Question

A specimen of steel 20 mm diameter with a gauge length of 200 mm is tested to destruction. It has an extension of 0.25 mm under a load of 80 kn and the load at elastic limit is 102 kn. The maximum load is 130 kn.

Answer 1

Explanation:

A specimen of steel 20 mm diameter with a gauge length of 200 mm is tested for destruction. It has an

extension of 0.25 mm under a load of 80 kN and the load at the elastic limit is 102 kN. The maximum

load is 130 kN. The total extension at fracture is 56 mm and the diameter at the neck is 15 mm. Find

(i) The stress at the elastic limit. (ii) Young’s modulus. (iii) Percentage elongation. (iv) Percentage

reduction in area. (v) Ultimate tensile stress

Answer 2

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[A specimen of steel 20 mm diameter with a gauge length of 200 mm is tested to destruction. It has an extension of 0.25 mm under a load of 80 kN and the load at elastic limit is 102 kN. The maximum load is 130 kN.  The total extension at fracture is 56 mm and diameter at neck is 15 mm. Find  (i) The stress at elastic limit.  (ii) Young’s modulus.  (iii) Percentage elongation.  (iv) Percentage reduction in area.  (v) Ultimate tensile stress.]

Concept

Elastic limits, maximum stresses or forces per unit area in a solid material that can occur before permanent deformation begins. When the stress is removed to the elastic limit, the material returns to its original size and shape. Stresses that exceed the elastic limit cause the material to yield or flow.

Given

We have given the diameter d=20mm,load at elastic limit=102kn, pressure = 80kn ,gauge length of 200 mm , maximum load 130kn,increment in length =0.25mm

Find

We are asked to find (i) The stress at elastic limit.  (ii) Young’s modulus.  (iii) Percentage elongation.  (iv) Percentage reduction in area.  (v) Ultimate tensile stress

Solution

As the diameter is 20mm then, area will be = $\frac{\pi d^{2} }{4} =314.16mm^{2}$

(i) Stress at elastic limit =load at elastic limit/Area

$=\frac{102\times10^{3} }{314.16} \\\\=324.675N/mm^{2}$

(ii) Young’s modulus (E) = $\frac{stress}{strain}$  within elastic limit

$=\frac{P/A}{\triangle L/L} \\\\=\frac{80\times10^{3} /314.16}{0.25/200} \\\\=203718Nmm^{2}$

(iii) Percentage elongation = final extension/ original length

$\frac{56}{200} \times100=28$

(iv) Percentage reduction in area = (Initial area - Final area/Initial area) × 100

$=\frac{\frac{\pi }{4} \times 20^{2}-\frac{\pi }{4} \times 15^{2} }{\frac{\pi\times 20^{2} }{4} } \times100$

$=43.75$

(v) Ultimate tensile stress = Ultimate load/Area

$=\frac{130\times 10^{3} }{314.16} \\\\=413.80N/mm^{2}$

Hence, stress is $324.675N/mm^{2}$, Young's modulus is $203718Nmm^{2}$ ,Percentage elongation is $28$, percentage reduction in area is $43.75$, ultimate stress is $413.80N/mm^{2}$

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