Lecture 11 Moment of Resistance

1. STRUCTURAL DESIGN I
STRUCTURAL MECHANICS

2. MOMENT OF RESISTANCE
DESIGN OF SIMPLE BEAMS

3. The cantilever cut at XX
shows a couple in which
the upper part of the
cantilever gets a tension
force of T and the lower
part gets a compressive
force of C with ‘a’ as the
arm.
This couple has to be
resisted by the material
of the cantilever. The
moment of the resisting
couple is called the
“moment of resistance”.

4. At every beam section subjected to bending moment, the
moment of resistance will equal the bending moment, when
the beam has deflected to its position of equilibrium.
The general problem of beam design is to determine a
suitable size and shape for the beam section so that the
beam particles/fibers are not excessively stressed and that
the material of the beam is used economically.
Definition of Moment of Resistance (MR):
The moment of resistance of a beam section is the moment
of the couple which is set up at the section by the
longitudinal forces created in the beam by its deflection.

5. The diagram shows the elevation of a simple beam. The lower diagram shows
the
beam bent under the action of bending moment. In every bent beam there is a
layer of material which gets neither longer nor shorter when the when the
beam deflects.
This is termed as the ‘neutral layer’ of the beam.
Neutral Axis
The ‘neutral axis’ of a beam section is the straight line in which the neutral
layer of the beam cuts that particular section. It represents the position in the
beam section where there is no stress or strain.

7. Variation of Stress and Strain in a Beam Section
The diagram clearly shows that beam layers above the neutral axis will be
shortened and those below the neutral layer will b lengthened. Therefore all the
material/fibers above the neutral axis will be in compressive strain and hence
compressive stress and Those below the neutral axis will be in tensile strain and
hence tensile stress.
In case of negative bending as in a cantilever the upper fibers will be in tension
and the lower in compression. It must be appreciated that a vertical plane cross
section of a beam before bending remains plane (i.e. flat) after bending.
Concave upward Convex upward
Positive bending
caused by a positive
BM
Negative bending
caused by negative
BM

8. In the diagram above ‘AB’ and ‘CD’ represent two cross sections of the beam .
These are assumed to be very close together so that the beam layers between
them may be assumed to deflect to the arcs of concentric circles. In the latter
figure ‘NA’ and ‘LC’ are straight lines The beam layers are subjected to a
shortening in length from length ‘NL to ‘AC’ in a proportionate manner as we
proceed upward from ‘NL’. The amount of shortening is directly proportionate
to the distance of the particular layer from the neutral layer.

9. As all the beam layers between section ‘AB’ and ‘CD’ were the same
length in the unbent beam, it will be clear the ‘strain’ will be proportional
to the distance from ‘NL’.
The ‘strain variation’ diagram will therefore be linear in character as
shown in the diagram.
Applying Hooks Law that the stress is proportional to strain the the ‘stress
variation diagram’ will also be linear.
This must be clearly understood.
The stress in any fiber in a beam cross section is proportional to its
distance from the neutral axis of the section.

10. The rectangular beam has breadth ‘b’ and depth ‘d’ cm. The applied bending
moment at section AB of the beam induces a maximum stress – in the extreme
upper and lower fibers of the section – of ‘f’ N/cm².
N A
b cm
d/2 cm
d cm
bd/2 cm²
bd/2 cm²
fN/cm²
2d/3
T
C
fN/cm²

11. N A
b cm
d/2 cm
d cm
bd/2 cm²
bd/2 cm²
fN/cm²
2d/3
T
C
fN/cm²
We may consider the beam section to be composed of very large number of thin
horizontal of equal width and depth. The stress acting on each strip will depend on the
position of the strip with reference to the Neutral Axis of the section. The load carried
by a strip will be stress x area. The load variation diagram will be the same as the
‘stress variation diagram’ as the elemental strips are of equal area. The system of loads
acting on the top half –section of the beam will have a resultant ‘C’ N and a resultant
pull ‘T’ N will act upon the bottom half-section. ‘C’ and ‘T’ are equal forces and they
form a couple of moment ‘C’ or ‘T’ x ‘arm of the couple’. The moment of this couple is
the moment of resistance of the beam section.

12. Value of C (or T)
The stress on the top half-section of the beam section varies uniformly
from ‘f’ N/mm² to ‘zero’. The total load ‘C’ can be calculated as the
average x area of half-section = f/2 N/mm² x bd/2 mm² = fbd/4 N = T.
Value of ‘arm of couple’ ‘C’ will act through the center of gravity of the
stress triangle which is 1/3 distance From the edge. Similarly ,T, will be
acting at d/6 from the bottom edge. Hence the distance between ‘C’ and
‘T’ will be 2d/3 = arm of the couple
Moment of couple
Moment = force x arm
= fbd/4 x 2d/3 = fbd²/6 Nmm
This moment is balanced by the moment of resistance for equilibrium
Or BM or MR = fbd²/6 in any units
This is generally stated as M= fbd²/6

13. Allowance for self weight of beams
The self weight of the beam needs to be always considered.
It is a UDL producing a bending moment of Wl/8at the
center of the span in the case of simply supported beams.
However, in problems it is sometimes overlooked as
it is very small compared to the superimposed loads and
also in the safety factors applied this can be covered by the
loading factors.
In the preliminary analysis the weight of the beam can only
be estimated as the size is not known but is to be
calculated. Nevertheless after the design the self weight of
the beam must be added and the calculations done once
again to check that the beam is safe.

14. A simply supported timber beam 5cm wide and 15 cm deep carries a
UDL of total value of 1kN. The effective span of the beam is 2.4m.
Calculate the maximum stress induced in the beam.
Max. stress will be at the mid span
M = fbd²/6 M= Wl/8
We have to find f
M= Wl/8 = (1000 x240) / 8 = 30000Ncm
30000 = (fx5x15x15)/6
f = 160N/cm²

15. Calculate the safe central point load for a simply supported timber beam of 10
cm by 30 cm and an effective span of 3m if the maximum permissible stress
Is 200N/cm²
Let the safe central point load be W
M = fbd²/6
= (200x10x30x30)/6 = 300000Ncm or 300kNcm
MR = BMMAX = Wl/4
= (W x 300)/4kNcm
W = (300x4) / 300 = 4kN

16. A concrete beam 15cm wide and 30 cm deep has an effective span
of 2m.
Assuming the density of concrete to be 2400N/m³ calculate the
central point load the beam can carry in addition to its own weight.
The maximum tensile stress in the concrete is to be limited to
20N/cm²
Here f = 20N/cm²
Self weight of beam = .15 x .30 x 2 x 2400 = 216N
Max BM by self weight = Wl/8 = (216 x 200) / 8 = 5400 Ncm
MR = fbd²/6 = (20 x 15 x 30 x 30) / 6 = 45000Ncm
BM that can be carried by the additional load = 45000 – 5400 Ncm
= 39600Ncm
BM by point load = Wl/4 or W = (BM x 4) / l = (39600 x4) / 200 = 792N