chapter 3 beam

TOPIC 3

STRUCTURAL THEORY RELATED TO

SIMPLE BEAMS

3.0 Beams

3.1 Failure of beams due to bending, shear and deflection

3.2 Bending moments and shear force

3.3 Laws of bending

3.4 Bending moments and shear force diagrams

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3.0 BEAM

Members that are slender and support loading that areapplied perpendicular to their longitudinal axis. Ingeneral , beams are long, straight bars having a constantcross-section area. Often they are classified as to howthey are supported. Example :-

i) Simply supported beam

ii) Cantilever beam

iii) Overhanging beam

iv) Fixed end beam

v) Continuous beam

Support the slabs. Its major function is to resist thebending moment and shear.

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Simply Supported Beam

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A simply supported beam is supported at both ends with loads

applied between the supports.

Cantilever Beam

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A cantilever is a beam anchored at only one end. The beam carries the load

to the support where it is forced against by a moment and shear stress.

Cantilever construction allows for overhanging structures without external

bracing. Cantilevers can also be constructed with trusses or slabs.

Overhanging Beam

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Fixed End Beam

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A fixed or a build in beam has both of its ends rigidly fixed.

Continuous Beam

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A continuous beam is a structural component that provides

resistance to bending when a load or force is applied. These beams

are commonly used in bridges. A beam of this type has more than

two points of support along its length.

3. 1 FAILURE OF BEAMS DUE TO SHEAR

SHEAR (V)

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Actually, max shear force (V), occurs at a beam

supports.

A cracking happens near the support.Shear failure

Moment failureV max V max

3.1 SHEAR STRESS

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v = V / bvd,

where v = shear stress in N/mm2.

V = shear force in N.

bv = width section of beam in mm.

d = effective depth in mm.

v

linkd

bv

MODE OF SHEAR FAILURE

Mechanisms of shear failure in concrete are verycomplicated and complex to understand.

Effected by a ratio of shear [av/d] give three situated:

Case I : av/d > 6

Case II : 2< av/d < 6

Case III : av/d < 2

When av increased, av/d become bigger, bendingmoment will increase, failure become closer to thecentre.

However, if d is increased, av/d become smaller,failure become closer to support.

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CASE I : av/d > 6

V

av

d

Bending moment (BM) > shear force (V)

Bending failure happened and stress tension steel close to tensile limit.

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CASE II : 2 < av/d < 6

V

d

av

Cracking happened by bending.

The increasing of loads yield angular cracking and

horizontal cracking at tension reinforced cause by a bonding

failure between concrete and reinforced.

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CASE III : av/d < 2

V

d

av

No bending cracking

crack by shear immediately happened at slope closed to 45°

from horizontal axis.

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3.2 BENDING MOMENTS AND SHEAR FORCE

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Support Conditions - Cable

One unknown. The reaction is a

tension force which acts away

from the member in the

direction of the cable.

F

3.2 BENDING MOMENTS AND SHEAR FORCE

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Support Conditions – Rocker

The rocker/roller support for this bridge

girder allows horizontal movement so

the bridge is free to expand and

contract due to temperature.

One unknown. The reaction is a force

which acts perpendicular to the surface

at the point of contact.

3.2 BENDING MOMENTS AND SHEAR FORCE

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Support Conditions- Roller

This concrete girder rests on the ledge

that is assumed to act as a smooth

contacting surface.

One unknown. The reaction is a force

which acts perpendicular to the surface

at the point of contact.

3.2 BENDING MOMENTS AND SHEAR FORCE

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Support Conditions - Pin

This utility building is pin supported at the

top of the column.

Two unknowns. The reactions are two

components of force, or which acts

perpendicular and horizontal to the

surface at the point of contact.

3.2 BENDING MOMENTS AND SHEAR FORCE

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Support Conditions - Fixed

The floor beam of this building are

welded together and thus form fixed

connections.

Three unknowns. The reactions are the

couple moment and the two force

components.

3.3 Laws of bending

Loaded beam has a bending tendency due to bending

moment (N.mm).

Cross-sectional shape measured in moment of Inertia, I

(mm4), contribute to bending resistance.

Materials strength, f (N/mm2), also effect bending.

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Bending Cross-section

Stress-diagram

fC = My/I

Strain-diagram

+ve

-ve

fT = My/I

n.a.

-y

y

3.3 Laws of bending

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Moment Resistance (Cracking Moment),

Mr = (I/ymax) =.Z

= permissible bending stress

M = bending moment

I = Moment of Inertia

y = distance from neutral axis (n.a.)

Section Modulus, Z = I/ymax

Example

A steel bar of 20 x 60 mm rectangular cross section is

subjected to two equal and opposite couples acting in

vertical plane of symmetry of the bar shown in Figure

below. Determine the value of the bending moment

which causes the bar to yield. Assume = 250 MPa

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M M

20 mm

60 mm

SOLUTION

The value of the bending moment which causes the bar to yield = Moment Resistance (Cracking Moment),

= 250 x 106 N/m2

I = bh3/12 = 0.02 x 0.063/12 = 3.6 x 10-7

y = 0.03 m

Mr = (I/ymax)

= 250 x 106 x (3.6 x 10-7/0.03)

= 3000 Nm = 3 kNm

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3.4 Bending moments and shear

force diagram

Example 1

Draw the shear and bending moment for the beam and

loading shown.

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3.0 m 1.0 m

40 kN

AC

B

Solution

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40 kN

3 m 1 m

VAVC

∑MA = 0; VC(4) – 40(3) =0

VC = 30 kN

∑FY = 0; VA +VC = 40

VA = 10 kN

Solution

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SFD

BMD

Example 2

Draw the shear and moment diagrams for the cantilevered

beam.

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2 kN

A C

1 kN

1.5 m 1.5 m

B

Solution

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3 kN

2 kN1 kN

SFD

Solution

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BMD

Example 3

Draw the shear and bending moment for the beam and

loading shown.

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5.0 m

10 kN/m

AB

Solution

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10 kN/m = 10(5) = 50 kN

5 m

VAVB

∑MA = 0; VB(5) – 50(2.5) =0

VB = 25 kN

∑FY = 0; VA +VB = 50

VA = 25 kN

Solution

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SFD

BMD

25 kN

25 kN31.25 kNm

By formula: Mmax = wl2/8

= 10x 52/8 = 31.25 kNm

Example 4

Draw the shear and bending moment for the beam and

loading shown.

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2.0 m

3.0 m

1.5 kN/m

AB

C

Solution

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1.5 kN/m = 1.5(1) = 1.5 kN

2 m

VAVC

∑MA = 0; VC(3) – 1.5(2.5) =0

VC = 1.25 kN

∑FY = 0; VA +VC = 1.5

VA = 1.5- 1.25

= 0.25 kN

1 m

Solution

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SFD

BMD

0.25 kN

1.25 kN

0.5 kNm

0.17 0.83

0.521 kNm

Exercise 1

Draw the shear and bending moment for the beam and

loading shown.

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2.5 m 3.0 m 2.0 m

20 kN40 kN

AB C

Exercise 2

Draw the shear and bending moment for the beam and

loading shown.

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1.8 m 0.9 m 1.8 m

1.5 kN/m65 kN

AB C

D

Exercise 3

Draw the shear and bending moment for the beam and

loading shown.

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1.8 m 2.5 m

1.0 kN/m40 kN

AB

C