Simply Supported And Cantilever Beams

A beam is a structural member which safely carries loads i.e. without failing due to the applied loads. We will be restricted to beams of uniform cross-sectional area.

Simply Supported Beam

A beam that rests on two supports only along the length of the beam and is allowed to deflect freely when loads are applied. Note – see section A of unit.

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Cantilever Beam

A beam that is supported at one end only. The end could be built into a wall, bolted or welded to another structure for means of support.

Point or Concentrated Load

A load which acts at a particular point along the length of the beam. This load is commonly called a force (F) and is stated in Newtons (N). A mass may be converted into a force by multiplying by gravity whose value is constant at 9.81 m/s2.

Uniformly Distributed Load (UDL)

A load which is spread evenly over a given length of the beam. This may be the weight of the beam itself. The UDL is quoted as Newtons per metre (N/m).

Beam Failure

If excessive loads are used and the beam does not have the necessary material properties of strength then failure will occur. Failure may occur in two ways:-

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Calculating Shear Forces (we must use the shear force rule).

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

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

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Starting at point A and looking left:

(note: the negative sign (-) means just to the left of the position and the positive sign (+) means just to the right of the position.)

SFA – = 0 kN

SFA + = 6 kN

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An alternative method of drawing the shear force diagram is to follow the directions of each force on the line diagram.SFB – = 6 kN

SFB + = 6 kN

SFC – = 6 kN

SFC + = 6 kN

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SFD – = 6 kN

SFD + = 6 – 12 = -6 kN

SFE – = 6 – 12 = -6 kN

SFE + = 6 – 12 = -6 kN

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SFF – = 6 – 12 = -6 kN

SFF + = 6 – 12 = -6 kN

SFG – = 6 – 12 = -6 kN

SFG + = 6 – 12 + 6 = 0 kN

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Note: the shear force at either end of a simply supported beam must equate to zero.

Calculating Bending Moments (we must use the bending moment rule).

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

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

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section

F –

F –

section

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F +

F +

Hogging Beam

Sagging Beam

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Starting at point A and looking left:

BMA = 0 kNm

BMB = (6 x 1) = 6 kNm

BMC = (6 x 2) = 12 kNm

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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

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Note: the bending moment at either end of a simply supported beam must equate to zero.

The following page shows the line, shear force and bending moment diagrams for this beam.

Simply Supported Beam with Point Load

6 m

F

E

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D

C

G

B

A

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6 kN

6 kN

F =12 kN

Shear Force Diagram (kN)

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0

-6

6

0

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Line Diagram

12

12

18

6

0

6

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Bending Moment Diagram (kNm)

Max Tensile Stress

SAGGING (+ve bending)

Max Compressive Stress

F

F

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A maximum bending moment of 18 kNm occurs at position D. Note the shear force is zero at this point.

Simply Supported Beam with Distributed Load

UDL = 2 kN/m

F

E

D

C

G

B

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A

6 m

RA

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

e.g. if we take moments about ‘D’ then the total 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 moments about ‘D’, then the moment would be: (-6 x 1.5) = -9 kNm

1.5m

UDL = 2 kN/m

D

C

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B

A

3 m

Taking moments about point D (looking left)

We must first calculate the reactions RA and RG. We take moments about one of the reactions to calculate the other, therefore to find RA:

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Take moments about RG

ΣClockwise moments (CM) = ΣAnti-clockwise moments (ACM)

RA x 6 = 2 x 6 x 3

RA = 6 kN now,

ΣUpward Forces = ΣDownward Forces

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RA + RG = 2 x 6

6 + RG = 12

RG = 6 kN

section

F +

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F –

F –

F +

Calculating Shear Forces (we must use the shear force rule).

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

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

Starting at point A and looking left:

(note: the negative sign (-) means just to the left of the position and the positive sign (+) means just to the right of the position.)

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 end of a simply supported beam must equate to zero.

Calculating Bending Moments (we must use the bending moment rule).

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

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

section

F –

F –

section

F +

F +

Hogging Beam

Sagging Beam

Starting 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 x 2.5 = 5 kNm

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

Note: the bending moment at either end of a simply supported beam must equate to zero.

The following page shows the line, shear force and bending moment diagrams for this beam.

Simply Supported Beam with Distributed Load

4

2

0

-2

-4

UDL = 2 kN/m

6 m

F

E

D

C

G

B

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 maximum bending moment of 9 kNm occurs at position D. Note the shear force is zero at this point.

Simply Supported Beam with Point Loads

6 m

F

E

D

C

G

B

A

RA

RG

F = 15 kN

F = 30 kN

We must first calculate the reactions RA and RG. We take moments about one of the reactions to calculate the other, therefore to find RA:

Take moments about RG

ΣClockwise moments (CM) = ΣAnti-clockwise moments (ACM)

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

RA = 20 kN now,

ΣUpward Forces = ΣDownward Forces

RA + RG = 15 + 30

20 + RG = 45

RG = 25 kN

section

F +

F –

F –

F +

Calculating Shear Forces (we must use the shear force rule).

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

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

Starting at point A and looking left:

(note: the negative sign (-) means just to the left of the position and the positive sign (+) means just to the right of the position.)

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 end of a simply supported beam must equate to zero.

Calculating Bending Moments (we must use the bending moment rule).

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

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

section

F –

F –

section

F +

F +

Hogging Beam

Sagging Beam

Starting 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 moment at either end of a simply supported beam must equate to zero.

The following page shows the line, shear force and bending moment diagrams for this beam.

0

20

-25

0

Shear Force Diagram (kN)

5Simply Supported Beam with Point Loads

6 m

F

E

D

C

G

B

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 maximum bending moment of 50 kNm occurs at position E. Note the shear force is zero at this point.

Simply Supported Beam with Point and Distributed Loads (1)

6 m

F

E

D

C

G

B

A

RA

RG

15 kN

30 kN

UDL = 10 kN/m

We must first calculate the reactions RA and RG. We take moments about one of the reactions to calculate the other, therefore to find RA:

Take moments about RG

ΣClockwise moments (CM) = ΣAnti-clockwise moments (ACM)

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

RA = 30 kN now,

ΣUpward Forces = ΣDownward Forces

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

30 + RG = 65

RG = 35 kN

section

F +

F –

F –

F +

Calculating Shear Forces (we must use the shear force rule).

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

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

Starting at point A and looking left:

(note: the negative sign (-) means just to the left of the position and the positive sign (+) means just to the right of the position.)

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 end of a simply supported beam must equate to zero.

Calculating Bending Moments (we must use the bending moment rule).

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

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

section

F –

F –

section

F +

F +

Hogging Beam

Sagging Beam

Starting 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 moment at either end of a simply supported beam must equate to zero.

The value of the maximum bending moment occurs where the shear force is zero and is therefore still unknown (see Shear Force diagram). The distance from point A to this zero SF point must be determined as follows:-

x = 2

15 20

x = 1.5 m Total distance from point A = 2 + 1.5 = 3.5 m

therefore,

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

The following page shows the line, shear force and bending moment diagrams for this beam.

70

71.25

35

30

60

70

0

0

Simply Supported Beam with Point and Distributed Loads (1)

2 m

x

30

-5

Shear Force Diagram (kN)

0

-35

15

0

6 m

F

E

D

C

G

B

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 maximum bending moment of 71.25 kNm occurs at a distance 3.5 m from position A.

Simply Supported Beam with Point and Distributed Loads (2)

1 m

RB

12 m

E

D

C

F

B

A

8 kN

RE

UDL = 6 kN/m

UDL = 4 kN/m

12 kN

We must first calculate the reactions RB and RE. We take moments about one of the reactions to calculate the other, therefore to find RB.

Take moments about RE

ΣClockwise moments (CM) = ΣAnti-clockwise moments (ACM)

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

RB = 26 kN

now,

ΣUpward Forces = ΣDownward Forces

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

26 + RE = 60

RE = 34 kN

Calculating Shear Forces

Starting 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

Starting at point A and looking left:

BMA = 0 kNm

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

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

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

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

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

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

(-6 x 1 x 0.5) = 41 kNm

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

(-6 x 2 x 1) = 22 kNm

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

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

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

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

Point of Contraflexure

At any point where the graph on a bending moment diagram passes through the 0-0 datum line (i.e. where the BM changes sign) the curvature of the beam will change from hogging to sagging or vice versa. Such a point is termed a Point of Contraflexure or Inflexion. These points are identified in the following diagram. It should be noted that the point of contraflexure corresponds to zero bending moment.

Turning Points

The mathematical relationship between shear force and corresponding bending moment is evidenced on their respective graphs where the change of slope on a BM diagram aligns with zero shear on the complementary shear force diagram. Thus, at any point on a BM diagram where the slope changes direction from upwards to downwards or vice versa, all such Turning Points occur at positions of Zero Shear. Turning points are also identified in the following diagram.

Simply Supported Beam with Point and Distributed Loads (2)

1 m

26 kN

12 m

E

D

C

F

B

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 maximum bending moment 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

E

D

C

G

B

A

RA

12 kN

Free End

Fixed End

In this case there is only one unknown reaction at the fixed end of the cantilever, therefore:

ΣUpward Forces = ΣDownward Forces

RA = 12 kN

Calculating Shear Forces –

Starting 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 end of a cantilever beam must equate to zero.

Calculating Bending Moments – NB for simplicity at this stage we shall always look towards the free end of the beam.

Starting at fixed end, point A, and looking right towards the free end:

(the same results may be obtained by starting 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 maximum bending moment in a cantilever beam occurs at the fixed end. In this case the 12kN force in the beam is trying to bend it downwards, (a clockwise moment). The support at the fixed end must therefore be applying an equal but opposite moment to the beam. This would be 72 kNm in an anti-clockwise direction. See the following diagram.

The value of the bending moment at the free end of a cantilever beam will always 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

E

D

C

G

B

A

12 kN

12 kN

The following shows the line, shear force and bending moment diagrams for this beam.

F

F

HOGGING (-ve bending)

Max Tensile Stress

Max Compressive Stress

A maximum bending moment of -72 kNm occurs at position A.

Cantilever Beam with Distributed Load

UDL = 2 kN/m

6 m

F

E

D

C

G

B

A

RA

To calculate the unknown reaction at the fixed end of the cantilever:

ΣUpward Forces = ΣDownward Forces

RA = 2 x 6

RA = 12 kN

Calculating Shear Forces

Starting 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 end of a cantilever beam must equate to zero.

Calculating Bending Moments

Starting at fixed end, point A, and looking right towards the free end:

(the same results may be obtained by starting at point G and looking right)

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

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

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

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

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

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

BMG = 0 kNm

The following page shows the line, shear force and bending moment 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

E

D

C

G

B

A

12 kN

UDL = 2 kN/m

F

F

HOGGING (-ve bending)

Max Tensile Stress

Max Compressive Stress

A maximum bending moment of -36 kNm occurs at position A.

Cantilever Beam with Point and Distributed Loads

RG

2 m

10 kN

B

C

D

E

A

F

G

4 m

UDL = 10 kN/m

To calculate the unknown reaction at the fixed end of the cantilever:

ΣUpward Forces = ΣDownward Forces

RG = (10 x 6) + 10

RG = 70 kN

Calculating Shear Forces

Starting 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 end of a cantilever beam must equate to zero.

Calculating Bending Moments

Starting at point A, and looking left from the free end:

(the same results may be obtained by starting at point G and looking left)

BMA = 0 kNm

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

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

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

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

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

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

The following page shows the line, shear force and bending moment diagrams for this beam.

70 kN

2 m

10 kN

B

C

D

E

A

F

G

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 maximum bending moment of -220 kNm occurs at position G.