Shear Force and Bending MomentDiagrams for a Beam

StevenVukazichSanJoseStateUniversity

1. Find all of the external forces and draw the external force diagram;2. Choose a sign convention for each diagram;3. If necessary, choose a reference coordinate system:4. Use equilibrium analysis or differential and integral relationships to

construct internal force functions;• Cut structure at appropriate sections,• The FBD on either side of the cut may be analyzed,• Indicate unknown internal forces consistent with the chosen sign

convention,• Plot the internal force function for each segment,

5. Check each diagram for errors;• Check discontinuities at location of applied forces in shear diagram,• Check discontinuities at location of applied moment in moment diagram,• Check differential and integral relationships between distributed load,

shear, and bending moment.

General procedure for the construction ofinternal force diagrams

Shear and Bending Moment Diagram Example

A beam is supported by a pin support at point A and extends over a roller support at point D. The beam is and subjected to a linearly varying load from A to B, a point moment at point D a point load at point E as shown.

Draw the diagram and the bending moment diagram for the beam. Label all local maximum and minimum values and their locations and show your sign convention for each diagram.

9 ft 2 ft 3 ft 4 ft

3 k

E

32 k-ft

2 k/ft

DCB

A

1. Find all of the external forces and draw the external force diagram;

FBD of beam

9 ft 2 ft 3 ft 4 ft

3 k

E

32 k-ft

2 k/ft

D

CB

A

Ax

Ay Dy

Find support reactions using equations of equilibrium

!𝑀#

�

�

= 0+

9 ft 2 ft 3 ft 4 ft

3 k

E

32 k-ft

D

CB

A

Ax

Ay Dy

12

2k/ft 9ft = 9k

6 ft 3 ft

Dy = 10 k

Find support reactions using equations of equilibrium

!𝐹1

�

�

= 0+

9 ft 2 ft 3 ft 4 ft

3 k

E

32 k-ft

D

CB

A

Ax

Ay Dy

12

2k/ft 9ft = 9k

6 ft 3 ft

Ay = 2 k

Find support reactions using equations of equilibrium

!𝐹2

�

�

= 0+

Ax = 0

9 ft 2 ft 3 ft 4 ft

3 k

E

32 k-ft

D

CB

A

Ax

Ay Dy

12

2k/ft 9ft = 9k

6 ft 3 ft

External force diagram

FBD of beam showing all known external forces

9 ft 2 ft 3 ft 4 ft

3 k

E

32 k-ft

2 k/ft

D

CB

A

2 k 10 k

9 ft 2 ft 3 ft 4 ft

3 k

E

32 k-ft

2 k/ft

D

CB

A

2 k 10 k

2. Choose a sign convention for each diagram;3. If necessary, choose a reference coordinate system

++Positive shear Positive bending moment

Reference coordinatex

4. Use equilibrium analysis or differential and integralrelationships to construct internal force functions;• Cut structure at appropriate sections,• The FBD on either side of the cut may be analyzed,• Indicate unknown internal forces consistent with the chosen

sign convention,• Plot the internal force function for each segment,

9 ft 2 ft 3 ft 4 ft

3 k

E

32 k-ft

2 k/ft

D

CB

A

2 k 10 k

x

a

a

b

b

c

c

d

d

Analysis of Segment AB FBD of sections on either side of cut at section a-a

3 k

E

32 k-ft

CB

10 k

2 k/fta

a

M

V

Notes• The FBD on either side of the cut may be analyzed;• Unknown internal forces are consistent with the

chosen sign convention;• For this segment the FBD to the left of section a-a is

probably the best choice to analyze.

a

a2 k

x

M

V

FBD of section to the left of cut a-a

Notes• Find the intensity of the distributed load as a function of x;• Replace the distributed load as an equivalent point load;• Two unknowns (M, V) and two equations of equilibrium available.

a

a2 k

x

M

V

𝑤 =2k/ft9ft

𝑥 =29𝑥

a

a2 k 𝑥3

M

V

𝑅 =12

𝑥29

𝑥 =19𝑥7

2𝑥3

Rw

x

Solve for V and M functions for segment AB

a

a2 k 𝑥3

M

V

19𝑥7

2𝑥3

x

!𝐹1

�

�

= 0+

𝑉 = 2 −19𝑥7

0 ≤ 𝑥 ≤ 9

𝑉 0 = 𝑉# = 2kV 9 = 𝑉< = −7k

Solve for V and M functions for segment AB

a

a2 k 𝑥3

M

V

19𝑥7

2𝑥3

x

!𝑀>

�

�

= 0+

𝑀 = 2𝑥 −127

𝑥?

0 ≤ 𝑥 ≤ 9

𝑀 0 = 𝑀# = 0𝑀 9 = 𝑀< = −9 k-ft

Relationships between w, V, and M

𝑀 = 2𝑥 −127

𝑥?

0 ≤ 𝑥 ≤ 9 ft

𝑉 = 2 −19𝑥7

𝑤 = −29𝑥

𝑑𝑉𝑑𝑥

= −192𝑥 = 𝑤

𝑑𝑀𝑑𝑥

= 2 −127

3𝑥7 = 𝑉

a

a2 k

x

M

V

𝑤 =29𝑥

9 ft 2 ft 3 ft 4 ft

3 k

E

32 k-ft

2 k/ft

DCBA

2 k 10 k

Plot V and M functions for segment AB

+

+V

M

2 k

-7 k3 2� = 4.24ft

-9 k-ft

𝑀C>2 = 4 2� = 5.66k−ft0

9 ft 3 ft 4 ft

3 k

E

32 k-ft

2 k/ft

D

CB

A

2 k 10 k

x

b

b

Analysis of Segment BC FBD of sections on either side of cut at section b-b

b

b

M

V

M

V

FBD of section to the left of cut b-b

9 ft

2 k/ft

B

A

2 k

x

b

b

M

V

9 ft

9 k

BA

2 k

x

b

b

M

V3 ft6 ft

Solve for V and M functions for segment BC

!𝐹1

�

�

= 0+

𝑉 = −7

9 ≤ 𝑥 ≤ 11

𝑉 9 = 𝑉<F = −7kV 11 = 𝑉GH = −7k

9 ft

9 k

BA

2 k

x

b

b

M

V3 ft6 ft

Solve for V and M functions for segment BC

!𝑀I

�

�

= 0+

𝑀 = 54 − 7𝑥

9 ≤ 𝑥 ≤ 11

𝑀 9 = 𝑀< = −9k−ft𝑀 11 = 𝑀GH = −23 k-ft

9 ft

9 k

BA

2 k

x

b

b

M

V3 ft6 ft

Relationships between w, V, and M

𝑤 = 0

𝑑𝑉𝑑𝑥

= 0 = 𝑤

𝑑𝑀𝑑𝑥

= −7 = 𝑉

𝑀 = 54 − 7𝑥

9 ft

2 k/ft

B

A

2 k

x

b

b

M

V

𝑉 = −7

9 ft 2 ft 3 ft 4 ft

3 k

E

32 k-ft

2 k/ft

DCBA

2 k 10 k

Plot V and M functions for segment BC

+

+V

M

2 k

-7 k3 2� = 4.24ft

-9 k-ft

𝑀C>2 = 4 2� = 5.66k−ft0

-23 k-ft

Analysis of Segment CD FBD of sections on either side of cut at section c-c

9 ft 2 ft 4 ft

3 k

E

32 k-ft

2 k/ft

D

CB

A

2 k 10 k

x

c

c

c

c

M

V

M

V

FBD of section to the left of cut c-c

9 ft

9 k

BA

2 k

x

3 ft6 ft

c

c

M

V

2 ft

32 k-ft

C

Solve for V and M functions for segment CD

!𝐹1

�

�

= 0+

𝑉 = −7

11 ≤ 𝑥 ≤ 14

𝑉 11 = 𝑉GF = −7kV 14 = 𝑉JH = −7k

9 ft

9 k

BA

2 k

x

3 ft6 ft

c

c

M

V

2 ft

32 k-ft

C

Solve for V and M functions for segment CD

!𝑀K

�

�

= 0+

𝑀 = 86 − 7𝑥

11 ≤ 𝑥 ≤ 14

𝑀 11 = 𝑀GF = 9k−ft𝑀 14 = 𝑀J = −12 k-ft

9 ft

9 k

BA

2 k

x

3 ft6 ft

c

c

M

V

2 ft

32 k-ft

C

Relationships between w, V, and M

𝑤 = 0

𝑑𝑉𝑑𝑥

= 0 = 𝑤

𝑑𝑀𝑑𝑥

= −7 = 𝑉

𝑀 = 86 − 7𝑥

𝑉 = −7

9 ft 2 ft

32 k-ft

2 k/ft

CB

A

2 k

x

c

c

M

V

9 ft 2 ft 3 ft 4 ft

3 k

E

32 k-ft

2 k/ft

DCBA

2 k 10 k

Plot V and M functions for segment CD

+

+V

M

2 k

-7 k3 2� = 4.24ft

-9 k-ft

𝑀C>2 = 4 2� = 5.66k−ft0

-23 k-ft

9 k-ft

-12 k-ft

-7 k

Analysis of Segment DE FBD of sections on either side of cut at section d-d

9 ft 2 ft 3 ft

3 k

E

2 k/ft

D

CB

A

2 k 10 k

x

d

d

32 k-ft M

V

M

V

18 - x

d

d

FBD of section to the right of cut d-d

3 k

E

M

V

18 - x

d

d

Solve for V and M functions for segment CD

!𝐹1

�

�

= 0+

𝑉 = 3

14 ≤ 𝑥 ≤ 18

𝑉 14 = 𝑉JF = 3kV 18 = 𝑉M = 3k

3 k

E

M

V

18 - x

d

d

Solve for V and M functions for segment DE

!𝑀N

�

�

= 0+

𝑀 = −54 + 3𝑥

14 ≤ 𝑥 ≤ 18

𝑀 14 = 𝑀JF = −12k−ft𝑀 18 = 𝑀M = 0

Relationships between w, V, and M

𝑤 = 0

𝑑𝑉𝑑𝑥

= 0 = 𝑤

𝑑𝑀𝑑𝑥

= 3 = 𝑉

𝑀 = −54 + 3𝑥

𝑉 = 3

9 ft 2 ft 3 ft

2 k/ft

D

CB

A

2 k 10 k

x

d

d

32 k-ft M

V

9 ft 2 ft 3 ft 4 ft

3 k

E

32 k-ft

2 k/ft

DCBA

2 k 10 k

Plot V and M functions for segment DE

+

+V

M

2 k

-7 k3 2� = 4.24ft

-9 k-ft

𝑀C>2 = 4 2� = 5.66k−ft0

-23 k-ft

9 k-ft

-12 k-ft

3 k

0

3 k

-7 k