of output pulley = 7 inches and D = diameter of input pulley = 2.5 inches
So, IMA = D'/D
= 7/2.5
= 2.8
So, the ideal mechanical advantage of the pulley IMA = 2.8
Answer:
The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm
Explanation:
For steel bolt
Stress = 210 MPa or 210 N/mm2
Pressure = Stress* Area
Pbolt = 210 N/mm2 * 16^2 *(pi)/4
Pbolt = 210 N/mm2 * 200.96 mm^2 = 42201.6 N
For Brass spacer
Pressure = 42201.6 N
Area of Brass spacer = Pressure/Stress
Area of Brass spacer = 42201.6 N/145 N/mm^2 = 291.044 mm^2
Area of Brass spacer = (pi) (d^2 - 16^2)/4 = 291.044 mm^2
d^2 - 16^2 = 291.044 mm^2* 4/(pi) = 370.758
d^2 = 370.758 + 16^2
d^2 = 626.758
d = 25.03 mm
The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm
Answer:
The value of Modulus of elasticity E = 85.33 × 
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
We know that moment of inertia is given as
I = 1.40625 
Deflection = 0.15 in
We know that deflection of the beam in this case is given as
Δ = 
E = 85.33 ×
This is the value of Modulus of elasticity.
Beam deflection is = 0.15 in
