lectures in theory of structure

1. Theory of Structures
3rd Lecture (9/10/2012):
Analysis of Determinate
Beams;
Shear & Moment Diagrams
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 1

2. Beams
• Members that are slender and support loads
applied perpendicular to their longitudinal axis.
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 2
Span, L
Distributed Load, w(x) Concentrated Load, P
Longitudinal
Axis

3. Types of Beams
• Depends on the support configuration
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 3
M
Fv
FH
Fixed
FV
FV
FH
Pin
Roller
PinRoller
FVFV
FH

4. Statically Indeterminate Beams
• Can you guess how we find the “extra”
reactions?
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 4
Continuous Beam
Propped Cantilever
Beam

5. Internal Loadings developed in a structure member
• in order to properly design structural components, we
must know the distribution of the internal forces within
the component.
• In this lecture, we will determine the normal (axial)
force, shear, and moment at a point in a structural
component.
• Internal loads at a point within a structural component.
For a coplanar structure, the internal load at a specified
point will consist of a normal (Axial) force, N, a shear
force, V, and a bending moment, M.
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org
5

6. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org
6
Positive normal (Axial) force, N, tends to elongate the
components. Again, note that the normal (axial) forces
act in opposite directions on either side of the cut.
Positive shear, V, tends to rotate the component
clockwise. Note that the shear is in opposite directions
on either side of a cut through the component.
Positive moment, M, tends to deform the component into
a dish-shaped configuration such that it would hold
water. Again, note that the moment acts in opposite
directions on either side of the cut.

7. Internal Reactions in Beams
• At any cut in a beam, there are 3 possible
internal reactions required for equilibrium:
– Normal (Axial) force,
– Shear force,
– Bending moment.
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org
7
L
P
a b

8. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 8

9. Internal Reactions in Beams
• At any cut in a beam, there are 3 possible
internal reactions required for equilibrium:
– normal force,
– shear force,
– bending moment.
Dr Hussein M. Al.Khuzaie;
husseinma@coe-muth.org
9
Pb/L
x
Left Side of Cut
V
M
N
Positive Directions
Shown!!!

10. Internal Reactions in Beams
• At any cut in a beam, there are 3 possible
internal reactions required for equilibrium:
– normal force,
– shear force,
– bending moment.
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 10
Pa/L
L - x
Right Side of CutVM
N
Positive Directions
Shown!!!

11. Finding Internal Reactions
• Pick left side of the cut:
– Find the sum of all the vertical forces to the left of
the cut, including V. Solve for shear, V.
– Find the sum of all the horizontal forces to the left
of the cut, including N. Solve for axial force, N.
It’s usually, but not always, 0.
– Sum the moments of all the forces to the left of
the cut about the point of the cut. Include M.
Solve for bending moment, M
• Pick the right side of the cut:
– Same as above, except to the right of the cut.
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org
11

12. Example: Find the internal reactions at points
indicated. All axial force reactions are zero. Points
are 2-ft apart.
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 12
20 ft
P = 20 kips
12 kips8 kips
12 ft
1
7
10
6
2 3 94 5 8
Point 6 is just left of P and Point 7 is just right of P.

13. 20 ft
P = 20 kips
12 kips8 kips
12 ft
1
7
10
6
2 3 94 5 8
V
(kips)
M
(ft-kips)
8 kips
-12 kips
96
48
64
48
72
24
80
16
32
x
xDr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 13

14. 20 ft
P = 20 kips
12 kips8 kips
12 ft
V
(kips)
M
(ft-kips)
8 kips
-12 kips
96 ft-kips
x
x
V & M Diagrams
What is the slope
of this line?
a
b
c
96 ft-kips/12’ = 8 kips
What is the slope
of this line?
-12 kips
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org
14

15. 20 ft
P = 20 kips
12 kips8 kips
12 ft
V
(kips)
M
(ft-kips)
8 kips
-12 kips
96 ft-kips
x
x
V & M Diagrams
a
b
c
What is the area of
the blue rectangle?
96 ft-kips
What is the area of
the green rectangle?
-96 ft-kips
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org
15

16. Draw Some Conclusions
• The magnitude of the shear at a point equals
the slope of the moment diagram at that
point.
• The area under the shear diagram between
two points equals the change in moments
between those two points.
• At points where the shear is zero, the moment
is a local maximum or minimum.
Dr Hussein M. Al.Khuzaie;
husseinma@coe-muth.org
16

17. 




dxxVxM
dxxwxV
functionloadthexw
)()(
)()(
)(
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 17
The Relationship Between Load, Shear and
Bending Moment

18. Shear Force & Bending Moment Diagram
Structure under loads
w
dx
dV

V
dx
dM

Bending moment diagram
Shear force diagram
Slope of V = load
Slope of M = V
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 18

19. Load
0 Constant Linear
Shear
Constant Linear Parabolic
Moment
Linear Parabolic Cubic
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org
19
Common Relationships

20. Load
0 0 Constant
Shear
Constant Constant Linear
Moment
Linear Linear Parabolic
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org
20
Common Relationships
M

21. Example: Draw Shear & Moment diagrams
for the following beam
Dr Hussein M. Al.Khuzaie;
husseinma@coe-muth.org
21
3 m 1 m1 m
12 kN 8 kN
A C
B
D
RA = 7 kN  RC = 13 kN 

22. 3 m 1 m1 m
12 kN
A C
B
D
V
(kN)
M
(kN-m)
7
-5
8
8 kN
7
-15
8
7
-82.4 m
Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org
22

23. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org
23
Because the form of the loading does not
change anywhere along the beam, single
equations will suffice for moment and shear:
∑Fy=0; 75-10*x-20/(2*9)*x^2
∑M=0; 75*x-5*x^2-(10/27)*x^3
The reactions are VL = 75kN, VR =
105kN, and HL = 0.
75 kN 105 kN

24. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org
24
-105
-85
-65
-45
-25
-5
15
35
55
75
0 1 2 3 4 5 6 7 8 9
Shear Diagram
0
20
40
60
80
100
120
140
160
180
200
220
0 1 2 3 4 5 6 7 8 9
Shear and
moment are
plotted :

25. Dr Hussein M. Al.Khuzaie; husseinma@coe-
muth.org
25
Examples

26. Dr Hussein M. Al.Khuzaie; husseinma@coe-
muth.org 26

27. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 27
4- Develop equations for shear and moment as a function
of position for the following structural components. Plot
the functions.
9 m

28. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 28
Any questions?