TAKORADI TECHNICAL UNIVERSITY
FACULTY OF ENGINEERING
MECHANICAL ENGINEERING DEPARTMENT
END OF SECOND SEMESTER EXAMINATION 2020 (B-TECH)
COURSE: BTM 412 MACHINE DESIGN
SEPTEMBER 2020 TIME: 3 HOURS
Answer Question ONE and any Other TWO Questions
Question 1
A steel solid shaft shown in Figure 1 transmitting 37.5 kW at 500 rpm is supported on two bearings
750 mm apart and has two gears keyed to it. The pinion having 30 teeth of 5 mm module is located
100 mm to the left of the right hand bearing and delivers power horizontally to the right. The gear
having 100 teeth of 5 mm module is located 150 mm to the right of the left hand bearing and
receives power in a vertical direction from below. The shaft is to be machined from
AISI11370QT1300 steel of 99.9% reliability and assume average diameter of 200 mm, a stationary
shaft with gradually applied load with factor of safety to be three. Determine the following
a) i) The maximum bending moment
ii) The shaft diameter
b) Design a key of the shaft using the same material as the shaft
Figure 1: Space Diagram of a Shaft
[𝟒𝟎 𝐌𝐀𝐑𝐊𝐒]
1
Answer Any TWO Questions
Question 2
a) i. Explain briefly a hydrodynamic lubrication?
ii. List four basic assumptions used in the theory of hydrodynamic lubrication.
iii) What are rolling contact bearings? State four advantages and four disadvantages of rolling
contact bearings over sliding contact bearings.
b) The load on the journal bearing is 120 kN due to turbine shaft of 300 mm diameter running at
1200 rpm. Determine the following:
i. Length of the bearing if the allowable bearing pressure is 1.5 N/mm2, and
ii. Amount of heat to be removed by the lubricant per minute if the bearing temperature is 60°C
and viscosity of the oil at 60°C is 0.02 kg/m-s and the bearing clearance is 0.25 mm.
c) Select a double row angular contact ball bearing for a radial load of 3000 N and applied thrust
load of 1800 N, operating at a speed of 2000 rpm for an average life of 3 years at 18 hours per day.
Assume uniform and steady load. Assuming 100 working days per year and uniform and steady
load. [Take Y = 1.50, X= 0.56 and V = 1]
[𝟐𝟎 𝐌𝐀𝐑𝐊𝐒]
Question 3
a) i. What is the function of a spring?
ii. Discuss the materials and practical applications for the various types of springs.
b) A helical compression spring for a maximum load of 1100 N for a deflection of 30 mm using
the value of spring index as 8. The maximum permissible shear stress for spring wire is 550 MPa
and modulus of rigidity is 75 kN/mm2. Calculate the following:
2
i) Diameter of the spring wire
ii) Mean coil diameter
iii) Number of active turns taking the ends of the coil as squared, and
iv) Pitch of the coil
[𝟐𝟎 𝐌𝐀𝐑𝐊𝐒]
Question 4
a) A double riveted lap joint is made between 15 mm thick plates. The rivet diameter and pitch are
25 mm and 75 mm respectively. If the ultimate stresses are 400 MPa in tension, 320 MPa in shear
and 640 MPa in crushing, find the minimum force per pitch which will rupture the joint. If the
above joint is subjected to a load such that the factor of safety is 4, find out the actual stresses
developed in the plates and the rivets.
b) A welded joint as shown in Fig. 2, is subjected to an eccentric load of 2 kN. Find the size of
weld, if the maximum shear stress in the weld is 25 MPa.
Fig. 2 A welded joint
[𝟐𝟎 𝐌𝐀𝐑𝐊𝐒]
3
FORMULA SHEET 𝐿 = 60𝑁 × 𝐿𝐻
1. The tangential force acting on the gear 10. The basic dynamic equivalent radial load,
W = X.V.WR + Y.WA
𝑇
𝐹𝑡𝑐 =
𝑅𝐶 11. Basic dynamic load rating,
2. The normal load on the tooth of gear 1
𝐿 𝑘
𝐹𝑡𝑐 𝐶=𝑊 10 6
𝑊𝑐 =
cos ∝𝑐
12. 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝜏 ,
3. Diameter of the shaft, D
8𝐾𝑊𝐶
1 𝜏=
3 𝜋𝑑 2
2
32𝑁 𝐾𝑡 𝑀 2 3 𝑇
𝐷= S𝜄 n
+4
𝜋 Sy 13. Maximum compression of the spring,
8𝑊𝐶 3 𝑛
9. Life of bearing δ= 𝐺𝑑
4. Length of key, L 14. Free length of the spring,
LF = n' d + δmax + 0.15 δmax
4𝑇𝑁
𝐿= 15. Ultimate tearing resistance of the plate
𝐷𝑊Sy
per pitch, 𝑃𝑡𝑢 = (𝑝 – 𝑑)𝑡 × 𝜎𝑡𝑢
5. Bearing pressure, p 16. Ultimate shearing resistance of the rivets
𝜋
per pitch, 𝑃𝑠𝑢 = 𝑛 × 4 × 𝑑 2 × 𝜏𝑢
𝑊
𝑝= 17. Ultimate crushing resistance of the rivets
𝑙𝑑
per pitch, 𝑃𝑐𝑢 = 𝑛 × 𝑑 × 𝑡 × 𝜎𝑐𝑢
6. Coefficient of friction,
18. Section modulus of the weld through the
33 𝑍𝑁 𝑑 𝑠×𝑙 2
𝜇= +𝑘 throat, 𝑍 = 4.242
108 𝑝 𝑐
7. Heat generated 19. Maximum shear stress
𝜋𝑑𝑁 1 2
𝑄𝑔 = 𝜇𝑊𝑉 = 𝜇𝑊 𝜏𝑚𝑎𝑥 = 𝜎𝑏 + 4 𝜏2
60 2
8. Heat dissipated,
𝑄𝑑 = 𝐶. 𝐴 𝑡𝑏 − 𝑡𝑎 = 𝐶. 𝑙. 𝑑 𝑡𝑏 − 𝑡𝑎
