The document provides information to calculate the width and depth of a rectangular beam subjected to a bending load. It is given that the beam carries a 400 N load at a 300 mm distance from its fixed end, and the maximum bending stress is 40 MPa. Using the bending stress formula and setting the section modulus equal to the product of depth and width/2, the width is calculated to be 16.5 mm and depth to be 2 * width = 33 mm.

1. 22564 EMD 2.0 Design of Joints, levers, offset links
2) A beam of uniform rectangular cross-section is fixed at one
end and carries an electric motor weighing 400 N at a distance of
300 mm from the fixed end. The maximum bending stress in the
beam is 40 MPa. Find the width and depth of the beam, if depth
is twice that of width.
Data : F= 400 N ; L = 300 mm ; σb = 40 MPa = 40 N/mm2 ;
h = 2b
Bending Stress =
Moment M = Force x Distance = 400 N x 300 mm
= 120000 N-mm

2. 22564 EMD 2.0 Design of Joints, levers, offset links
Section modulus of rectangular section is
h = 2b
Therefore b = 16.5 mm
h = 2 x b = 2 x 16.5 = 33mm
Neutral Axis
Tensile stress
Compressive stress

3. 22564 EMD 2.0 Direct & Bending Stresses
Revision on Direct & Bending Stresses
Consider a simple case shown in figure, axial force ‘F’ is
subjected to tensile stress at point A and B
A
B
F
A
B
F
e= eccentric distance
Now again the Force ‘F’ is shifted with eccentric distance ‘e’,
than the resultant stresses at point A and B is not only direct
tensile but bending stress also.

4. 22564 EMD 2.0 Direct & Bending Stresses
Revision on Direct & Bending Stresses
Failure and stresses are shown in figure
A
B
F
e= eccentric distance
Direct tension
+ Tension Bending
Direct tension
+ Compression
Bending

5. 22564 EMD 2.0 Direct & Bending Stresses
Calculations of direct and bending
Consider equal and opposite force ‘F’ as shown in figure
A
B
F
e= eccentric distance
F
F
Two equal and opposite forces ‘F’ separated by distance ‘e’
creates a Moment, so equivalent figure can be
A
B
Moment = F x e
e= eccentric distance
Force
Total Stress = Direct Stress + Bending Stress
t + b (Tensile +
Tensile)
t - b (Tensile +
Compression)

6. 22564 EMD 2.0 Direct & Bending Stresses
Revision on Direct & Bending Stresses
Total Stress = Direct Stress + Bending Stress

7. 22564 EMD 2.0 Direct & Bending Stresses
1) A steel bracket is subjected to a load of 4.5 kN, as shown in
Figure. Determine the required thickness of the section at A-A in
order to limit the tensile stress to 70 MPa.
Given Data : Force = 4.5 KN, Tensile stress = 70 Mpa

8. 22564 EMD 2.0 Direct & Bending Stresses
t = Force / Area
Area = 50 x t = 50t
t = 4.5 x 1000 N / 50t = 90 / t
bt = M / Z and bc = - M / Z
Moment = F x e = 4.5 x 1000 x 50 = 225000 N-mm
Z = b x h2 / 6 = t x 502 /6 = 416.67 t
bt/c = 225000 / 416.67 t = 539.99/ t
Neutral
Axis

t
+

bt

t
+

bc

9. 22564 EMD 2.0 Direct & Bending Stresses
t = 4.5 x 1000 N / 50t = 90 / t
bt/c =  337500 / 16.67 t =  539.99 / t (Tensile + and
Compression -)
Maximum total stress = t + bt
total = 70 Mpa = 90 / t + 539.99 / t
t= 8.99 = 9mm
Safe thickness = 9mm