nor azah binti aziz kolej matrikulasi teknikal kedah 1 2.0 analysis and design 2.1 introduction to...
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NOR AZAH BINTI AZIZ
KOLEJ MATRIKULASI TEKNIKAL KEDAH
1
2.0 ANALYSIS AND DESIGN
2.1 INTRODUCTION TO ANALYSIS AND DESIGN
Statics of structural supports
TA027
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TYPES OF FORCES
i) External Forces
- actions of other bodies on the structure
under consideration.
- are classified as;
i) applied forces(loads)
- e.g live loads and wind loads
- able to move the structure
- usually known in the analysis
ii) reaction forces(reactions)
- the forces exerted by the support on the structure
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TYPES OF FORCES
ii) Internal Forces
- forces and couples exerted on a member
or portion of the structure by the rest
of the structure.
- Internal forces always occur in equal but
opposite pairs
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SUPPORT REACTIONS
Support reactions
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Constraints Type and direction of forces produced
The connection point on the bar can not move downward.
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SUPPORT REACTIONS5
Constraints Type and direction of forces produced
The support prevents translation in vertical and horizontal directions and also rotation,Hence a couple moment is developed on the body in that direction as well.
Constraints Type and direction of forces produced
The joint can not move in vertical and horizontal directions.
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TYPES OF LOADING
A beam may be loaded in a variety of ways. For the analysis purpose it may be splitted in three categories:
i) Concentrated or point load ii) Distributed load
Uniformly distributed loadUniformly varying load
iii) Couple
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TYPES OF LOADING
i) Concentrated load:
A concentrated load is the one which acts over so small length that it is assumed to act at a point.
Practically, a point load can not be places as knife edge contact but for calculation purpose we consider that load is being transmitted at a point.
Figure represents point loading at points A and B.
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TYPES OF LOADING
ii) Distributed load:
A distributed load acts over a finite length of the beam.
Such loads are measured by their intensity which is expressed by the force per unit distance along the axis of the beam.
Figure represents distributed loading between point A and B.
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TYPES OF LOADING
a)) Uniformly Distributed load(UDL)
A uniformly distributed load implies a constant intensity of loading (w).
Figure represents a U.D.L. between points A and B.
Such loads are measured by their intensity which is expressed by the force per unit distance along the axis of the beam.
Figure represents distributed loading between point A and B.
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TYPES OF LOADING
ii) Uniformly Varying load(triangularly distributed load):
A uniformly varying load implies that the intensity of loading increases or decreases at a constant rate along the length.
w = w0 = k . x
where k is the rate of change of the loading intensity, w0 being the loading at the reference point.
Such a loading is also known as triangularly distributed load. Figure represents such a loading between points A and B. Sometimes, the distributed loading may be parabolic, cubic or a higher order curve for non-uniformly varying load i.e.,
w = w0 + k1x +k2x2 (Parabolic)
w = w0 + k1x + k2x2 + k3x3
(Cubic) and so on.
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TYPES OF LOADING
ii) Couple
A beam may also b subjected to a couple ‘µ’ at any point. As shown in figure.
Note:
In general, the load may be a combination of various types of loadings.
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Free body diagram
considering supports:
Example:
Draw the FBD for the
Following beam
(the trusses are imposing
the same forces on the
beam):
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FBD:
Solution:
A pin can be considered for left support (A); no motion in 2 directions,
A roller can be considered for right support (B); no vertical motion,
Weight of the beam is generally neglected (when not mentioned and) when it is small compared to the load the beam supports.
Replace each truss with force F:
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F F F(FA)y
(FA)x
(FB)ya b c d
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POINT LOAD
A B
RAY RBY
A B
RAY RBY
UNIFORM LY DISTRIBUTED LOAD
POINT LOAD
Types Of Load14
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The diagram shows shear force and bending moment diagram
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Example10 kN
A BC
4m
2m
+VE
-VE
0
5
0
55
5
10
+VE
0 0
SHEAR FORCE DIAGRAM
BENDING MOMENT DIAGRAM
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Calculation of Reaction Force, Shear Force and
Bending Moment
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Determine reaction RAy and RBy:
MA = 0 RBy(4) – 10(2) = 0
4 RBy = 20
RBy = 20 = 5kN 4
Fy = Fy RBy + RAy = 10kN
RAy = 10 – 5 = 5kN
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RBy
10kN
RAy
RAx
2m 2m
Example 1
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Determine shear force at A, C and B:
At A, VA = 5 = 5kN
At C, VC = 5 – 10= -5kN
At B, VB = 5 – 10= -5kN
Determine bending moment at A, C and B
At A, MA = 5(0) = 0kNm
At C, MC = 5(2) – 10(0) =10kNm
At B, MB = 5(4) -10(2) + 5(0) = 0kNm
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RBy
10kN
RAy
RAx
2m 2m
Example 1
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RBy
10kN
RAy
RAx
2m 2m
+ve
-ve
+ve
SFD (kN) 0
5 5
5 5
BMD (kNm) 0
0
10
0
Example 1
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RBy = 25kN
10kN
Ray = 15kN
0
1m 1m 1m 1m
P Q
MC = 0M = M 15(1) + 10(1) + Q(2) = 25(3)
15 + 10 + 2Q = 75 2Q = 75 – 25
= 50 2
= 25kNFy = Fy P + 10 + Q = 15 + 25
P = 15 + 25 – 10 – 25 P = 5kN
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Example 2
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RBy = 25kN
10kN
Ray = 15kN
0
1m 1m 1m 1m
P Q
Determine shear force at A, C and B:At A,
VA = 15kNAt C,
VC = 15 – 5 = 10kNAt D,
VD = 15 – 5 – 10 = 0kNAt E,
VE = 15 – 5 – 10 – 25 = -25kNAt B,
VB = 15 – 5 – 10 – 25 = -25kN
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Example 2
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RBy = 25kN
10kN
Ray = 15kN
0
1m 1m 1m 1m
P Q
Determine bending moment at A, C and BAt A,
MA = 15(0) = 0kNmAt C,
MC = 15(1) – 5(0) = 15kNmAt D,
MD = 15(2) – 5(1) – 10(0) = 25kNmAt E,
MD = 15(3) – 5(2) – 10(1) – 25(0) = 25kNmAt B,
MB = 15(4) – 5(3) – 10(2) – 25(1) + 25(0) = 0kNm
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Example 2
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RBy = 25kN
10kN
Ray = 15kN
0
1m 1m 1m 1m
P Q
+ve
-ve
+ve
SFD (kN) 0
BMD (kNm) 0
0
0
15
25 25
15 15
10
25 25
Example 2
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Equilibrium of forces;
Fx = 0 RAx = 0
Fy = 0 RAy + RBy - wL = 0
RAy + RBy = wLEquilibrium of moments;
MA = 0 RBy(L) – wL(L/2) = 0
RBy(L) – wL2/2 = 0
RBy = wL/2
RAy + wL/2 = wL
RAy = wL/2
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w kN/m
L
A B
RByRAy
RAx
w kN/m
LL/2
Example 3
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w kN/m
Mx = 0 Mx + wx(x) – wL(x) = 0 2 2Mx + wx2 – wLx = 0 2 2Mx = wx(L-x) = 0 2
When x = L/2 M = wL2
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A B
L/2 L/2
+ve
-veSFD (kN)
0
BMD (kNm) 0
wL/2
wL/2
0
0
wL2
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Example 3
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8 kN/m
5m
A B
Solution:
P = wL
= 8kN/m x 5m
= 40kN
RAy = RBy = 40/2
= 20kN
Example 4
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Determine shear forces at A, mid span (C) and BAt A, FA = 20kNAt mid span (2.5m), F2.5 = 20 – 8(2.5) = 0kNAt B, FB = 20 – 8(5) = -20kN
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RByRAy
RAx
40 kN/m
2.5m2.5m
Example 4
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Determine bending moments at A, 1m, 2m, 2.5m, 3m, 4m and at BAt A,
MA = 20(0) = 0 kNmAt 1,
M1 = 20(1) – 8(1)(0.5) = 20 – 4 = 16kNm At 2,
M2 = 20(2) – 8(2)(1) = 40 – 16 = 24kNmAt mid span (2.5m),
M2.5 = 20(2.5) – 8(2.5)(1.25) = 50 – 25 = 25kNmAt 3,
M3 = 20(3) – 8(3)(1.5) = 60 – 36 = 24kNmAt 4,
M4 = 20(4) – 8(4)(2) = 80 – 64 = 16kNmAt B,
MB = 20(5) – 8(5)(2.5) = 100 – 100 = 0kNm
29 RByRAy
RAx
40 kN/m
2.5m2.5m
Example 4
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Free Body Diagram (FBD)
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Equilibrium Conditions
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Calculate Support reactions
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SHEAR FORCESHEAR FORCE DIAGRAM
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BENDING MOMENTBENDING MOMENT DIAGRAM
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Simply supported beam with point load
at center
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Simply supported beam with eccentric
load