A bar has a length of 8 in. and cross-sectional area of 12 in2. Part A Determine the modulus of elasticity of the material if it is subjected to an axial tensile load of 10 kip and stretches 0.005 in . The material has linear-elastic behavior. Express your answer to three significant figures and include the appropriate un

The modulus of elasticity of the material is 3.82 N/m^2

Explanation:

Modulus of elasticity = stress/strain

Stress = load/area

Load = 10 kip = 10×4448.2216 = 44482.216 N

Area = 12 in^2 = 12 in^2 × (1 m/0.0254 in)^2 = 18600 m^2

Stress = 44482.216/18600 = 2.39 N/m^2

Strain = extension/length = 0.005 in/8 in = 6.25×10^-4

Modulus of elasticity = 2.39/6.25×10^-4 = 3.82 N/m^2

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )

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Explanation: