Structural Analysis (II)
Stresses and Deformations
Chapter 1
Strain
Gauge
Cantilever
Beam
A cantilever beam with strain gauges to measure strain (and so stress) near the fixed end
Straining
Actions
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Dr. M. Abdel-Kader
Structural Analysis (II)
Stresses and Deformations
Failure due to Bending
Failure due to Shear
Failure due to Normal force (compression) - very few stirrups
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Dr. M. Abdel-Kader
Structural Analysis (II)
Stresses and Deformations
Chapter 1
Straining Actions
1.1 Introduction
To design any structural member, it is necessary to determine the
straining actions (internal forces) that act at some critical sections.
Straining actions are generated within loaded structural members
(elements). These forces are generated within every type of element;
if they were not developed, the structure would fail. These internal
forces are known as Moment, Shear, and Normal forces. The
normal force is found in columns and beams with an axial load.
Shear and moment are found in beams and frames.
The internal loads can be determined by the method of sections. A
member is cut at the point of interest, and the internal forces are
revealed as equivalent external loads. Since the member was in
equilibrium before being cut, these equivalent external loads must
keep the member in equilibrium.
y
In three dimensions (3D), internal
forces have three components (one
normal N and two shear Vx and Vy
or Qx and Qy or Sx and Sy) and three
moment components (one twisting
Mz or Mt or T and two bending Mx
and My). These forces and
moments are shown in Fig.1.1.
For the two dimensions (2D) case,
there are two forces, an axial force
N and a shear force Vy or Qy, plus
one moment Mx (Fig.1.2).
Vy
y V
x
x
x
N z
z
Fig.1.1 Internal forces in 3D
y
Vy
x
x
N
Fig. 1.2 Internal forces in 2D
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Dr. M. Abdel-Kader
z
Structural Analysis (II)
Stresses and Deformations
In order to quantify the internal forces of any part of a structure, a
FBD (Free Body Diagram) must be drawn for only that specific part
of the structure.
For beams, the internal forces (N, V, and M) actually represents the
resultant of the stress distribution acting over the cross-sectional
area of the beam at the cut section. Once the resultant internal forces
are known the magnitude of the stress can be determined. This
chapter will discuss briefly how to determinate these internal forces
at specified sections along beam's axis and how to draw the
variation graphically.
If the beam shown in Fig.1.3 is cut into two parts by a section s-s.
F3
F2
F1
s
s
RA
RB
Figure 1.3 Beam subjected to several loads.
Since the beam as a whole is in equilibrium then every part of it
must be in equilibrium. The left part must be in equilibrium under
forces F1, RA and the internal forces N, V (or Q), and M which are
the action of the right part on the left part (Fig.1.4).
F2
F1
s
M
N
RA
M
F3
s
N
V
V
RB
Figure 1.4 Internal forces at a section of a beam
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Dr. M. Abdel-Kader
Structural Analysis (II)
Stresses and Deformations
1.2 Normal forces
The normal force N at a section is the sum of all the
horizontal forces acting to the left (or to the right) of that
section. N is positive when the sum is out of the section.
Example 1.1:
For the shown cantilever which
carries a horizontal force at a
and an inclined force at c,
determine the normal force N at
the sections b and d.
10 kN
+
Normal Sign
3
4
2 kN
a
1m
Solution:
First resolve the inclined force
(10 kN) to horizontal component
(10 x 3 / 5 = 6 kN) and vertical
component (10x4/5 = 8 kN), then
determine the reactions.
c
b
1m
e
d
1m
1m
8 kN
2 kN
4 kN
6 kN
a
c
b
1m
1m
e
d
1m
1m
16 kN.m
8 kN
At the section b: (Note that the out is positive)
Nb = the sum of all the horizontal
forces acting to the left of b
= +2 kN

Nb
2 kN
a
Nb = 2 kN Tension
b
1m
At the section d: (Note that the out is positive)
Nd = the sum of all the horizontal
Left Part
forces acting to the left of d
= +2 – 6 = – 4 kN
 Nd = 4 kN Compression
or,
Nd = the sum of all the horizontal
forces acting to the right of d
= – 4 kN
 Nd = 4 kN Compression
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8 kN
2 kN
Nd
6 kN
a
c
b
1m
1m
d
1m
Left Part
Dr. M. Abdel-Kader
Structural Analysis (II)
Stresses and Deformations
1.3 Shear forces
The shear force V at a section is the sum of all the vertical
forces acting to the left (or to the right) of that section. V is
positive when the sum from left is up (or when it rotates
clockwise).
Example 1.2:
For the shown simple beam
which
carries
triangular
distributed load, determine
the internal shear force at the
section c.
+
Shear Sign
w = 9 kN/m
a
b
c
2m
4m
27 kN
Solution:
To determine the reaction,
replace the distributed load by
its equivalent point force =
(9) (6) / 2 = 27kN downward.
The reactions are as shown in
the figure.
a
c
b
4m
9 kN
18 kN
4m
2m
3 kN
To determine the internal
wx =3 kN/m
a
shear force at the section c,
w
c
wx
keep the original distributed
9 kN
x
load on the beam and replace
L
2m
only the segment to the left
Note that the intensity of the
(or to the right) of c by its
triangular load at the section c
is found by proportion, i.e.;
equivalent point force =(3)(2) / 2
wx/x = w /L = 9/6 = 1.5
= 3 kN.
or wx = 1.5x = 1.5(2) =3 kN/m
Vc = the sum of all the vertical
forces acting to the left of c
=+9–3=6
Vc = 6 kN
4(9+3)/2 =24kN
+
w = 9 kN/m
Shear Sign
or,
Vc = the sum of all the vertical
wx = 3 kN/m
b
c
forces acting to the right of c
= + 24 – 18 = 6
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Vc = 6 kN
18 kN
4m
6
Dr. M. Abdel-Kader
Structural Analysis (II)
Stresses and Deformations
1.4 Bending moments
The bending moment M at a section is the sum of all the
couples and moments of all the forces acting to the left (or
to the right) of the section. M is positive when the sum from
left rotates clockwise (or when the sum from right rotates
anti-clockwise).
Example 1.3:
For the shown cantilever
which carries three point
loads, calculate the bending a
moments at the points a, c, d,
and b.
4 kN
6 kN
c
2m
+
Moment Sign
5 kN
b
d
2m
2m
Solution:
- At point a: the bending moment is equal to zero.
At point a Ma = 0
- At point c: only the 4kN force is to the left of point c.
The 4kN force is at a distance = 2 m from point c.
So, the bending moment = force x distance = -4(2) = 8 kN.m 
At point c Mc = 8 kN.m
- At point d: the 4kN force and the 6kN force are to the left of
point d.
The 4kN force is at a distance = 4 m from point d
The 6kN force is at a distance = 2 m from point d.
So, the bending moment = -4(4) - 6(2) = - 28 kN.m = 28 kN.m 
At point d Md = 28 kN.m
- At point b: the 4kN force, the 6kN force, and the 5kN force are to
the left of point b.
The 4kN force is at a distance = 6 m from point b
The 6kN force is at a distance = 4 m from point b.
The 5kN force is at a distance = 2 m from point b.
So, the bending moment = -4(6) - 6(4) - 5(2) = 58 kN.m 
At point b Mb = 58 kN.m
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Dr. M. Abdel-Kader
Structural Analysis (II)
Stresses and Deformations
1.5 Relations between load, shear and moment
There are differential relations that exist between the load, shear,
and moment.
To drive these relations,
consider the beam ab shown
in Fig.1.5, which is subjected a
to varying distributed load w.
The first relation is:
dV
 w
dx
w
b
∆x
x
Figure 1.5 Beam subjected to varying
distributed load
….. (1.1)
(w)(∆x)
V
which states that the slope of
shear force diagram at any
section (dV/dx) is equal to the
negative intensity of the
distributed load w acting on
the beam at that section.
M+∆M
M
O
∆x
V+∆V
Figure 1.6 small segment of
the beam
Equation 1.1 can be integrated from one section to another between
concentrated forces or couples, in which case
V   w dx
………….. (1.2)
which states that the change in the shear between any two sections
on a beam equals the negative area under the distributed load
diagram between the two sections. If w is a curve of degree n, then
V will be a curve of degree n+1. For example, if w is uniform
(constant), V will be linear (straight line).
The second relation is:
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dM
V
dx
………….. (1.3)
8
Dr. M. Abdel-Kader
Structural Analysis (II)
Stresses and Deformations
which states that the slope of bending moment diagram (dM/dx) at
any section is equal to the shear at that section. If the shear is zero at
a section, then dM/dx = 0, and the moment will be a maximum or
minimum at that section.
Also, Equation 1.3 can be integrated from one section to another,
M   V dx
………….. (1.4)
Which states that the change in the moment between the two
sections equals the area under the shear force diagram between the
two sections. If V is a curve of degree n, then M will be a curve of
degree n+1. For example, if V is linear (straight line), M will be
second degree curve (parabolic).
1.6 Internal forces diagrams
The ability to construct normal force, shear force and bending
moment diagrams is an essential tool for the construction engineer.
These diagrams are a graphical representation of N(x), V(x) and
M(x) functions.
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Dr. M. Abdel-Kader
Structural Analysis (II)
Stresses and Deformations
Example 1.4:
For the shown simple beam which
carries a point load at b and an
anti-clockwise moment at c, draw
the S.F.D and B.M.D.
Solution:
The reactions are calculated and
shown in the figure.
30 kN
40 kN.m
a
4m
d
c
b
2m
2m
30 kN
40 kN.m
a
20 kN
b
4m
d
c
2 m 10 kN
2m
Shear Force Diagram, S.F.D:
- Beginning at a the 20 kN force acts
20 kN
upward, so Va = +20.
- No load acts between a and b, so the
shear remains constant.
- At b the 30 kN force is down, so the
0
shear jumps down 30, from 20 to 10.
- Again the shear is constant (no load)
and ends at -10, point d, which
closes the diagram back to zero since
0
the 10 kN force at d acts upward.
20 kN
+
0 S.F.D
-10 kN
-
+
+
-10 kN
0 B.M.D
20 kN.m
Bending Moment Diagram, B.M.D:
- The moment at each end of the beam
60 kN.m
is zero, then
Ma = 0
and
Md = 0
80 kN.m
- The value of the moment at b can be determined from left or right,
From left: Mb = 20(4) = + 80 kN.m = + 80 kN.m 
+
From right: Mb = 10(4) + 40 = + 80 kN.m = + 80 kN.m 
Sign Convention
- The value of the moment at c:
Just to the left of c:
From left: MJust to the left of c = 20(6) – 30(2) = 60 kN.m 
From right: MJust to the left of c = 10(2) + 40 = 60 kN.m 
Just to the right of c:
From left: MJust to the right of c = 20(6) – 30(2) – 40 = 20 kN.m 
From right: MJust to the right of c = 10(2)
= 20 kN.m 
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Dr. M. Abdel-Kader