Shear Force and Bending Moment

Loading condition on the element: Lateral load on the member

y
x
z
Lateral load on a member cause bending in the member
Acting along the
Axial force on the Produces axial deformation
element Force Nx longitudinal axis i.e.,
i.e., deformation in x-direction
x – axis
Force Ny Acting along the
Lateral loading on y - axis Produces bending in
the element Acting about the xy-plane
Moment Mz
z - axis
Acting along the
Lateral loading on Force Nz z - axis Produces bending in
the element Acting about the xz-plane
Moment My y - axis
Axial moment on
Acting about the Produces twisting in
the element Moment Mx x - axis yz-plane
(Torsion)

S I
Prismatic member with R
Rectangular cross section

K L
D N
G C
P J
Q
E F
M

A H B

Prismatic member
ABCD & PQRS represent cross section of member
M-N line is the longitudinal axis (X - axis direction)
GMH and INJ is the lateral direction i.e., in Y – axis direction
EMF and KNL is the lateral direction i.e., in Z – axis direction
GMHJNI plane is the central plane of the member (applied load is in this plane)
GMHJNI i.e., GHJI plane represents the member subjected lateral loading
a) Forces in Y – direction
b) Couple i.e., moment vector about Z – direction

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Beam: A member subjected to lateral loading is called Beam
Geometry of beam: It is represented by length and cross section dimensions
Loading on beam: Coplanar general force system i.e.,
All forces are in the plane of beam (central plane of beam)
Equilibrium equations: ∑ 𝑭𝒙 = 𝟎 , ∑ 𝑭𝒚 = 𝟎 and ∑ 𝑴𝒛 = 𝟎
Types of Loading:
Load Diagram Notation Units
a) Point Load P or N
(Concentrated load) W or or
Q kN
b) Distributed Load N/m
i) Uniformly distributed load (udl) w or
kN/m

ii) Uniformly varying load (uvl)

w
Triangular loading N/m
or
kN/m
Trapezoidal loading w1 and w2

Distributed load is replaced by equivalent concentrated

a) magnitude of equivalent concentrated load

= area of distributed loading diagram
b) location of equivalent concentrated load
= centroid of distributed loading diagram

c)
Couple
N-m
Couple is represented by its
anticlockwise M or
moment vector. This moment vector
or kN-m
is normal to the plane of beam
m
clockwise

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Types of supports:
• all supports are in the xy – plane

W1 W2 W3

A B

W1 X W2 W3

A B
Va x X Rb

a) Translation in z – direction
Degrees of freedom absent: b) Rotation in yz - plane
c) Rotation in zx – plane

a) Translation in x – direction
Degrees of freedom present: b) Translation in y – direction
c) Rotation in xy – plane

Degrees of freedom
a) Fixed Support restrained = 3 Degrees of freedom
(Reactions: allowed = 0
Horizontal, a) Translation in x – direction
Vertical, b) Translation in y – direction
Moment) c) Rotation in xy - plane

b) Simple Support

a) Hinged Degrees of freedom Degrees of freedom

Support restrained = 2 allowed = 1

(Pinned a) Translation in x – direction Rotation in xy –

support) b) Translation in y – direction plane

(Reactions:
Horizontal,
Vertical)
b) Roller support Degrees of freedom Degrees of freedom
restrained = 1 allowed = 2
Translation in y – direction a) Translation in
x-direction
(Reaction: Normal to base
b) Rotation in
of roller)
xy – plane

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Types of Beams:
Coplanar general force system ∑ 𝑭𝒙 = 𝟎 ∑ 𝑭𝒚 = 𝟎 ∑ 𝑴𝒛 = 𝟎
Equilibrium equations
By using the above equations, maximum of 3 unknowns can be solved

1) Statically determinate beams

• All three degrees of freedom are
a) restrained at fixed support A
Cantilever A
B • All three degrees of freedom are
beam allowed at free end B
One end fixed • 3 reactions at fixed end A
another end free VA, HA and MA
• No reactions at free end B
b) • One degree of freedom is allowed at
A B hinged support A
• Two degrees of freedom are allowed
Simply at roller support B
supported One end hinged • Two degrees of freedom are
beam support restrained at hinged support A
Other end roller
• One degree of freedom is restrained
support
at roller support B
• 2 reactions at hinged support
VA and HA
• 1 reaction at roller support RB

Simply Not permitted

supported
beam

Simply Unstable beam

supported Not permitted
beam

c) One hinged support

Overhang One roller support • 2 reactions at hinged support
Beam • 1 reaction at roller support RB

• All three unknowns (support reactions) in a beam can be solved

by using the equilibrium equations

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 2) Statically Indeterminate beams
Static Indeterminacy (static redundancy) of beam =
number of unknowns - number of available equilibrium equations
• 3 reactions at fixed end A
a) Propped VA, HA and MA
Cantilever • 1 reaction at roller support RB
beam
One fixed support • 4 unknowns and three equilibrium
and equations
One roller support
(at any point) • S.I. = 1

b) • Load is inclined
Fixed beam • All three degrees of freedom are
restrained at fixed supports A and B
• 3 reactions at each fixed end

Both ends are with • 6 unknowns and three equilibrium

fixed supports equations
• S.I. = 3

• Load is vertical (Horizontal

component of load is zero).
• HA = HB = 0 therefore equilibrium
equation ∑ 𝑭𝒙 = 𝟎 is not useful.
• Available equilibrium equations = 2
Both ends are with • 2 reactions at each fixed end
fixed supports • 4 unknowns and three equilibrium
equations
• S.I. = 4 – 2 = 2

c) Continuous Beam supported at more than two points

beam Examples:

Static indeterminacy of given beam can be found from the given

supports

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 Shear Force:
Algebraic sum of forces acting normal to the longitudinal axis of beam, either
left or right of section
Bending moment:
Algebraic sum of moments of all forces and couples about the section, either
left or right of section
Shear Force Diagram:
Variation of the shear force along the length of beam is represented
graphically. The ordinate in the shear force diagram at any section represents
the shear force in the beam at that section.
Bending moment Diagram:
Variation of the bending moment along the length of beam is represented
graphically. The ordinate in the bending moment diagram at any section
represents the bending moment in the beam at that section.
Sign convention:
Shear Force Bending moment

Positive sagging Positive

Negative hogging Negative

Procedure to find S.F. and B.M. and to draw the S.F.D. and B.M.D.
a) Consider the beam subjected to the given loading.
b) Introduce the support reactions (unknowns). By using the available
static equilibrium equations, determine these unknown support
reactions.
c) Divide the beam into segments, based on the discontinuity of the
given loading on the beam.
d) Consider a section X1-X1 in the first segment. Cut the beam along
the section X1-X1 and draw the free body diagram either to the left or
right of the section X1-X1. Formulate the shear force and bending
moment expressions at that section. Evaluate the discrete values (at
the end points of segment) of shear force and bending moment. Draw
the S.F.D. and B.M.D. in this segment. The variation of Shear force
and Bending moment in the segment is based on the corresponding
expressions.

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 e) Consider the next section X2-X2 in the 2nd segment. Find the shear
force and bending moment values in this segment by following the
procedure in the above step i.e., (d).
f) Analyse all segments by considering a section in each segment and
find the shear force and bending moment values and complete the
S.F.D. and B.M.D. of the given the beam.

W1 X W2 W3

A B
Va x X Rb

W1
X

Mx

A Vx
Va X

X W2 W3

Mx
Vx
B
X Rb

W1
X
C

T
Vx
Va X

X W2 W3

T
Vx
Rb

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 Loading diagram – Shear force diagram – Bending moment diagram
S. Load Shear Force Bending moment diagram
No. diagram
1) at a point
A) Point load Sudden change at Slope changes at that
(concentrated load) that point point
B) Couple No change Sudden change at that
(moment) point
2) Over a segment
A) No load constant Linear variation
B) Uniformly distributed Linear variation 2nd order variation
load
(u d l)
C) Uniformly varying load 2nd order variation 3nd order variation
(u v l)

Note:
𝒅𝑴
1) = 𝑽𝒙  𝑴 = ∫ 𝑽𝒙 𝒅𝒙
𝒅𝒙

Rate of change of Bending Moment (slope of B.M.D.) is equal to the Shear Force

Area of Shear Force Diagram between two points is the change in Bending Moment
between these two points

𝒅𝑽
2) = −𝒘𝒙 = rate of loading  𝑽 = ∫ 𝒘𝒙 𝒅𝒙
𝒅𝒙

Rate of change of Shear Force (slope of S.F.D.) is equal to rate of loading

Area of loading diagram between two points is the change in S.F. between these
two points

oOo

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