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JJ310 STRENGTH OF MATERIAL
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1
STRENGTH OF MATERIAL
JJ 310
CHAPTER 3 (a)
Shear Force & Bending Moment
2
LEARNING OUTCOMES
At the end of this lecture, student should be able to;
Understand the types of beams
Understand the types of beam load
Understand the shear force and bending moment
Calculate the magnitude of the shear force and bending moment
Draw the shear force diagram (SFD)
3
DEFINITION OF BEAM
- Members that are slender and support loads applied perpendicular to their longitudinal axis.
- A beam has a characteristic feature that internal forces called shear forces, V and the internal moments called bending moments, M.
- V and M are developed in beam to resists the external loads.
- For the beams, the distance (L) between the supports is called a span.
4
EXAMPLE OF BEAMS
5
TYPES OF BEAM
FV
FH
Pin
FV
Roller
M
Fv
FH
Fixed
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TYPES OF BEAM LOAD
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TYPES OF BEAM LOAD (CONT.)
i) Concentrated Load (CL), F
- A load is applied at a single point, or over a very small area of a beam.
(unit: N)
ii) Uniformly Distributed Load (UDL), ω
- A load distributed uniformly over a portion or over the entire length of a beam.
(unit: N/m)
iii) Applied Couple
- Combination of various types of loadings.
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SHEAR FORCE & BENDING MOMENT
Shear Force, V : is the sum of the vertical forces acting to the left or right of a cut section of the beam.
Bending Moment, M : is the sum of the
moment of the forces to the left or to the right of the cut section of the beam.
For CT,
BUT... for UDL
Moment = Force x Distance @
M = FD (unit: Nm)
M = F(D/2) (unit: Nm)
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LAW OF EQUILIBRIUM Stated that ….
i) To find the sum of forces, ∑ F
The sum of upward forces = The sum of downward forces
ii) To find the sum of moments, ∑ M
The sum of clockwise moments = The sum of counter clockwise moments
∑ F = ∑ F (unit N)
∑ M = ∑ M (unit Nm)
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Load
F W
Linear
Shear
Constant Linear Parabolic
Moment
Linear Parabolic Cubic
SFD & BMD TABLE
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Load F W
Shear
Constant Constant Linear
Moment
Linear Linear Parabolic
M
SFD & BMD TABLE
12
STEPS TO FIND SHEAR FORCES
Step 1 – Draw FBD (Free Body Diagram)
Step 2 – Calculate ∑F
Step 3 – Calculate ∑M
Step 4 – Draw SFD
13
EXAMPLE 1: (SIMPLY SUPPORTED BEAM)CALCULATE THE REACTION FORCE AT EACH
SUPPORT
4 m
1 m
1 m
5 KN 10 KN
ACB
D
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STEP 1: DRAW FBD
5 KN
10 KN
RA
C B
RD
4 m 1 m
1 m
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STEP 2: CALCULATE ∑F
We know; ↑∑ F = ↓∑ F
RA + RD = 5 + 10
RA + RD = 15
5 KN 10 KN
RA
CB
RD4 m 1
m1 m
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STEP 3: CALCULATE ∑M
5 KN 10 KN
RA
CB
RD4 m 1
m1 m
We know; ∑M = ∑M 5(1) + 10(5) = RD (6)
55 = 6RD
RD = 9.2 KN # Thus;
RA + RD = 15 RA = 15 – 9.2 = 5.8 KN #
Take point A as
reference point
M = FD
17
STEP 4: DRAW SFD
5.8 KN
5 KN
V Value:
VA = 0 KN, 5.8 KN
VB = 5.8 KN, 0.8 KN
VC = 0.8 KN, -9.2 KN
VD = -9.2 KN, 0 KN
1m4m1m
10 KN
FBD
SFD
5.8 KN
0.8 KN
5.8 -5=0.8
9.2KN
00
0.8-10= -9.2
-9.2 + 9.2= 0
0+5.8=0.8
-9.2 KN
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EXAMPLE 2: (UNIFORMLY DISTRIBUTED LOAD)
CALCULATE THE REACTION FORCE AT EACH SUPPORT
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15 K/Nm
BA
4 m
19
STEP 1: DRAW FBD
W = 15 K/Nm
BA
4m RA RB
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STEP 2: CALCULATE ∑F
↑∑ F = ↓∑ F
RA + RB = 60 KN
15 KN/m @ 60 KN
BA
4m RA RB
F = WDF = 15 x 4F = 60 KN
Remember!!For UDL, must change W into F
first.F = WD
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STEP 3: CALCULATE ∑M
15 KN/m @ 60 KN
BA
4m RA RB
∑M = ∑M 60(4/2) = RB (4)
120 = 4RB
RB = 30 KN #Thus;
RA + RB = 60 RA = 60 - 30
RA = 30 KN #
Remember…
For UDL,M = F x (D/2)
Take point A as
reference point
22
STEP 4: DRAW SFD
15 KN/m @ 60 KN
B A
4m 30 KN 30 KN
FBD
SFD0 0
30 - 60 = -30
0+30 = 30 30 KN
- 30 KN
-30 + 30 = 0
V Value:
VA = 0 KN, 30 KN
VB = -30KN, 0KN