Simplified engineering for architecture students

1. TOPIC 3
STRUCTURAL THEORY RELATED TO
SIMPLE BEAMS
3.0 Beams
3.1 Failure of beams due to bending, shear and deflection
3.2 Bending moments and shear force
3.3 Laws of bending
3.4 Bending moments and shear force diagrams
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2. 3.0 BEAM
 Members that are slender and support loading that are
applied perpendicular to their longitudinal axis. In
general , beams are long, straight bars having a constant
cross-section area. Often they are classified as to how
they are supported. Example :-
i) Simply supported beam
ii) Cantilever beam
iii) Overhanging beam
iv) Fixed end beam
v) Continuous beam
 Support the slabs. Its major function is to resist the
bending moment and shear.
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3. Simply Supported Beam
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A simply supported beam is supported at both ends with loads
applied between the supports.

4. Cantilever Beam
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A cantilever is a beam anchored at only one end. The beam carries the load
to the support where it is forced against by a moment and shear stress.
Cantilever construction allows for overhanging structures without external
bracing. Cantilevers can also be constructed with trusses or slabs.

5. Overhanging Beam
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6. Fixed End Beam
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A fixed or a build in beam has both of its ends rigidly fixed.

7. Continuous Beam
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A continuous beam is a structural component that provides
resistance to bending when a load or force is applied. These beams
are commonly used in bridges. A beam of this type has more than
two points of support along its length.

8. 3. 1 FAILUREOF BEAMSDUETO SHEAR
SHEAR (V)
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 Actually, max shear force (V), occurs at a beam
supports.
 A cracking happens near the support.
Shear failure
Moment failureV max V max

9. 3.1 SHEAR STRESS
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v = V / bvd,
where v = shear stress in N/mm2.
V = shear force in N.
bv = width section of beam in mm.
d = effective depth in mm.
v
link
d
bv

10. MODE OF SHEAR FAILURE
 Mechanisms of shear failure in concrete are very
complicated and complex to understand.
 Effected by a ratio of shear [av/d] give three situated:
Case I : av/d > 6
Case II : 2< av/d < 6
Case III : av/d < 2
 When av increased, av/d become bigger, bending
moment will increase, failure become closer to the
centre.
 However, if d is increased, av/d become smaller,
failure become closer to support.
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11. CASE I : av/d > 6
V
av
d
 Bending moment (BM) > shear force (V)
 Bending failure happened and stress tension steel close to
tensile limit.
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12. CASE II : 2 < av/d < 6
V
d
av
 Cracking happened by bending.
 The increasing of loads yield angular cracking and
horizontal cracking at tension reinforced cause by a bonding
failure between concrete and reinforced.
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13. CASE III : av/d < 2
V
d
av
 No bending cracking
 crack by shear immediately happened at slope closed to 45°
from horizontal axis.
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14. 3.2 BENDING MOMENTSAND SHEARFORCE
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Support Conditions - Cable
One unknown. The reaction is a
tension force which acts away
from the member in the
direction of the cable.

F

15. 3.2 BENDING MOMENTSAND SHEARFORCE
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Support Conditions – Rocker
The rocker/roller support for this bridge
girder allows horizontal movement so
the bridge is free to expand and
contract due to temperature.
One unknown. The reaction is a force
which acts perpendicular to the surface
at the point of contact.

16. 3.2 BENDING MOMENTSAND SHEARFORCE
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Support Conditions- Roller
This concrete girder rests on the ledge
that is assumed to act as a smooth
contacting surface.
One unknown. The reaction is a force
which acts perpendicular to the surface
at the point of contact.

17. 3.2 BENDING MOMENTSAND SHEARFORCE
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Support Conditions - Pin
This utility building is pin supported at the
top of the column.
Two unknowns. The reactions are two
components of force, or which acts
perpendicular and horizontal to the
surface at the point of contact.

18. 3.2 BENDING MOMENTSAND SHEARFORCE
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Support Conditions - Fixed
The floor beam of this building are
welded together and thus form fixed
connections.
Three unknowns. The reactions are the
couple moment and the two force
components.

19. 3.3 Laws of bending
Loaded beam has a bending tendency due to bending
moment (N.mm).
Cross-sectional shape measured in moment of Inertia, I
(mm4), contribute to bending resistance.
Materials strength, f (N/mm2), also effect bending.
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Bending Cross-section
Stress-diagram
fC = My/I
Strain-diagram
+ve
-ve
fT = My/I
n.a.
-y
y

20. 3.3 Laws of bending
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 Moment Resistance (Cracking Moment),
Mr = (I/ymax) =.Z
 = permissible bending stress
M = bending moment
I = Moment of Inertia
y = distance from neutral axis (n.a.)
Section Modulus, Z = I/ymax

21. Example
A steel bar of 20 x 60 mm rectangular cross section is
subjected to two equal and opposite couples acting in
vertical plane of symmetry of the bar shown in Figure
below. Determine the value of the bending moment
which causes the bar to yield. Assume  = 250 MPa
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M M
20 mm
60 mm

22. SOLUTION
 The value of the bending moment which causes the bar to
yield = Moment Resistance (Cracking Moment),
 = 250 x 106 N/m2
I = bh3/12 = 0.02 x 0.063/12 = 3.6 x 10-7
y = 0.03 m
Mr = (I/ymax)
= 250 x 106 x (3.6 x 10-7/0.03)
= 3000 Nm = 3 kNm
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23. 3.4 Bending moments and shear
force diagram
Example 1
Draw the shear and bending moment for the beam and
loading shown.
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3.0 m 1.0 m
40 kN
A
C
B

24. Solution
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40 kN
3 m 1 m
VA
VC
∑MA = 0; VC(4) – 40(3) =0
VC = 30 kN
∑FY = 0; VA +VC = 40
VA = 10 kN

25. Solution
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SFD
BMD

26. Example 2
Draw the shear and moment diagrams for the cantilevered
beam.
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2 kN
A C
1 kN
1.5 m 1.5 m
B

27. Solution
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3 kN
2 kN1 kN
SFD

28. Solution
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BMD

29. Example 3
Draw the shear and bending moment for the beam and
loading shown.
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5.0 m
10 kN/m
A
B

30. Solution
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10 kN/m = 10(5) = 50 kN
5 m
VA
VB
∑MA = 0; VB(5) – 50(2.5) =0
VB = 25 kN
∑FY = 0; VA +VB = 50
VA = 25 kN

31. Solution
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SFD
BMD
25 kN
25 kN
31.25 kNm
By formula: Mmax = wl2/8
= 10x 52/8 = 31.25 kNm

32. Example 4
Draw the shear and bending moment for the beam and
loading shown.
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2.0 m
3.0 m
1.5 kN/m
A
B
C

33. Solution
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1.5 kN/m = 1.5(1) = 1.5 kN
2 m
VA
VC
∑MA = 0; VC(3) – 1.5(2.5) =0
VC = 1.25 kN
∑FY = 0; VA +VC = 1.5
VA = 1.5- 1.25
= 0.25 kN
1 m

34. Solution
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SFD
BMD
0.25 kN
1.25 kN
0.5 kNm
0.17 0.83
0.521 kNm

35. Exercise 1
Draw the shear and bending moment for the beam and
loading shown.
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2.5 m 3.0 m 2.0 m
20 kN
40 kN
A
B C

36. Exercise 2
Draw the shear and bending moment for the beam and
loading shown.
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1.8 m 0.9 m 1.8 m
1.5 kN/m
65 kN
A
B C
D

37. Exercise 3
Draw the shear and bending moment for the beam and
loading shown.
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1.8 m 2.5 m
1.0 kN/m
40 kN
A
B
C