Mechanics of solids:Simple stresses and strains

NUMERICALS ON MECHANICS OF SOLIDS

Q-1 Two solid cylindrical roads AB and BC are welded together at B and loaded as shown. Knowing that d1 = 30 mm and d2 = 50 mm, find the average normal stress in the mid section of (a) rod AB, (b) rod BC.

Sol.1:

Diameter (d₁) = 30 mm

Then, Area (a₁) = 3.14 x (d₁/2)^² =3.14/4 x (30)^² = 0.785 x 900 = 706.5 m㎡.

We know, stress = load/ area

Also, Stress (S₁) = 60 x 1000/706.5

= 84.926 N/m㎡

Again, Diameter (d₂) = 50 mm.

Then, Area (a₂) = 3.14 x (d/2)^² =3.14/4 x (50)= 0.785 x 2500= 1962.5 m㎡.

Also, Stress (S₂) = 190 x 1000/1962.5 {because, 125 + 125 – 60 = 190 }.

= 96.8 N/m㎡.

Q-2 Calculate the elongation of a mild steel rod 25 mm in diameter and 2 m long when subjected to an axial pull of 80 KN. Take for steel E = 200 KN/m㎡.

Sol.2:

Diameter of the mild steel (d) = 25 mm

Length = 2 m = 2 x 1000 = 2000 mm

Axial pull = 80 kN= 80 x 1000 =80000 N.

E = 200 k N/m㎡ = 200 x 1000 = 200000 N/m㎡ .

Now, Area of rod = 3.14 x (d₁/2)² =3.14/4 x (25)² = 0.785 x 625 = 490.625 m㎡

We know, stress = load/ area

= 80000/ 490.625 = 163.057 163.057 N/m㎡.

Also, E = STRESS / STRAIN

Then, STRAIN = STRESS / E

= 163.057 / 200000 = 0.0008152

Now , Strain = dl / l

dl = Strain * length

= 0.0008152 x 2000 = 1.629 mm

Thus, the elongation ( dl ) = 1.629 mm

Q-3 A steel rod is 2.2 m long and must not stretch more than 1.2 mm when a 8.5 kN load is applied to it. Knowing that E=200 GPa, determine (a) the smallest diameter rod which should be used, (b) the corresponding normal stress caused by the load

Sol 3:

Length of the rod ( l ) = 2.2 m = 2.2 x 10^³ mm

Elongation ( dl ) = 1.2 mm

Load = 8.5 k N = 8.5 x 10^³ N

E = 200 GPa = 200 x 10^⁹ N/m㎡.

Now , Strain = dl / l

= 1.2 / 2.2 x 10^³

= 0.5454 x 10^^-3

Now, E = STRESS / STRAIN

Then, STRESS = STRAIN x E

= 200 x 10^⁹ x 0.5454 x 10^^-3

= 109.09 x 10^⁶ N/m㎡

We know, stress = load/ area

Then, Area = load/ stress

= 8.5 x 10^³ / 109.09 x 10^⁶

= 0.0779 x 10^^-3 m㎡

But, Area of steel rod = 3.14 x (d/2)^² = 0.0779 x 10^^-3

3.14/4 * (d)^² = 0.0779 x 10^^-3

(d)^² = 0.0779 x 10^^-3 x 4 / 3.14

(d)^²= 0.3116 x 10^^-3 / 3.14

(d)^² = 0.09923 x 10^^-3

(d) = 3.15 mm

Thus, smallest diameter a rod should have = 3.15 mm

Q-4 A 1.5 m long steel wire of 6 mm diameter steel wire is subjected to a 3.4 kN tensile load. Knowing that E = 200 GPa, determine (a) the elongation of the wire, (b) the corresponding normal stress.

SOL 4 :

Length of the steel wire = 1.5 m = 1.5 x 10^³ mm

Diameter of the steel wire = 6 mm

Tensile load = 3.4 k N = 3400 N

E = 200 GPa = 200 x 10^⁹N/m㎡

Now, Area of the steel rod = 3.14 x (d/2)^²

= 3.14/4 x (6)^² = 113.04/4

= 28.26 m㎡

We know, stress = load/ area

= 3400 / 28.26

= 120.31 N/m㎡

Also, E = STRESS / STRAIN

Then, STRAIN = STRESS / E

= 120.31 / 200 x 10^⁹

= 0.60155 x 10^^-9

Now , Strain = dl / l

dl = Strain x length

= 0.60155 x 10^^-9x 1.5 x 10^³

dl = 0.9023 x 10^^-6 mm

Thus, the elongation of the wire = 0.9023 x 10^^-6 mm

Q-5 A rod of length 2 m and diameter 50 mm is elongated by 5mm when an axial force of 400 KN is applied to it. Determine the stress and strain in the rod. Also calculate the value of modulus of elasticity of the material of the rod.

SOL 5 :

Length of the rod = 2m = 2 x 10^³ mm

Diameter of the rod = 50 mm

Elongation (dl) = 5mm

Axial force = 400 k N = 400 x 1000 = 400000 N.

Now, Area of the rod = 3.14 x (d/2)^² = 3.14/4 x (50)^²

= 3.14/4 x 2500

= 7850/4

= 1962.5 m㎡

We know, stress = load/ area

= 400000 / 1962.5 = 203.82 N/m㎡

Thus , stress in the rod = 203.82 N/m㎡

Now , Strain = dl / l

= 5 / 2 x 10^³

= 2.5 x 10^^-3

Thus , the strain in the rod = 2.5 x 10^^-3

Also, E = STRESS / STRAIN

= 203.82 / 2.5 x 10^^-3

= 81.53 x 10^³ N/m㎡

Thus, modulus of elasticity of the rod = 81.53 x 10^³ N/m㎡

Q-6 A compound tube consists of a steel tube 200 mm internal diameter and 10 mm thickness and an outer brass tube 220 mm inner diameter and 10 mm thickness. The two tubes are of same length.The compound tube carries an axial load of 1500 kN.Find the stresses and loads transmitted to the two tubes. Find also the decrease in length of the compound tube, if each tube is 200 mm long.( Take E_s=2x10⁵N/m㎡and E_b=1x10⁵N/m㎡)