A beam is a structural member which safely carries tonss i.e. without neglecting due to the applied tonss. We will be restricted to beams of unvarying cross-sectional country.

## Simply Supported Beam

A beam that rests on two supports merely along the length of the beam and is allowed to debar freely when tonss are applied. Note – see subdivision A of unit.

Roentgen

Roentgen

Force ( Load )

Force ( Load )

Force ( Load )

Beam Cross Section

## Cantilever Beam

A beam that is supported at one terminal merely. The terminal could be built into a wall, bolted or welded to another construction for agencies of support.

Rw – reaction at wall

Force ( Load )

Beam Cross Section

## Point or Concentrated Load

A burden which acts at a peculiar point along the length of the beam. This burden is normally called a force ( F ) and is stated in Newtons ( N ) . A mass may be converted into a force by multiplying by gravitation whose value is changeless at 9.81 m/s2.

Bacillus

A

F

Rubidium

RA

F = m x g ( N )

## Uniformly Distributed Load ( UDL )

A burden which is dispersed equally over a given length of the beam. This may be the weight of the beam itself. The UDL is quoted as Newtons per meter ( N/m ) .

UDL ( N/m )

Bacillus

A

RA

Rubidium

## Beam Failure

If inordinate tonss are used and the beam does non hold the necessary stuff belongingss of strength so failure will happen. Failure may happen in two ways: –

By shearing across the cross sectional country of the beam due to inordinate shear force.

Roentgen

Excessive

Shear Force

## Shear Failure

Roentgen

Excessive

Shear Force

By flexing of the beam due to inordinate tensile and/or compressive emphasiss set up by flexing minutes.

Top of beam in compaction

## Bending Failure

Bottom of beam in tenseness

Roentgen

Roentgen

## Simply Supported Beam with Point Load

6 m

F

Tocopherol

Calciferol

C

A

RA

Roentgenium

F =12 kN

We must foremost cipher the reactions RA and RG. We take minutes about one of the reactions to cipher the other, hence to happen RA:

Take minutes about RG

?Clockwise minutes ( CM ) = ?Anti-clockwise minutes ( ACM )

RA x 6 = 12 x 3

RA = 6 kN now,

?Upward Forces = ?Downward Forces

RA + RG = 12

6 + RG = 12

## RG = 6 kN

subdivision

F +

F –

F –

F +

Calculating Shear Forces ( we must utilize the shear force regulation ) .

When looking right of a subdivision: downward forces are positive and upward forces are negative.

When looking left of a subdivision: downward forces are negative and upward forces are positive.

Get downing at point A and looking left:

( note: the negative mark ( – ) means merely to the left of the place and the positive mark ( + ) means merely to the right of the place. )

SFA – = 0 kN

SFA + = 6 kN

An alternate method of pulling the shear force diagram is to follow the waies of each force on the line diagram.SFB – = 6 kN

SFB + = 6 kN

SFC – = 6 kN

SFC + = 6 kN

SFD – = 6 kN

SFD + = 6 – 12 = -6 kN

SFE – = 6 – 12 = -6 kN

SFE + = 6 – 12 = -6 kN

SFF – = 6 – 12 = -6 kN

SFF + = 6 – 12 = -6 kN

SFG – = 6 – 12 = -6 kN

SFG + = 6 – 12 + 6 = 0 kN

Note: the shear force at either terminal of a merely supported beam must compare to zero.

Calculating Bending Moments ( we must utilize the bending minute regulation ) .

When looking right of a subdivision: downward forces are negative and upward forces are positive.

When looking left of a subdivision: downward forces are negative and upward forces are positive.

subdivision

F –

F –

subdivision

F +

F +

Hoging Radio beam

Saging Radio beam

Get downing at point A and looking left:

BMA = 0 kNm

BMB = ( 6 x 1 ) = 6 kNm

BMC = ( 6 x 2 ) = 12 kNm

BMD = ( 6 x 3 ) = 18 kNm

BME = ( 6 x 4 ) + ( -12 x 1 ) = 12 kNm

BMF = ( 6 x 5 ) + ( -12 x 2 ) = 6 kNm

BMG = ( 6 x 6 ) + ( -12 x 3 ) = 0 kNm

Note: the bending minute at either terminal of a merely supported beam must compare to zero.

The undermentioned page shows the line, shear force and flexing minute diagrams for this beam.

## Simply Supported Beam with Point Load

6 m

F

Tocopherol

Calciferol

C

Gram

Bacillus

A

6 kN

6 kN

F =12 kN

Shear Force Diagram ( kN )

0

0

-6

6

0

Line Diagram

12

12

18

6

0

6

Bending Moment Diagram ( kNm )

Max Tensile Stress

SAGGING ( +ve bending )

Max Compressive Stress

F

F

A maximal bending minute of 18 kNm occurs at place D. Note the shear force is zero at this point.

## Simply Supported Beam with Distributed Load

UDL = 2 kN/m

F

Tocopherol

Calciferol

C

Gram

Bacillus

A

6 m

RA

The force from a UDL is considered to move at the UDL mid-point.

e.g. if we take minutes about ‘D ‘ so the entire force from the UDL ( looking to the left ) would be: ( 2 x 3 ) = 6 kN. This force must be multiplied by the distance from point ‘D ‘ to the UDL mid point as shown below.

e.g. Take minutes about ‘D ‘ , so the minute would be: ( -6 x 1.5 ) = -9 kNm

1.5m

UDL = 2 kN/m

Calciferol

C

Bacillus

A

3 m

Taking minutes about point D ( looking left )

We must foremost cipher the reactions RA and RG. We take minutes about one of the reactions to cipher the other, hence to happen RA:

Take minutes about RG

?Clockwise minutes ( CM ) = ?Anti-clockwise minutes ( ACM )

RA x 6 = 2 ten 6 ten 3

RA = 6 kN now,

?Upward Forces = ?Downward Forces

RA + RG = 2 tens 6

6 + RG = 12

## RG = 6 kN

subdivision

F +

F –

F –

F +

Calculating Shear Forces ( we must utilize the shear force regulation ) .

When looking right of a subdivision: downward forces are positive and upward forces are negative.

When looking left of a subdivision: downward forces are negative and upward forces are positive.

Get downing at point A and looking left:

( note: the negative mark ( – ) means merely to the left of the place and the positive mark ( + ) means merely to the right of the place. )

SFA – = 0 kN

SFA + = 6 kN

SFB – = 6 – ( 2×1 ) = 4 kN

SFB + = 6 – ( 2×1 ) = 4 kN

SFC – = 6 – ( 2×2 ) = 2 kN

SFC + = 6 – ( 2×2 ) = 2 kN

SFD – = 6 – ( 2×3 ) = 0 kN

SFD + = 6 – ( 2×3 ) = 0 kN

SFE – = 6 – ( 2×4 ) = -2 kN

SFE + = 6 – ( 2×4 ) = -2 kN

SFF – = 6 – ( 2×5 ) = -4 kN

SFF + = 6 – ( 2×5 ) = -4 kN

SFG – = 6 – ( 2×6 ) = -6 kN

SFG + = 6 – ( 2×6 ) + 6 = 0 kN

Note: the shear force at either terminal of a merely supported beam must compare to zero.

Calculating Bending Moments ( we must utilize the bending minute regulation ) .

When looking right of a subdivision: downward forces are negative and upward forces are positive.

When looking left of a subdivision: downward forces are negative and upward forces are positive.

subdivision

F –

F –

subdivision

F +

F +

Hoging Radio beam

Saging Radio beam

Get downing at point A and looking left:

BMA = 0 kNm

BMB = ( 6 x 1 ) + ( -2 x 1 x 0.5 ) = 5 kNm

BMC = ( 6 x 2 ) + ( -2 x 2 x 1 ) = 8 kNm

BMD = ( 6 x 3 ) + ( -2 x 3 x 1.5 ) = 9 kNm

BME = ( 6 x 4 ) + ( -2 x 4 x 2 ) = 8 kNm

BMF = ( 6 x 5 ) + + ( -2 x 5 ten 2.5 = 5 kNm

BMG = ( 6 x 6 ) + + ( -2 x 6 ten 3 ) = 0 kNm

Note: the bending minute at either terminal of a merely supported beam must compare to zero.

The undermentioned page shows the line, shear force and flexing minute diagrams for this beam.

## Simply Supported Beam with Distributed Load

4

2

0

-2

-4

UDL = 2 kN/m

6 m

F

Tocopherol

Calciferol

C

Gram

Bacillus

A

Shear Force Diagram ( kN )

0

0

-6

6

0

Line Diagram

8

8

9

5

0

Bending Moment Diagram ( kNm )

5

6 kN

6 kN

Max Tensile Stress

SAGGING ( +ve bending )

Max Compressive Stress

F

F

A maximal bending minute of 9 kNm occurs at place D. Note the shear force is zero at this point.

## Simply Supported Beam with Point Loads

6 m

F

Tocopherol

Calciferol

C

Gram

Bacillus

A

RA

Roentgenium

F = 15 kN

F = 30 kN

We must foremost cipher the reactions RA and RG. We take minutes about one of the reactions to cipher the other, hence to happen RA:

Take minutes about RG

?Clockwise minutes ( CM ) = ?Anti-clockwise minutes ( ACM )

RA x 6 = ( 15 x 4 ) + ( 30 x 2 )

RA = 20 kN now,

?Upward Forces = ?Downward Forces

RA + RG = 15 + 30

20 + RG = 45

## RG = 25 kN

subdivision

F +

F –

F –

F +

Calculating Shear Forces ( we must utilize the shear force regulation ) .

When looking right of a subdivision: downward forces are positive and upward forces are negative.

When looking left of a subdivision: downward forces are negative and upward forces are positive.

Get downing at point A and looking left:

( note: the negative mark ( – ) means merely to the left of the place and the positive mark ( + ) means merely to the right of the place. )

SFA – = 0 kN

SFA + = 20 kN

SFB – = 20 kN

SFB + = 20 kN

SFC – = 20 kN

SFC + = 20 -15 = 5 kN

SFD – = 20 -15 = 5 kN

SFD + = 20 -15 = 5 kN

SFE – = 20 -15 = 5 kN

SFE + = 20 -15 – 30 = -25 kN

SFF – = 20 -15 – 30 = -25 kN

SFF + = 20 -15 – 30 = -25 kN

SFG – = 20 -15 – 30 = -25 kN

SFG + = 20 -15 – 30 + 25 = 0 kN

Note: the shear force at either terminal of a merely supported beam must compare to zero.

Calculating Bending Moments ( we must utilize the bending minute regulation ) .

When looking right of a subdivision: downward forces are negative and upward forces are positive.

When looking left of a subdivision: downward forces are negative and upward forces are positive.

subdivision

F –

F –

subdivision

F +

F +

Hoging Radio beam

Saging Radio beam

Get downing at point A and looking left:

BMA = 0 kNm

BMB = ( 20 x 1 ) = 20 kNm

BMC = ( 20 x 2 ) = 40 kNm

BMD = ( 20 x 3 ) + ( -15 x 1 ) = 45 kNm

BME = ( 20 x 4 ) + ( -15 x 2 ) = 50 kNm

BMF = ( 20 x 5 ) + ( -15 x 3 ) + ( -30 x 1 ) = 25 kNm

BMG = ( 20 x 6 ) + ( -15 x 4 ) + ( -30 x 2 ) = 0 kNm

Note: the bending minute at either terminal of a merely supported beam must compare to zero.

The undermentioned page shows the line, shear force and flexing minute diagrams for this beam.

## 0

## 20

## -25

## 0

## Shear Force Diagram ( kN )

## 5Simply Supported Beam with Point Loads

6 m

F

Tocopherol

Calciferol

C

Gram

Bacillus

A

20 kN

25 kN

F = 15 kN

F = 30 kN

Bending Moment Diagram ( kNm )

0

0

45

40

20

50

25

Max Tensile Stress

SAGGING ( +ve bending )

Max Compressive Stress

F

F

A maximal bending minute of 50 kNm occurs at place E. Note the shear force is zero at this point.

## Simply Supported Beam with Point and Distributed Loads ( 1 )

6 m

F

Tocopherol

Calciferol

C

Gram

Bacillus

A

RA

Roentgenium

15 kN

30 kN

UDL = 10 kN/m

Take minutes about RG

?Clockwise minutes ( CM ) = ?Anti-clockwise minutes ( ACM )

RA x 6 = ( 15 x 4 ) + ( 10 x 2 ten 3 ) + ( 30 x 2 )

RA = 30 kN now,

?Upward Forces = ?Downward Forces

RA + RG = 15 + ( 10 x 2 ) + 30

30 + RG = 65

## RG = 35 kN

subdivision

F +

F –

F –

F +

Calculating Shear Forces ( we must utilize the shear force regulation ) .

When looking right of a subdivision: downward forces are positive and upward forces are negative.

When looking left of a subdivision: downward forces are negative and upward forces are positive.

Get downing at point A and looking left:

SFA – = 0 kN

SFA + = 30 kN

SFB – = 30 kN

SFB + = 30 kN

SFC – = 30 kN

SFC + = 30 – 15 = 15 kN

SFD – = 30 – 15 – ( 10 x 1 ) = 5 kN

SFD + = 30 – 15 – ( 10 x 1 ) = 5 kN

SFE – = 30 – 15 – ( 10 x 2 ) = -5 kN

SFE + = 30 – 15 – ( 10 x 2 ) – 30 = -35 kN

SFF – = 30 – 15 – ( 10 x 2 ) – 30 = -35 kN

SFF + = 30 – 15 – ( 10 x 2 ) – 30 = -35 kN

SFG – = 30 – 15 – ( 10 x 2 ) – 30 = -35 kN

SFG + = 30 – 15 – ( 10 x 2 ) – 30 + 35 = 35 kN

Note: the shear force at either terminal of a merely supported beam must compare to zero.

Calculating Bending Moments ( we must utilize the bending minute regulation ) .

When looking right of a subdivision: downward forces are negative and upward forces are positive.

When looking left of a subdivision: downward forces are negative and upward forces are positive.

subdivision

F –

F –

subdivision

F +

F +

Hoging Radio beam

Saging Radio beam

Get downing at point A and looking left:

BMA = 0 kNm

BMB = ( 30 x 1 ) = 30 kNm

BMC = ( 30 x 2 ) = 60 kNm

BMD = ( 30 x 3 ) + ( -15 x 1 ) + ( -10 x 1 x 0.5 ) = 70 kNm

BME = ( 30 x 4 ) + ( -15 x 2 ) + ( -10 x 2 x 1 ) = 70 kNm

BMF = ( 30 x 5 ) + ( -15 x 3 ) + ( -10 x 2 x 2 ) + ( -30 x 1 ) = 35 kNm

BMG = ( 30 x 6 ) + ( -15 x 4 ) + ( -10 x 2 x 3 ) + ( -30 x 2 ) = 0 kNm

Notes:

the bending minute at either terminal of a merely supported beam must compare to zero.

The value of the maximal bending minute occurs where the shear force is zero and is hence still unknown ( see Shear Force diagram ) . The distance from point A to this nothing SF point must be determined as follows: –

ten = 2

15 20

ten = 1.5 m Entire distance from point A = 2 + 1.5 = 3.5 m

hence,

BM soap = ( 30 x 3.5 ) + ( -15 x 1.5 ) + ( -10 X 1.5 x 0.75 ) = 71.25 kNm

The undermentioned page shows the line, shear force and flexing minute diagrams for this beam.

70

71.25

35

30

60

70

0

0

## Simply Supported Beam with Point and Distributed Loads ( 1 )

2 m

ten

30

-5

Shear Force Diagram ( kN )

0

-35

15

0

6 m

F

Tocopherol

Calciferol

C

Gram

Bacillus

A

30 kN

35 kN

15 kN

30 kN

UDL = 10 kN/m

20 kN

Bending Moment Diagram ( kNm )

Max Tensile Stress

SAGGING ( +ve bending )

Max Compressive Stress

F

F

A maximal bending minute of 71.25 kNm occurs at a distance 3.5 m from place A.

## Simply Supported Beam with Point and Distributed Loads ( 2 )

1 m

Rubidium

12 m

Tocopherol

Calciferol

C

F

Bacillus

A

8 kN

Rhenium

UDL = 6 kN/m

UDL = 4 kN/m

12 kN

We must foremost cipher the reactions RB and RE. We take minutes about one of the reactions to cipher the other, hence to happen RB.

Take minutes about RE

?Clockwise minutes ( CM ) = ?Anti-clockwise minutes ( ACM )

( RBx10 ) + ( 6x1x0.5 ) = ( 4 x 4 ten 9 ) + ( 8 x 7 ) + ( 12 x 3 ) + ( 6 x 3 ten 1.5 )

## RB = 26 kN

now,

?Upward Forces = ?Downward Forces

RB + RE = ( 4 x 4 ) + 8 + 12 + ( 6 x 4 )

26 + RE = 60

## RE = 34 kN

## Calculating Shear Forces

Get downing at point A and looking left:

SFA – = 0 kN

SFA + = 0 kN

SFB – = -4 x 1 = -4 kN

SFB + = ( -4 x 1 ) + 26 = 22 kN

SFC – = ( -4 x 4 ) + 26= 10 kN

SFC + = ( -4 x 4 ) + 26 – 8 = 2 kN

SFD – = ( -4 x 4 ) + 26 – 8 = 2 kN

SFD + = ( -4 x 4 ) + 26 – 8 – 12 = -10 kN

SFE – = ( -4 x 4 ) + 26 – 8 – 12 – ( 6 x 3 ) = -28 kN

SFE + = ( -4 x 4 ) + 26 – 8 – 12 – ( 6 x 3 ) + 34 = 6 kN

SFF – = ( -4 x 4 ) + 26 – 8 – 12 – ( 6 x 4 ) + 34 = 0 kN

SFF + = ( -4 x 4 ) + 26 – 8 – 12 – ( 6 x 4 ) + 34 = 0 kN

## Calculating Bending Moments

Get downing at point A and looking left:

BMA = 0 kNm

BMB = ( -4 x 1 ten 0.5 ) = -2 kNm

BM 2m from A = ( -4 x 2 ten 1 ) + ( 26 x 1 ) = 18 kNm

BM 3m from A = ( -4 x 3 ten 1.5 ) + ( 26 x 2 ) = 34 kNm

BMC = ( -4 x 4 ten 2 ) + ( 26 x 3 ) = 46 kNm

BMD = ( -4 x 4 ten 6 ) + ( 26 x 7 ) + ( -8 x 4 ) = 54 kNm

BM 9m from A = ( -4 x 4 ten 7 ) + ( 26 x 8 ) + ( -8 x 5 ) + ( -12 x 1 ) +

( -6 x 1 ten 0.5 ) = 41 kNm

BM 9m from A = ( -4 x 4 ten 8 ) + ( 26 x 9 ) + ( -8 x 6 ) + ( -12 x 2 ) +

( -6 x 2 ten 1 ) = 22 kNm

BME = ( -4 x 4 ten 9 ) + ( 26 x 10 ) + ( -8 x 7 ) + ( -12 x 3 ) +

( -6 x 3 ten 1.5 ) = -3 kNm

BMF = ( -4 x 4 ten 10 ) + ( 26 x 11 ) + ( -8 x 8 ) + ( -12 x 4 ) +

( -6 x 4 ten 2 ) + ( 34 x 1 ) = 0 kNm

## Point of Contraflexure

At any point where the graph on a bending minute diagram passes through the 0-0 data point line ( i.e. where the BM alterations mark ) the curvature of the beam will alter from hogging to drooping or frailty versa. Such a point is termed a Point of Contraflexure or Inflexion. These points are identified in the undermentioned diagram. It should be noted that the point of contraflexure corresponds to zero flexing minute.

## Turning Points

The mathematical relationship between shear force and matching flexing minute is evidenced on their several graphs where the alteration of incline on a BM diagram aligns with zero shear on the complementary shear force diagram. Therefore, at any point on a BM diagram where the incline alterations way from upwards to downwards or frailty versa, all such Turning Points occur at places of Zero Shear. Turning points are besides identified in the undermentioned diagram.

## Simply Supported Beam with Point and Distributed Loads ( 2 )

1 m

26 kN

12 m

Tocopherol

Calciferol

C

F

Bacillus

A

8 kN

34 kN

UDL = 6 kN/m

UDL = 4 kN/m

12 kN

2

6

2

-4

22

-10

Shear Force Diagram ( kN )

0

-28

10

0

F

F

SAGGING ( +ve bending )

-3

22

41

54

46

34

18

-2

Bending Moment Diagram ( kNm )

0

0

F

F

HOGGING ( -ve bending )

Points of Contraflexure

The maximal bending minute is equal to 54 kNm and occurs at point D where the shear force is zero. Turning points occur at -2 kNm and -3 kNm.

## Cantilever Beam with Point Load

6 m

F

Tocopherol

Calciferol

C

Gram

Bacillus

A

RA

12 kN

Free End

Fixed End

In this instance there is merely one unknown reaction at the fixed terminal of the cantilever, therefore:

?Upward Forces = ?Downward Forces

## RA = 12 kN

## Calculating Shear Forces –

Get downing at point A and looking left:

SFA – = 0 kN

SFA + = 12 kN

SFB – = 12 kN

SFB + = 12 kN

SFC – = 12 kN

SFC + = 12 kN

SFD – = 12 kN

SFD + = 12 kN

SFE – = 12 kN

SFE + = 12 kN

SFF – = 12 kN

SFF + = 12 kN

SFG – = 12 kN

SFG + = 12 – 12 = 0 kN

Note: the shear force at either terminal of a cantilever beam must compare to zero.

## Calculating Bending Moments – Niobium for simpleness at this phase we shall ever look towards the free terminal of the beam.

Get downing at fixed terminal, point A, and looking right towards the free terminal:

( the same consequences may be obtained by get downing at point G and looking right )

BMA = -12 x 6 = -72 kNm

BMB = -12 x 5 = -60 kNm

BMC = -12 x 4 = -48 kNm

BMD = -12 x 3 = -36 kNm

BME = -12 x 2 = -24 kNm

BMF = -12 x 1 = -12 kNm

BMG = 0 kNm

Notes:

the maximal bending minute in a cantilever beam occurs at the fixed terminal. In this instance the 12kN force in the beam is seeking to flex it downwards, ( a clockwise minute ) . The support at the fixed terminal must hence be using an equal but opposite minute to the beam. This would be 72 kNm in an anti-clockwise way. See the undermentioned diagram.

The value of the bending minute at the free terminal of a cantilever beam will ever be zero.

-12

-24

-36

-48

-60

-72

Bending Moment Diagram ( kNm )

0

0

12

125

Shear Force Diagram ( kN )

0

0

72 kNm

72 kNm

6 m

F

Tocopherol

Calciferol

C

Gram

Bacillus

A

12 kN

12 kN

The followers shows the line, shear force and flexing minute diagrams for this beam.

F

F

HOGGING ( -ve bending )

Max Tensile Stress

Max Compressive Stress

A maximal bending minute of -72 kNm occurs at place A.

## Cantilever Beam with Distributed Load

UDL = 2 kN/m

6 m

F

Tocopherol

Calciferol

C

Gram

Bacillus

A

RA

To cipher the unknown reaction at the fixed terminal of the cantilever:

?Upward Forces = ?Downward Forces

RA = 2 tens 6

## RA = 12 kN

## Calculating Shear Forces

Get downing at point A and looking left:

SFA – = 0 kN

SFA + = 12 kN

SFB – = 12 – ( 2 x 1 ) = 10 kN

SFB + = 12 – ( 2 x 1 ) = 10 kN

SFC – = 12 – ( 2 x 2 ) = 8 kN

SFC + = 12 – ( 2 x 2 ) = 8 kN

SFD – = 12 – ( 2 x 3 ) = 6 kN

SFD + = 12 – ( 2 x 3 ) = 6 kN

SFE – = 12 – ( 2 x 4 ) = 4 kN

SFE + = 12 – ( 2 x 4 ) = 4 kN

SFF – = 12 – ( 2 x 5 ) = 2 kN

SFF + = 12 – ( 2 x 5 ) = 2 kN

SFG – = 12 – ( 2 x 6 ) = 0 kN

SFG + = 12 – ( 2 x 6 ) = 0 kN

Note: the shear force at either terminal of a cantilever beam must compare to zero.

## Calculating Bending Moments

Get downing at fixed terminal, point A, and looking right towards the free terminal:

( the same consequences may be obtained by get downing at point G and looking right )

BMA = -2 x 6 ten 3 = -36 kNm

BMB = -2 x 5 ten 2.5 = -25 kNm

BMC = -2 x 4 ten 2 = -16 kNm

BMD = -2 x 3 ten 1.5 = -9 kNm

BME = -2 x 2 ten 1 = -4 kNm

BMF = -2 x 1 ten 0.5 = -1 kNm

BMG = 0 kNm

The undermentioned page shows the line, shear force and flexing minute diagrams for this beam.

Cantilever Beam with Distributed Load8

6

4

2

36 kNm

36 kNm

12

105

Shear Force Diagram ( kN )

0

0

-1

-4

-9

-16

-25

-36

Bending Moment Diagram ( kNm )

0

0

6 m

F

Tocopherol

Calciferol

C

Gram

Bacillus

A

12 kN

UDL = 2 kN/m

F

F

HOGGING ( -ve bending )

Max Tensile Stress

Max Compressive Stress

A maximal bending minute of -36 kNm occurs at place A.

## Cantilever Beam with Point and Distributed Loads

Roentgenium

2 m

10 kN

Bacillus

C

Calciferol

Tocopherol

A

F

Gram

4 m

UDL = 10 kN/m

To cipher the unknown reaction at the fixed terminal of the cantilever:

?Upward Forces = ?Downward Forces

RG = ( 10 x 6 ) + 10

## RG = 70 kN

## Calculating Shear Forces

Get downing at point A and looking left:

SFA – = 0 kN

SFA + = 0 kN

SFB – = -10 x 1 = -10 kN

SFB + = -10 x 1 = -10 kN

SFC – = -10 x 2 = -20 kN

SFC + = ( -10 x 2 ) + ( -10 ) = -30 kN

SFD – = ( -10 x 3 ) + ( -10 ) = -40 kN

SFD + = ( -10 x 3 ) + ( -10 ) = -40 kN

SFE – = ( -10 x 4 ) + ( -10 ) = -50 kN

SFE + = ( -10 x 4 ) + ( -10 ) = -50 kN

SFF – = ( -10 x 5 ) + ( -10 ) = -60 kN

SFF + = ( -10 x 5 ) + ( -10 ) = -60 kN

SFG – = ( -10 x 6 ) + ( -10 ) = -70 kN

SFG + = ( -10 x 6 ) + ( -10 ) + 70 = 0 kN

Note: the shear force at either terminal of a cantilever beam must compare to zero.

## Calculating Bending Moments

Get downing at point A, and looking left from the free terminal:

( the same consequences may be obtained by get downing at point G and looking left )

BMA = 0 kNm

BMB = -10 x 1 ten 0.5 = -5 kNm

BMC = -10 x 2 ten 1 = -20 kNm

BMD = ( -10 x 3 ten 1.5 ) + ( -10 x 1 ) = -55 kNm

BME = ( -10 x 4 ten 2 ) + ( -10 x 2 ) = -100 kNm

BMF = ( -10 x 5 ten 2.5 ) + ( -10 x 3 ) = -155 kNm

BMG = ( -10 x 6 ten 3 ) + ( -10 x 4 ) = -220 kNm

The undermentioned page shows the line, shear force and flexing minute diagrams for this beam.

70 kN

2 m

10 kN

Bacillus

C

Calciferol

Tocopherol

A

F

Gram

4 m

UDL = 10 kN/m

## 0

## 0

## Shear Force Diagram ( kN )

## -60

## -70

## -10

## -20

## -40

## -50

## 220 kNm

## 220 kNm

## -30Cantilever Beam with Point and Distributed Loads

0

0

Bending Moment Diagram ( kNm )

-220

-5

-20

-55

-100

-155

F

F

HOGGING ( -ve bending )

Max Tensile Stress

Max Compressive Stress

A maximal bending minute of -220 kNm occurs at place G.