STRESS ANALYSIS (AOC) SOLUTION

This Article shows Area of concentration (AOC) Questions and Answer workings in Stress Analysis for Higher National Diploma (HND) Students of Mechanical Engineering Technology Auchi Polytechnic, Auchi.

This (AOC) Questions and Answers is helpful to those currently writing their on-going examination the Department of Mechanical Engineering Technology.

N.B – These questions and answers are uploaded only as a guide and ease for your studies.

Below are the Questions:

AUCHI POLYTECHNIC, AUCHI
SCHOOL OF ENGINEERING TECHNOLOGY
DEPARTMENT OF MECHANICAL ENGINEERING TECHNOLOGY

Class: HND 1 Mech                  Time: 3hrs
Course: Stress Analysis MEC 313

QUESTION 1

A flat bar 35mm wide and 13mm thick is loaded by extending tensile load of 80KN. The material is mild steel with yield point stress of 320 MN/m2. Find the factor of safety based on the yield point.

ANSWER

Data:

Width, b = 35mm = 0.035m

Thickness, t = 13mm = 0.013m

Load, p = 80KN

Yield point stress, Ծ= 3201N/m²

Let N be the factor of safety. Ny = (yield stress)/(working stress)

Working stress = p/A

Where A = bt

= (0.035 x 0.013)m²

Ծw = (80 x 10˄3)/(0.035 x 0.13) N/m²

QUESTION 2

A continuous and aligned glass reinforce composite consists of 30% by volume of glass fibre, having modulus of elasticity 81GN/m2. If the area of the composite is 210mm2 and a stress of 65MN/m2 is applied in the longitudinal direction.

Calculate: (i) The magnitude of the load carried by each of the fibres and the matrix phases.

ANSWER

Data:

Volume of Fibre, Vf = 30% = 0.3

Vmatrix, Vm = 70% = 0.7

Ef = 81GN/m²

Em = 22.3GN/m²

Ac = 210mm² = 210 x 10-⁶m²

Ծc = 65MN/m²

Let Pf be the load carried by the fibre

Let Pm be the load carried by the matrix

Pc = Pf + Pm – – – – – – – – – (1)

Pf = Vf Ef = 0.3 x 81 x 10⁹            = 1.556

Pm  Vm Em 0.7 x 22.3 x 10⁹

Pf  =    0.3 x 81 x 109               x  Pm  ————- (2)

                  0.7 x 22.3 x 10⁹

= 1.556

Pc = Ac x ∏c = (210 x 10-⁶ x 65 x 10⁹)N

= 13650

Therefore, Pc = Pf + Pm

= 1.556Pm + Pm = 2.556Pm

Recall, Pm = 1Pm

13650 = 2.556Pm

Now, Pm =  13650/2.556 = 5340.3755

Pm ≈ 5340.38

QUESTION 3

A mild steel material of diameter 140mm and length 340mm when subjected to an axial compressive load of 230KN resulted to an increase of diameter 0.129mm and a decrease length of 0.21mm. Calculate the value of the poisons ratio and the values of the Poissons ratio and the modulus of elasticity.

 If E = 2G (1 + 1/m) and E = 9KG

                                                    3K + G

Show that G/k = 3/2 ((m-2)/(m+1)  )and hence find the value of G and K

Data:

Original diameter, d = 140mm = 0.14m

L = 320mm = 0.32m

P = 230KN

∆ d =0.129mm = 0.129 x 10-³m

∆ L = 0.21mm = 0.21 x 10-³m

Poisson’s ratio = (Lateral Strain)/(Linear Strain)

Lateral strain =  ( ∆ d)/d = 0.129/140 = 6.5625 x 10-⁴

Therefore, 1/m = 9.214 x 10-⁴ = 1.4

                            6.5625 x 10-⁴

1/m = 1.4

M = 1/1.4 = 0.714

Recall, Ծ =E£

E = Ծ/£

Where Ծ = P/A

A = πd2/4 = π/4 (0.14)²

= 0.01539m²

Therefore, Ծ = 230 x 103 = 14944769.33

  0.01539

= 14.94MN/m²

E = 14.94 x 106 N/m² = 

       6.56 x 10-⁴

If E = 2G (1 + 1/m) and E = 9KG/3K + G

 G/k = 3/2 ((m-2)/(m+1)  ) and hence find the values of G and K

QUESTION 4

A thin cylindrical tube 75mm internal diameter is closed at the ends and subjected to an internal pressure of 5.5 Mpa. A torque 500π N/M is also applied to the tube. Calculate: The maximum and minimum principal stresses and maximum sharing stress in the tube.

ANSWER

Data:

Internal diameter, di = 75mm = 0.75m

Thickness, t = 5mm

External diameter, do = di + 2t = 0.075 + 2 x 0.005 = 0.085m

Pi = 5.5mpa = 5 x 106 N/m²

Torque, T = 500πNm

           The maximum and minimum principal stresses

Ծmax = ½ (Ծx + Ծy) ± √((Ծx- Ծy))2 + 4y²

QUESTION 5
(a) define (i) stress (ii) strain (iii) Young modulus

I. Stress – refers to the internal resistance that a material offers to deformation. It’s a measure of the internal forces within a material that resist deformation or change in shape.

II. Strain – is the measure of deformation or change in shape of a material in response to an applied force. It quantifies how much a material deforms under the influence of stress.

III. Young’s Modulus, also known as the modulus of elasticity, is a measure of the stiffness of a material. It quantifies how much a material deforms under a given amount of force or stress.

Gooluck!