1. NOR AZAH BINTI AZIZ KOLEJ MATRIKULASI TEKNIKAL KEDAH
2.0 ANALYSIS AND DESIGN
2.1 INTRODUCTION TO ANALYSIS AND DESIGN
Statics of structural supports
NOR AZAH BINTI AZIZ
KOLEJ MATRIKULASI TEKNIKAL KEDAH
TA027

2. 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

3. 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

4. SUPPORT REACTIONS Support reactions Constraints
Type and direction of forces produced
The connection point on the bar can not move downward.
If a support prevents translation
of a body in a given direction,
a force is developed on the body
in that direction.

5. SUPPORT REACTIONS Constraints Type and direction of forces produced
The joint can not move in vertical and horizontal directions.
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.

6. 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 load
Uniformly varying load
iii) Couple

7. 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.

8. 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.

9. 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.

10. 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.

11. 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.

12. considering supports: Example: Draw the FBD for the Following beam
Free body diagram
considering supports:
Example:
Draw the FBD for the
Following beam
(the trusses are imposing
the same forces on the
beam):

13.  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: F
(FA)y
(FA)x
(FB)y a b c d

14. Types Of Load POINT LOAD UNIFORM LY DISTRIBUTED LOAD POINT LOAD A B
RAY
RBY
UNIFORM LY DISTRIBUTED LOAD

15. The diagram shows shear force and bending moment diagram

16. Example 10 kN 2m C A B 4m 5 10 +VE SHEAR FORCE DIAGRAM -VE5 10
SHEAR FORCE DIAGRAM
BENDING MOMENT DIAGRAM

17. Calculation of Reaction Force, Shear Force and Bending Moment

18. Example 1
10kN
RAx
RAy
RBy 2m 2m
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

19. Example 1
10kN
RAy
RAx 2m
RBy
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

20. Example 1 +ve -ve +ve 10kN RAx RAy RBy 2m 2m 5 5 SFD (kN) 0 5 5 10
SFD (kN) 0
-ve 5 5 10
+ve
BMD (kNm) 0

21. Example 2
10kN
Ray = 15kN 1m P Q
RBy = 25kN
MC = 0 M = M 15(1) + 10(1) + Q(2) = 25(3) Q = 75 2Q = 75 – 25 = 50 2 = 25kN Fy  = Fy P Q = P = – 10 – 25 P = 5kN

22. Example 2
10kN
Ray = 15kN 1m P Q
RBy = 25kN
Determine shear force at A, C and B: At A, VA = 15kN At C, VC = 15 – 5 = 10kN At D, VD = 15 – 5 – 10 = 0kN At E, VE = 15 – 5 – 10 – 25 = -25kN At B, VB = 15 – 5 – 10 – 25 = -25kN

23. Example 2
RBy = 25kN
10kN
Ray = 15kN 1m P Q
Determine bending moment at A, C and B At A, MA = 15(0) = 0kNm At C, MC = 15(1) – 5(0) = 15kNm At D, MD = 15(2) – 5(1) – 10(0) = 25kNm At E, MD = 15(3) – 5(2) – 10(1) – 25(0) = 25kNm At B, MB = 15(4) – 5(3) – 10(2) – 25(1) + 25(0) = 0kNm

24. Example 2 +ve -ve 10kN Ray = 15kN 1m P Q SFD (kN) 0 BMD (kNm) 0 15 251m P Q
+ve
-ve
SFD (kN) 0
BMD (kNm) 0 15 25 10
RBy = 25kN

25. Example 3
w kN/m L A B
RBy
RAy
RAx
L/2
Equilibrium of forces;  Fx = 0 RAx = 0  Fy = 0 RAy + RBy - wL = 0 RAy + RBy = wL Equilibrium 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

26. Example 3 A B
L/2
+ve
-ve
SFD (kN) 0
BMD (kNm) 0
wL/2
wL2 8
w kN/m
 Mx = 0 Mx + wx(x) – wL(x) = Mx + wx2 – wLx = 0 Mx = wx(L-x) = 0 2 When x = L/2 M = wL2 8

27. Example 4 Solution: P = wL = 8kN/m x 5m = 40kN RAy = RBy = 40/2 = 20kN

28. Example 4
RBy
RAy
RAx
40 kN/m
2.5m
Determine shear forces at A, mid span (C) and B At A,  FA = 20kN At mid span (2.5m),  F2.5 = 20 – 8(2.5) = 0kN At B,  FB = 20 – 8(5) = -20kN

29. Example 4
RBy
RAy
RAx
40 kN/m
2.5m
Determine bending moments at A, 1m, 2m, 2.5m, 3m, 4m and at B At A,  MA = 20(0) = 0 kNm At 1,  M1 = 20(1) – 8(1)(0.5) = 20 – 4 = 16kNm At 2,  M2 = 20(2) – 8(2)(1) = 40 – 16 = 24kNm At mid span (2.5m),  M2.5 = 20(2.5) – 8(2.5)(1.25) = 50 – 25 = 25kNm At 3,  M3 = 20(3) – 8(3)(1.5) = 60 – 36 = 24kNm At 4,  M4 = 20(4) – 8(4)(2) = 80 – 64 = 16kNm At B,  MB = 20(5) – 8(5)(2.5) = 100 – 100 = 0kNm

30. Free Body Diagram (FBD)

31. Equilibrium Conditions

32. Calculate Support reactions

33. SHEAR FORCE SHEAR FORCE DIAGRAM

34. BENDING MOMENT BENDING MOMENT DIAGRAM

35. Simply supported beam with point load at center

36. Simply supported beam with eccentric load