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Solid Mechanics - Mock Test 1

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Question 1

The ultimate stress, for a hollow steel column which carries an axial load of 1.9 MN is 480N/Sq(mm). If the external diameter of the column is 200mm. determine the internal diameter. Take factor of safety as 4?

permissible stress = ultimate stess / factor of safety = 480/4 = 120N/Sq(mm),,also stress= force/area ,,, hence 120N/Sq(mm) = 1.9 x 1000000 N/ area of hollow steel column,, hence ,, Area of column= 1.9 x 1000000/120 = 15833.33 Sq(mm) ,, also Area = 0.785(external diameter^2 - internal diameter^2),,,hence by comparing area we get,,,15833.33 = 0.785 x (200 x 200 - internal diameter^2),, hence by solving we get,, 20169.85 =40000 - internal diameter^2,, also Internal diameter = (40000-20169.85)^0.5 =140.82 mm

Question 2

The safe stress, for a Hollow steel column which carries an axial load of 2.1 x 10^3 kN is 125MN/sq(m). If the external diameter of the column is 30cm, determine the internal diameter?

given;- force = 2.1 x 10^3 kN = 2.1 x 10^6 N,,,,stress = 125 x 10^6 N/sq(m) =125 N/Sq(mm),,, now,, stress = force/Area,,, hence Area = Force/Stress = 2100000/125 =16800 Sq(mm),,, Also Area of Hollow cylinder = 0.785 x (external diameter^2 -internal diameter^2)= 0.785(30x 30 - Internal diameter^2),,,,hence by comparing Area = 16800= 0.785(300x300 - internal diameter^2),, hence,,, 90000- internal diameter^2 = 16800/0.785 =21401.274,,hence Internal diameter^2 = 90000-21401.274=68598.726 ,,hence internal diameter=(68598.726)^0.5 =261.91 mm = 26.2 cm

Question 3

A tensile test is conducted on a mild steel bar. The following data was obtained from the test : diameter of the steel bar= 3cm, Gauge Length of the bar = 20cm, Load at Elastic limit= 250kN, Extension at a load of 150kN= 0.21 mm, Maximum Load = 380kN, Total Extension= 60mm, Diameter of the rod at failure=2.25cm, determine the percentage decrease in Area?

Percentage decrease in Area = ((Original Area - Area at failure)/(original Area)) x 100 = ((0.785 x 30 x 30 - 0.785 x 22.5 x 22.5)/(0.785 x 30 x 30)) x 100 = 43.75%

Question 4

A tensile test is conducted on a mild steel bar. The following data was obtained from the test : diameter of the steel bar= 3cm, Gauge Length of the bar = 20cm, Load at Elastic limit= 250kN, Extension at a load of 150kN= 0.21 mm, Maximum Load = 380kN, Total Extension= 60mm, Diameter of the rod at failure=2.25cm, determine percentage elongation?

percentage elongation = (total increase in length/ original length) x 100 % = (60/200) x 100 = 30%

Question 5

A tensile test is conducted on a mild steel bar. The following data was obtained from the test : diameter of the steel bar= 3cm, Gauge Length of the bar = 20cm, Load at Elastic limit= 250kN, Extension at a load of 150kN= 0.21 mm, Maximum Load = 380kN, Total Extension= 60mm, Diameter of the rod at failure=2.25cm, determine the stress at elastic limit?

stress at elastic limit = load at elastic limit/area = (250x 1000)/(0.785 x 30 x 30)= 353.68N/sq(mm) = 353.68 x 10^6 N/Sq(m) =353.68 GN/Sq(m)

Question 6

A tensile test is conducted on a mild steel bar. The following data was obtained from the test : diameter of the steel bar= 3cm, Gauge Length of the bar = 20cm, Load at Elastic limit= 250kN, Extension at a load of 150kN= 0.21 mm, Maximum Load = 380kN, Total Extension= 60mm, Diameter of the rod at failure=2.25cm, determine young's modulus?

from the data given Area of bar = 0.785 x 30 x 30 =706.5 sq(mm), now,, within elastic limit load is 150kN, stress = 150000/706.5= 212.314N/sq(mm) strain = 0.21/200 = 0.00105 young modulus = stress/strain = 212.314/0.00105 = 2.02 x 10^5 N/sq(mm) =2.02x 10^11 N/sq(m) = 202 x 10^9 N/sq(m) = 202GN/Sq(m)

Question 7

consider a two dimensional stress system, in which body is subjected to stress 1 and 2 in x and y directions respectively, then the total strain in x direction is equal to?

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