The document contains 6 numerical problems related to calculating stresses, moments of inertia, and section moduli for beams and shafts with various cross-sectional geometries under different loading conditions. Solutions are provided for rectangular, circular, square, and hollow cross-sections. Bending stresses, axial stresses, radii of curvature, and polar moments of inertia are calculated using relevant stress and bending formulas.

2. 1. A uniformly distributed load of 20 kN/m acts on a simply supported beam of rectangular cross section
of width 20 mm and depth 60 mm. What is the maximum bending stress acting on the beam of 5m?
solution:-
1) Position of neutral axis = y = Depth/2 = 60/2 = 30 mm
2) Bending moment = M = (wL2) / 8 = (20 x 25) / 8 = 62.5 kN
3) Moment of inertia (I) = (BD3) / 12 = (20 x 603) / 12 = 0.36 x 106 Mpa
The maximum bending stress is given as σmax = (M/I) y
= [(62.5 x 103) / (0.36 x 106)] x 30
= 5208 Mpa ANS.
UDL=20KN/m
L=5m
D=60mm
B=20mm

3. 2. Calculate the maximum stress due to Bending in a steel strip of 30
mm thick and 60 mm wide is bent around a circular drum of 3.6 m
diameter [Take Young’s modulus = 200kN/m2].
Solution:- Thickness of steel strip = 30 mm; b = 60 mm; d = 3.6m
R = 3.6/2 = 1.8 m
E = 200 kN/m2
y = 30/2 = 15 mm
E/R = fb/y ;
fb=Ey/R
fb= 200000×15/1800
= 1666.67 N/mm2.

4. 3.The steel plate is bent into a circular path of radius 10 metres. If the
plate section be 120 mm wide and 20 mm thick, then calculate the
maximum bending stress. [Consider Young’s modulus = 200000
N/mm2].
Solution:-
R = 10000 mm;
y = 20/2 = 10 mm;
E = 200000 N/mm2
By bending equation we have E/R = fb/y
fb=Ey/R
fb = 200000×10 / 10000
= 200 N/mm2. ANS.

5. 4. Calculate the modulus of section of rectangle beam of size 240 mm × 400
mm.
Solution:-
b = 240 mm & d = 400 mm
Moment of inertia (I) = bd3/12;
Max. corner Distance y = d/2
Section modulus (Z) = I/y
=(bd3/12)/(d/2)
= bd2/ 6
= 1/6 × 240 × 4002
= 6.4 × 106 mm3.answer.

6. 5. Determine section modulus for beam of 100mm diameter.
Solution: d = 300mm
For circular sections; I = π / 64 × d4
y= d/2
Z =I/Y
Z = π/32 × d3 (d = 100 mm)
Z = 98.17 × 103mm3.

7. NUMERICALS FOR PRACTICE:-
1. Find the modulus of section of square beam of size 300×300 mm.
2. Find the modulus of section of rectangle beam of size 300×600 mm.
3. To what radius an Aluminium strip 300 mm wide and 40mm thick can be
bent, if the maximum stress in a strip is not to exceed 40 N/mm2. Take
young’s modulus for Aluminium is 7×105 N/mm2.
4. A steel rod of 25 mm diameter and 600 mm long is subjected to an axial
pull of 40000. The intensity of stress is?
5. A Steel rod 200 mm diameter is to be bent into a circular arc section.
Find radius of curvature. Take f = 120N/mm2 & E = 2×105 N/mm2
6. A hollow shaft has outside diameter 120 mm and thickness 20 mm.
Find the polar moment of inertia (J).