A titanium alloy bar elongates 0.500 in. when subjected to an axial tensile load of 100 k. The bar is 16 ft long and has a cross-sectional area of 2.25 in2. Compute the modulus of elasticity E. Use a proportional limit of 125 ksi.

E= 17066.67 k per inch square

Explanation:

given  that

Si = 0.5 in,

P= 100k

l = 16ft = 192in

A= 2.25 in²

rp = 125ksi

recalll that strain

E = EL/L

= 0.5/192

= 0.0026

also, stress

r = P/A

=100/2.25

= 44.44ksi

since the stress(r)  is less than the proportionality limit

(rp) , Hence the stress will be proportional to  strain, from hook;s law and written as

 T∝ E

T =E∑

therefore,

Elastic modulus E= (P/A)/(d/l)

E= (100/2.25)/(0.5/192)

E= 17066.67 k per inch square

answer:

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Explanation:

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