ENT151/4
STATICS
CHAPTER 7
INTERNAL FORCES
7.1 Internal Forces Developed in Structure Members
Consider the simply supported beam which is subjected to the external forces F1 and F2
and support reactions AX, AY and BY.
If the internal loadings acting on the cross section at C are to be determined, then an
imaginary sections is passed through the beam, cutting it into two segments. By doing
this, the internal loadings at the section become external on the free body diagram.
Since both segment (AC and CB) were in equilibrium before the beam was sectioned,
equilibrium of each segment is maintained provided rectangular force components, Nc
and Vc, and a resultant couple moment MC are develop at the section.
Note that these loading must be equal in magnitud and opposite in directions on each of
the segment.
Direct solution – Nc : ∑Fx = 0 ;
Vc : ∑Fy = 0 ;
Mc : ∑Mo = 0
MC is determined by summing moments about point C,
∑MC = 0 ;
in order to eliminate the moments of unknowns NC and VC.
1
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
- where
V = shear force
N = Normal force
M = Bending Moment
My = torsional / twisting moment
All resultant force /moment acts on the center of centroid (c) of the section cross area.
Free Body Diagram
Frame and machine
- multi force members
-subjected to internal normal, shear and bending loadings.
Instead, we must dismember the frame and determine the reaction.
Note : We cannot apply the three equation of equilibrium to obtain these nine unknown.
2
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
Analysis
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Before the member is ‘cut’, determined the all members support reaction.
Keep all the forces, moments at exact locations.
Draw FBD of the segment that has the least number of loads on it (x,y,z components)
Coplaner system only N, V & M
Moments should summed at section passing through centre
EXAMPLES
Example 7-1 (pp 327)
Example 7-2 (pp 328)
Example 7-3 (pp 329)
Example 7-4 (pp 330)
Example 7-5 (pp 331)
Example 7-6 (pp 332)
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Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
7.2
Shear and moment equations and diagram
Beams are structural members which are design to support loadings applied
perpendicular to their axes. In general, beams are long, straight bars, constant crosssectional areas. Often they are classified as to how they are supported.
Simply supported beam – is pinned at one end and roller support at the other.
Centilever beam – is fixed at one end and free at the other.
The actual design of a beam - the variation of the
V
- internal shear force / shear diagram
M
- bending moment / bending moment diagram, acting at each point along the axis
of the beam.
The variation of V and M as functions of the position x along the beam’s axis can be
obtain by using the method of sections. It is necessary to section the beam at an arbitrary
distance x from one end rather then at a specified point.
If the result are plotted, the graphical variation of V and M as functions of x are termed
the shear diagram and bending moment diagram. Because of the internal shear and
bending moment functions will be discontinous, the functions must be determined for
each segment of the beam located between any two discontinuities of loading.
4
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
Procedure for analysis
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Determined all the reactive forces and couple moment acting on the beam, and
resolve all the forces into components acting perpendicular and parallel to the
beam axis.
Specify separate coordinates and having an origin at the beam’s left-end
Section the beam perpendicular to the axis.
Shear V-summing forces perpendicular to the axis
Moment M-slimming moments about section end.
Plot the shear diagram (V vs x) and moment diagram (M vs x)
EXAMPLES
Example 7-7 (pp 342)
Example 7-8 (pp 343)
7.3 Relations Betwen Distributed Load, Shear and Moment
Shear and moment function
Specify separates cordinates x having an origin at the beam’s left end and extending to
region of the beam between the concentrated force and/or couple moment, or where there
is no discontinuity of distributed loading.
Section the beam if its axis at each distance x and draw the free body diagram of one of
the segments. Be sure V and M are shown acting in their positive sense.
The shear V obtained by summing forces perpendicular to the beam’s axis.
The moment M is obtained by summing moments about the sectioned end of the
segment.
Example :
∑Fy = 0 ;
V = 2.5 kN
∑M = 0 ;
M = 2.5 x kN.m
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Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
A free body diagram for the left segment of the shaft extending a distance x within the
region BC
∑Fy = 0 ;
∑M = 0 ;
V = 2.5 kN
M = (10 - 2.5 x )kN.m
Shear and bending moment diagram
Plot the shear diagram ( V versus x) and the moment diagram ( M versus x). If compute
values of the function describing V and M are positive, the values are plotted above the x
axis, whereas the negative values are plotted below the x axis.
Generally, it is convenient to plot the shear and bending-moment diagrams directly below
the free body diagram of the beam.
Example:
The shear diagram indicates that the internal shear force always 2.5 kN (positive) within
shaft segment AB. Just to the right of point B, the shear force changes sign and remains
at the constant value of -2.5 kN for segment BC.
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Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
The moment diagram start at zero, increase linearly to point B at x = 2 m, where Mmax =
2.2 kN (2 m ) = 5kN.m, then thereafter decrease back to zero.
EXAMPLES
Example 7- 9 (pp 351)
Example 7-10 (pp 352)
Example 7-11 (pp 353)
Example 7-12 (pp 354)
EXERCISES
Exercise 7 - 2
Exercise 7 - 4
Exercise 7 - 10
Exercise 7 - 33
Exercise 7 - 49
Exercise 7 - 59
Exercise 7 - 86
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Zol Bahri Razali
School of Mechatronics, KUKUM