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STRENGTH OF MATERIAL

JJ 310

CHAPTER 3 (a)

Shear Force & Bending Moment

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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)

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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.

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EXAMPLE OF BEAMS

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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

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STEPS TO FIND SHEAR FORCES

Step 1 – Draw FBD (Free Body Diagram)

Step 2 – Calculate ∑F

Step 3 – Calculate ∑M

Step 4 – Draw SFD

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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

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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

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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

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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