Diploma Strength of Materials Important Questions Unit 4 | Diploma Strength of Materials Important Questions pdf

Diploma Strength of Materials Important Questions

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Subject : Strength of Materials   Scheme :N

Unit 4 – THEORY OF TORSION AND SPRINGS

NOTE :

1. The highlighted Questions are asked more than two times in the Board Exam
2. These Questions are collected from Board Examination Question Papers ( M Scheme)

3 marks questions.

1. Define polar modulus.
2. Write down the advantages of hollow shafts over solid shafts.
3. A closely coiled helical spring made of steel sire of 10 mm diameter has 10 coils of 120mm mean diameter. Calculate the deflection of the spring under an axial load of 100N. Take modulus of rigidity as 1.2×10^5 N/mm^2.
4. State the applications of springs.
5. Write down the assumptions made in theory of pure torsion.
6. Write down the various types of springs.
7. Write torsion equation.
8. Define polar modulus for (i) solid shaft (ii) hollow shaft.
9. State the difference between open and closely coiled helical spring.
10. Define stiffness of spring.
11. A closely coiled helical spring has the stiffness of 40N/mm. determine its number of turns, when the diameter of the wire of the spring is 10mm and diameter of the coil is 80mm. take C=0.8 x 10^5 N/mm^2
12. Find the torque transmitted by the solid shaft of diameter 0.4m. the angle of twist is not to exceed 1 degree in a length of 10m. take C = 0.8×10^5 N/mm^2.
13. What is laminated spring? List its applications.

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14 marks questions.

1. A hollow shaft having inner diameter 0.6 times the outer diameter is to replace a solid shaft of same material to transmit 550kW at 220 rpm. Calculate the diameters of the hollow and solid shafts. Also calculate the percentage of savings in material. The allowable shear stress is 80N/mm^2.
2. A weigh of 15ON is dropped on to a compression spring with 10 coils of 12mm diameter closely coiled to a mean diameter of 150mm. if the instantaneous contraction is 140mm, calculate the height of drop. Take C = 0.8×10^5 N/mm^2.
3. A solid shaft is transmitting. 100kW power at 180 rpm. If the allowable shear stress is 60N/mm^2, find the necessary diameter for the shaft. The shaft is not to twist more than 1 degree in a length of 3m. take C=80kN/mm^2.
4. Calculate the power transmitted by a shaft of 100mm diameter running at 250 rpm, if the shear stress in the shaft material is not to exceed 75N/mm^2 (5 marks)
5. A closely coiled helical spring made of steel wire of 100mm diameter has 10 coils of 120mm mean diameter. Calculate the deflection under an axial load of 100N and the stiffness of the spring. Take C =1.2 N/mm^2.
6. Hollow circular shaft of 25mm outside diameter and 20mm inside diameter is subjected to a torque of 50Nm. Find the shear stress induced at the outside and inside layer of shaft.
7. A truck weighing 30kN and moving at 5km/hr has to be brought to rest by buffer. Find how many springs each of 18 coils will be required to the energy of motion during a compression of 200mm. the spring is made of 25 mm diameter steel rod coiled to a mean diameter of 240mm. take N=0.84×10^5 N/mm^2.
8. With neat sketches, explain the various types of springs.
9. A solid shaft 20 mm diameter transmits 10kW at 1200 rpm. Calculate the maximum intensity of shear stress induced and the angle of twist in degrees in a length of 1m. if modulus of rigidity for the material of the shaft is 8×10^4 N/mm^2.
10. A truck weighing 20kN and moving at 6km/hr has to be brought to rest by buffer. Find how many springs each of 15 coils will be required to the energy of motion during a compression of 200mm. the spring is made of 25 mm diameter steel rod coiled to a mean diameter of 200mm. take N=0.945 x 10^5 N/mm^2.
11. A hollow shaft of 200mm external diameter, thickness of the metal 200mmm is transmitting power at 80 rpm. Angle of twist in a length 3m was found to be 0.7 degree. Calculate the power transmitted and shear stress developed. Take C=0.8 x 10^5 N/mm^2.
12. Prove the torsion equation.
13. The mean diameter of a closely coiled helical spring is 5 times the diameter of wire. If elongates 8mm under an axial pull of 120N, if the permissible shear stress is 40N/mm^2. Find the size of wire and number of coils in the spring. Take N = 0.8 X 10^5 N/mm^2