NOR AZAH BINTI AZIZ KOLEJ MATRIKULASI TEKNIKAL KEDAH 1 2.0 ANALYSIS AND DESIGN 2.1 INTRODUCTION TO...

NOR AZAH BINTI AZIZ

KOLEJ MATRIKULASI TEKNIKAL KEDAH

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2.0 ANALYSIS AND DESIGN

2.1 INTRODUCTION TO ANALYSIS AND DESIGN

Statics of structural supports

TA027

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.

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.

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

POINT LOAD

A B

RAY RBY

A B

RAY RBY

UNIFORM LY DISTRIBUTED LOAD

POINT LOAD

Types Of Load14

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

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

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

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

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

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

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

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

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