Chapter 6:
Internal Forces
Engineering Mechanics: Statics
Chapter Objectives
To show how to use the method of sections
for determining the internal loadings in a
member.
 To generalize this procedure by formulating
equations that can be plotted so that they
describe the internal shear and moment
throughout a member.

Chapter Outline
 Internal
Forces Developed in
Structural Members
 Shear and Moment Equations and
Diagrams
 Relations between Distributed
Load, Shear and Moment
7.1 Internal Forces Developed in
Structural Members
The design of any structural or
mechanical member requires the
material to be used to be able to resist
the loading acting on the member
 These internal loadings can be
determined by the method of sections

7.1 Internal Forces Developed in
Structural Members



Consider the “simply supported” beam
To determine the internal loadings acting on the cross
section at C, an imaginary section is passed through
the beam, cutting it into two
By doing so, the internal loadings become external on
the FBD
7.1 Internal Forces Developed in
Structural Members
Since both segments (AC and CB) were in
equilibrium before the sectioning, equilibrium of
the segment is maintained by rectangular force
components and a resultant couple moment
 Magnitude of the loadings is determined by the
equilibrium equations

7.1 Internal Forces Developed in
Structural Members
Force component N, acting normal to the
beam at the cut session and V, acting t
angent to the session are known as normal
or axial force
and the shear force
 Couple moment M is
referred as the bending
moment

7.1 Internal Forces Developed in
Structural Members
For 3D, a general internal force and couple
moment resultant will act at the section
 Ny is the normal force, and Vx and Vz are
the shear components
 My is the torisonal or
twisting moment, and
Mx and Mz are the
bending moment
components

7.1 Internal Forces Developed in
Structural Members
For most applications, these resultant
loadings will act at the geometric
center or centroid (C) of the section’s
cross sectional area
 Although the magnitude of each
loading differs at different points along
the axis of the member, the method of
section can be used to determine the
values

7.1 Internal Forces Developed in
Structural Members
Free Body Diagrams


Since frames and machines are composed of
multi-force members, each of these members will
generally be subjected to internal shear, normal
and bending loadings
Consider the frame with the blue
section passed through to
determine the internal loadings
at points H, G and F
7.1 Internal Forces Developed in
Structural Members
Free Body Diagrams



FBD of the sectioned frame
At each sectioned member, there is an unknown
normal force, shear force and bending moment
3 equilibrium equations cannot be used
to find 9 unknowns, thus dismember
the frame and determine
reactions at each connection
7.1 Internal Forces Developed in
Structural Members
Free Body Diagrams
Once done, each member may be sectioned at its
appropriate point and apply the 3 equilibrium
equations to determine the unknowns
Example
 FBD of segment DG can be used to determine
the internal loadings at G
provided the reactions of
the pins are known

7.1 Internal Forces Developed in
Structural Members
Procedure for Analysis
Support Reactions
 Before the member is cut or sectioned,
determine the member’s support reactions
 Equilibrium equations are used to solve for
internal loadings during sectioning of the
members
 If the member is part of a frame or machine, the
reactions at its connections are determined by
the methods used in 6.6
7.1 Internal Forces Developed in
Structural Members
Procedure for Analysis
Free-Body Diagrams
 Keep all distributed loadings, couple
moments and forces acting on the member
in their exact locations, then pass an
imaginary section through the member,
perpendicular to its axis at the point the
internal loading is to be determined
 After the session is made, draw the FBD of
the segment having the least loads
7.1 Internal Forces Developed in
Structural Members
Procedure for Analysis
Free-Body Diagrams
 Indicate the z, y, z components of the force and
couple moments and the resultant couple
moments on the FBD
 If the member is subjected to a coplanar system
of forces, only N, V and M act at the section
 Determine the sense by inspection; if not,
assume the sense of the unknown loadings
7.1 Internal Forces Developed in
Structural Members
Procedure for Analysis
Equations of Equilibrium
 Moments should be summed at the section about
the axes passing through the centroid or
geometric center of the member’s cross-sectional
area in order to eliminate the unknown normal
and shear forces and thereby, obtain direct
solutions for the moment components
 If the solution yields a negative result, the sense
is opposite that assume of the unknown loadings
7.1 Internal Forces Developed in
Structural Members
The link on the backhoe is a
two force member
 It is subjected to both
bending and axial load at its
center
 By making the member
straight, only an axial force
acts within the member

7.1 Internal Forces Developed in
Structural Members
Example 7.1
The bar is fixed at its end and is
loaded. Determine the internal normal
force at points B and C.
7.1 Internal Forces Developed in
Structural Members
Solution
Support Reactions
 FBD of the entire bar
 By inspection, only normal force Ay
acts at the fixed support
 Ax = 0 and Az = 0
+↑∑ Fy = 0; 8kN – NB = 0
NB = 8kN
7.1 Internal Forces Developed in
Structural Members
Solution
 FBD of the sectioned bar
 No shear or moment act on
the sections since they are
not required for equilibrium
 Choose segment AB and
DC since they contain the
least number of forces
7.1 Internal Forces Developed in
Structural Members
Solution
Segment AB
+↑∑ Fy = 0;
Segment DC
+↑∑ Fy = 0;
8kN – NB = 0
NB = 8kN
NC – 4kN= 0
NC = 4kN
7.1 Internal Forces Developed in
Structural Members
Example 7.2
The circular shaft is subjected to three
concentrated torques. Determine the internal
torques at points B and C.
7.1 Internal Forces Developed in
Structural Members
Solution
Support Reactions
 Shaft subjected to only collinear torques
∑ Mx = 0;
-10N.m + 15N.m + 20N.m –TD = 0
TD = 25N.m
7.1 Internal Forces Developed in
Structural Members
Solution
 FBD of shaft segments AB and CD
7.1 Internal Forces Developed in
Structural Members
Solution
Segment AB
∑ Mx = 0;
0
-10N.m + 15N.m – TB =
TB = 5N.m
Segment CD
∑ Mx = 0;
TC – 25N.m= 0
TC = 25N.m
7.2 Shear and Moment
Equations and Diagrams




Beams – structural members designed to support
loadings perpendicular to their axes
Beams – straight long bars with constant crosssectional areas
A simply supported beam is pinned at one end
and roller supported at
the other
A cantilevered beam is
fixed at one end and free
at the other
7.2 Shear and Moment
Equations and Diagrams
For actual design of a beam, apply
- Internal shear force V and the bending moment
M analysis
- Theory of mechanics of materials
- Appropriate engineering code to determine
beam’s required cross-sectional area
 Variations of V and M obtained by the method of
sections
 Graphical variations of V and M are termed as
shear diagram and bending moment diagram

7.2 Shear and Moment
Equations and Diagrams
Internal shear and bending moment
functions generally discontinuous, or their
slopes will be discontinuous at points
where a distributed load changes or where
concentrated forces or couple moments
are applied
 Functions must be applied for each
segment of the beam located between any
two discontinuities of loadings
 Internal normal force will not be
considered

7.2 Shear and Moment
Equations and Diagrams
Load applied to a beam act
perpendicular to the beam’s axis and
hence produce only an internal shear
force and bending moment
 For design purpose, the beam’s
resistance to shear, and particularly to
bending, is more important than its
ability to resist a normal force

7.2 Shear and Moment
Equations and Diagrams
Sign Convention
To define a positive and negative shear
force and bending moment acting on the
beam
 Positive directions are denoted by an
internal shear force that causes clockwise
rotation of the member on which it acts
and by an internal moment that causes
compression or pushing on the upper part
of the member

7.2 Shear and Moment
Equations and Diagrams
Sign Convention
A positive moment
would tend to bend the
member if it were
elastic, concave upwards
 Loadings opposite to the
above are considered
negative

7.2 Shear and Moment
Equations and Diagrams
Procedure for Analysis
Support Reactions
 Determine all the reactive forces and
couple moments acting on the beam’
 Resolve them into components acting
perpendicular or parallel to the beam’s
axis
7.2 Shear and Moment
Equations and Diagrams
Procedure for Analysis
Shear and Moment Reactions
 Specify separate coordinates x having an origin
at the beam’s left end and extending to regions
of the beams between concentrated force and/or
couple moments or where there is no continuity
of distributed loadings
 Section the beam perpendicular to its axis at
each distance x and draw the FBD of one of the
segments
7.2 Shear and Moment
Equations and Diagrams
Procedure for Analysis
Shear and Moment Reactions
 V and M are shown acting in their positive sense
 The shear V is obtained by summing the forces
perpendicular to the beam’s axis
 The moment M is obtained by summing
moments about the sectioned end of the
segment
7.2 Shear and Moment
Equations and Diagrams
Procedure for Analysis
Shear and Moment Diagrams
 Plot the shear diagram (V versus x) and the
moment diagram (M versus x)
 If computed values of the functions describing V
and M are positive, the values are plotted above
the x axis, whereas negative values are plotted
below the x axis
 Convenient to plot the shear and the bending
moment diagrams below the FBD of the beam
7.2 Shear and Moment
Equations and Diagrams
Example 7.7
Draw the shear and bending moments
diagrams for the shaft. The support at A is a
thrust bearing and the support at C is a
journal bearing.
7.2 Shear and Moment
Equations and Diagrams
Solution
Support Reactions
 FBD of the shaft
7.2 Shear and Moment
Equations and Diagrams
Solution
   Fy  0;V  2.5kN
 M  0; M  2.5 xkN.m
7.2 Shear and Moment
Equations and Diagrams
Solution
   Fy  0;2.5kN  5kN  V  0
V  2.5kN
 M  0; M  5kN ( x  2m)  2.5kN ( x)  0
M  (10  2.5 x)kN.m
7.2 Shear and Moment
Equations and Diagrams
Solution
Shear diagram
 internal shear force is always
positive within the shaft AB
 Just to the right of B, the shear
force changes sign and remains
at constant value for segment
BC
Moment diagram
 Starts at zero, increases linearly
to B and therefore decreases to
zero
7.2 Shear and Moment
Equations and Diagrams
Solution
 Graph of shear and moment
diagrams is discontinuous at
points of concentrated force
ie, A, B, C
 All loading discontinuous are
mathematical, arising from
the idealization of a
concentrated force and
couple moment
7.3 Relations between Distributed
Load, Shear and Moment
Distributed Load
Consider beam AD subjected to an arbitrary
load w = w(x) and a series of concentrated
forces and moments
 Distributed load assumed positive when loading
acts downwards

7.3 Relations between Distributed
Load, Shear and Moment
Distributed Load


A FBD diagram for a small
segment of the beam having a
length ∆x is chosen at point x
along the beam which is not
subjected to a concentrated force
or couple moment
Any results obtained will not apply
at points of concentrated loadings
7.3 Relations between Distributed
Load, Shear and Moment
Distributed Load


The internal shear force and bending
moments shown on the FBD are
assumed to act in the positive sense
Both the shear force and moment
acting on the right-hand face must
be increased by a small, finite
amount in order to keep the
segment in equilibrium
7.3 Relations between Distributed
Load, Shear and Moment
Distributed Load

The distributed loading has been replaced by a
resultant force ∆F = w(x) ∆x that acts at a
fractional distance k (∆x) from the right end,
where 0 < k <1
   Fy  0;V  w( x)x  (V  V )  0
V   w( x)x
 M  0;Vx  M  w( x)xk x   ( M  M )  0
M  Vx  w( x)k (x) 2
7.3 Relations between Distributed
Load, Shear and Moment
Distributed Load
dV
  w( x)
dx
Slope of the
shear diagram
Slope of
=
dM
V
dx
=
Negative of
distributed load intensity
Shear moment diagram
7.3 Relations between Distributed
Load, Shear and Moment
Distributed Load




At a specified point in a beam, the slope of the
shear diagram is equal to the intensity of the
distributed load
Slope of the moment diagram = shear
If the shear is equal to zero, dM/dx = 0, a point
of zero shear corresponds to a point of maximum
(or possibly minimum) moment
w (x) dx and V dx represent differential area
under the distributed loading and shear diagrams
7.3 Relations between Distributed
Load, Shear and Moment
Distributed Load
VBC    w( x)dx
Change in =
shear
Area under
shear diagram
M BC   Vdx
Change in =
moment
Area under
shear diagram
7.3 Relations between Distributed
Load, Shear and Moment
Distributed Load



Change in shear between points B and C is
equal to the negative of the area under the
distributed-loading curve between these
points
Change in moment between B and C is equal
to the area under the shear diagram within
region BC
The equations so not apply at points where
concentrated force or couple moment acts
7.3 Relations between Distributed
Load, Shear and Moment
Force


FBD of a small segment of
the beam
   Fy  0; V  F
Change in shear is negative
thus the shear will jump
downwards when F acts
downwards on the beam
7.3 Relations between Distributed
Load, Shear and Moment
Force

FBD of a small segment of the
beam located at the couple
moment
 M  0; M  M O

Change in moment is positive
or the moment diagram will
jump upwards MO is clockwise
7.3 Relations between Distributed
Load, Shear and Moment
Example 7.9
Draw the shear and moment diagrams for the
beam.
7.3 Relations between Distributed
Load, Shear and Moment
Solution
Support Reactions
 FBD of the beam
7.3 Relations between Distributed
Load, Shear and Moment
Solution
Shear Diagram
V = +1000 at x = 0
V = 0 at x = 2
Since dV/dx = -w = -500, a straight negative sloping
line connects the end points
7.3 Relations between Distributed
Load, Shear and Moment
Solution
Moment Diagram
M = -1000 at x = 0
M = 0 at x = 2
dM/dx = V, positive yet linearly decreasing from
dM/dx = 1000 at x = 0 to dM/dx = 0 at x = 2