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Sunday, 4 February 2018

STRENGTH OF MATERIALS Top 10 Numercial

Assume g = 10 m/s2 for all questions
1. a. A round steel bar with an initial diameter of 20 mm and length of 2 m is placed in tension, supporting a load of 2000 kg. If the Young’s modulus of the bar is 200 GPa what is the length of the bar when supporting the load (assuming it does not yield)?
b. If the Poisson’s ratio of the steel is 0.4, what will the diameter of the bar be when supporting the load?
Solution:                     
a. Cross-sectional area = 3.14 × 104 m2
Load = 2000 kg × g = 2 × 104 N
Stress = 2 × 104/3.14 × 104 = 6.4 × 107 Pa
Strain = 6.4 × 107/200× 109 = 3.2 × 104

New length = (3.2 × 104 × 2) + 2 = 2.00064 m
b. Lateral strain = 3.2 × 104 × 0.4 = 1.28 × 104
Change in diameter = 1.28 × 104 × 0.02 = 2.56 × 106m
New diameter = 0.02 2.56 × 106= 1.999744 × 102m = 19.99744 mm


2. A 2 m long tie in a steel frame is made of circular hollow section steel with an outer diameter of 100 mm and a wall thickness of 5 mm. The properties of the steel are as follows:
Young’s modulus: 200 GPa
Yield stress: 300 MPa
Poisson’s ratio: 0.15
a. What is the extension of the tie when the tensile load in it is 200 kN?
b. What is the reduction in wall thickness when this load of 200 kN is applied?
Solution:
a. Area = π(0.052 0.0452) = 1.491 ×103m2
Stress = 200 × 103/1.491 × 103= 1.34 × 108 Pa
This stress is 134 MPa, which is well below the yield stress.
Strain = 1.34 × 108/2 × 1011 = 6.7 × 104
Extension = 2 × 6.7 × 104= 0.00134 m = 1.34 mm
b. Strain = 6.7 × 104× 0.15 = 1.005 × 104
Reduction = 1.005 × 104× 0.005 = 5.025 × 107m = 0.5 mm
3. A water tank is supported by four identical timber posts that all carry an equal load. Each post measures 50 mm by 75 mm in cross-section, and is 1.0 m long. When 0.8 m3 of water is pumped into the tank, the posts get 0.07 mm shorter.
a. What is the Young’s modulus of the timber in the direction of loading?
b. If the cross-sections measure 75.002 mm by 50.00015 mm, after loading, what are the relevant Poisson’s ratios?
c. If 300 L of water are now pumped out of the tank, what are the new dimensions of the post?
(Assume that the strain remains elastic.)
Solution:
Normally, all calculations should be carried out in base units, but these calculations of strain are an example of where it is clearly easier to work in mm.

a. 0.8 m3 of water has a mass of 800 kg and thus a weight of 800 × g = 8000 N
Stress = 8000/(4 × 0.075 × 0.05) = 5.33 × 105 Pa
Strain = 0.07/1000 = 7 × 105
Modulus = 5.33 × 105/7 × 105= 7.6 × 109 Pa

b. Change in width on long side = 75.002 75 = 0.002 mm
Strain on long side = 0.002/75 = 2.66 × 105
Thus, Poisson ratio = 2.66 × 105/7 × 105= 0.38
Strain on short side = 0.00015/50 = 3 × 106
Thus, Poisson ratio = 3 × 106/7 × 105= 0.043

c. 300 L of water has a mass of 300 kg.
Thus, the mass of water now in the tank is 800 300 = 500 kg
The changes in dimensions are proportional to the load, and thus they are 500/800 = 0.625 of those in part (b).
0.07 × 0.625 = 0.044, thus length = 1000 0.044 = 999.956 mm
0.002 × 0.625 = 0.00125, thus long side = 75.00125 mm
0.00015 × 0.625 = 0.000094, thus short side = 50.000094 mm
4. The below figure shows observations that were made when a 100 mm length of 10 mm diameter steel bar was loaded in tension. The equation given is for a trendline that has been fitted to the straight portion of the graph. Calculate the following:
a. the Young’s modulus,
b. the estimated yield stress,
c. the ultimate stress.


Solution:
Cross-section area = π × 0.012/4 = 7.85 × 105m2

a. The equation given on the graph has been generated by the Excel compute package, and is a best fit to the linear part of the data (this is real experimental data). This technique is discussed in Chapter 15. The gradient taken from the equation is 157.76 kN/mm = 1.5776 × 108 N/m
E = 1.5776 × 108 × 0.1/7.85 × 105= 2.01 × 1011 Pa = 201 GPa
b. Load = 33 kN (estimated from the end of the linear section on the graph)
Thus, stress = 420 MPa
c. Load = 42 kN at failure (the highest point on the graph)
Thus, stress = 535 MPa
5. a. A 100 mm diameter concrete cylinder 200 mm long is loaded on the end with 6 tonnes. What is the stress in it?

b. The Young’s modulus of the cylinder is 25 GPa. What is its height after loading?

c. The Poisson’s ratio of the cylinder is 0.17. What is its diameter after loading?

d. A column 2 m high, measuring 300 mm by 500 mm on plan, is made with the same concrete as in the cylinder and supports a bridge. In order to prevent damage to the deck, the maximum permitted compression of the column is 0.3 mm. Assuming that the column is not reinforced, calculate the compression of the column when it is supporting the full weight of a 40 tonne lorry, and state whether the deck will fail.
Solution:
Area = π × 0.052 = 0.0078 m2
a. Load = 6 tonne = 6000 kg
Weight = 6,000 × g = 60,000 N
Stress = 60,000/0.0078 = 7.64 × 106 Pa
b. Strain = 7.64 × 106/25 × 109 = 3.06 × 104
New height = 0.2 (0.2 × 3.06 × 104) = 0.19994 m c. Lateral strain = 3.06 × 104 × 0.17 = 5.2 ×104
0.1 + (0.1 × 5.2 × 105) = 0.1000052 m
d. Mass of lorry = 40 tonne, thus Weight = 4 × 105 N
Stress = 4 × 105/(0.3 × 0.5) = 2.66 × 106 Pa
Compression = 2.66 × 106 × 2/(25 × 109) = 2 × 104m = 0.2 mm and
Deck will not fail.

6. a. How many Poisson’s ratios does timber have?
b. A timber sample measuring 50 mm × 50 mm on plan is supporting a mass of 2 tonnes. If
the relevant Young’s modulus and Poisson’s ratio are 1.3 GPa and 0.4, what is the lateral
expansion caused by the load?
Solution:
a. It has 6 Poisson’s ratios (see Section 33.5.2):
longitudinal–tangential
tangential–longitudinal
radial–tangential
tangential–radial
radial–longitudinal
longitudinal–radial

b. Load = 2 tonnes = 2000 kg
Weight = 2,000 × g = 20,000 N
Stress = 20,000/(0.05 × 0.05) = 8 × 106 Pa
Longitudinal strain = 8 × 106/1.3 × 109 = 6.15 × 103
Lateral strain = 0.4 × 6.15 × 103 = 2.46 × 103
Expansion = 2.46 × 103× 0.05 = 1.23 × 104m = 0.123 mm

7. A steel specimen measuring 1 in. by 0.5 in. in section is found to have yielded at a load of 38 kips, and failed at 50 kips. Assuming that the modulus of elasticity is 29,000 ksi, and the Poisson’s ratio is 0.3, calculate:
a. The tensile stress at yield and failure.
b. The strain when the stress is half the yield stress.
c. The dimensions of the section at half the yield stress.
Solution:
a. Yield stress = 38,000/(1 × 0.5) = 76,000 psi
Failure stress = 50,000/(1 × 0.5) = 105 psi
b. Longitudinal strain = (0.5 × 76,000)/(2.9 × 107) = 1.31 × 103
c. Lateral strain = 1.31 × 103× 0.3 = 3.4 × 104
New width = 1 (1 × 3.4 × 104) = 0.99966 in.
New thickness = 0.5 (0.5 × 3.4 × 104) = 0.49983 in.

8. a. A round steel bar with an initial diameter of ¾ in. and length of 6 ft. is placed in tension, supporting a load of 4,400 lb. If the Young’s modulus of the bar is 29,000 ksi, what is the length of the bar when supporting the load (assuming it does not yield)?
b. If the Poisson’s ratio of the steel is 0.4, what will the diameter of the bar be when supporting the load?
Solution:
a. Cross-sectional area = 0.442 in.2
Stress = 4,400/0.442 = 9,954 psi
Strain = 9,954/29,000 × 103 = 3.43 ×104
New length = (3.43 × 104 × 6) + 6 = 6.002 ft.

b. Lateral strain = 3.43 × 104× 0.4 = 1.37 × 104
Change in diameter = 1.37 × 104 × 0.75 = 1.027 × 104in.
New diameter = 0.75 1.027 × 104= 0.749897 in.

9. A 6 ft. long tie in a steel frame is made of circular hollow section steel with an outer diameter of4 in., and a wall thickness of 0.2 in. The properties of the steel are as follows:
Young’s modulus: 29,000 ksi
Yield stress: 43 ksi
Poisson’s ratio: 0.15
a. What is the extension of the tie when the tensile load in it is 44 kips?
b. What is the reduction in wall thickness when this load of 44 kips is applied?

Solution:
a. Area = π(22 1.82) = 2.387 in.2
Stress = 44 × 103/2.387= 18.43 ksi
This stress is well below the yield stress.
Strain = 18.43 × 103/2.9 × 107 = 6.35 × 104
Extension = 6 × 12 × 6.35 × 104 = 0.0457 in.
b. Strain = 6.35 × 104 × 0.15 = 9.5 × 105
Reduction = 9.5 × 105 × 0.2 = 1.9 × 105in.

10. The figure below shows observations which were made when a 4 in. length of 3/8 in. diameter steel bar was loaded in tension. The equation given is for a trendline that has been fitted to the straight portion of the graph. Calculate the following:
a. the Young’s modulus,
b. the estimated yield stress,
c. the ultimate stress.
Solution:
Cross-section area = π × 0.3752/4 = 0.11 in.2
a. The equation given on the graph has been generated by the Excel compute package, and is a best fit to the linear part of the data. The gradient taken from the equation is 816.4 kips/in.
E = 816.4 × 4/0.11 = 29,687 ksi
b. Load = 7.4 kips (estimated from the end of the linear section on the graph)
Thus stress = 67.3 ksi from
c. Load = 8.5 kips at failure (the highest point on the graph)
Thus stress = 77.3 ksi from

NOTATION
a Acceleration (m/s2)
A Area (m2)
B Width (m)
E Young’s modulus (Pa)
f Force (N)
g Gravitational constant (=9.81 m/s2)
L Length (m)
m Mass (kg)
W Load (N)
x Displacement (m)
g Shear angle (radians) or shear strain
σ Stress (Pa)

τ Shear stress (Pa)

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