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CE 160 Notes: Shear and Moment Diagrams for Beams

Shear and moment diagrams are plots of how the internal bending moment and shear vary along the length of the beam.

Sign Convention for V and M

Consider the arbitrarily loaded simple beam:

Cut the beam loose from its pin support at the left, roller support at the right and through the beam at section a-a

Free Body Diagrams (F. B. D.) of portion of beam to the left of a-a and portion of the beam to the right of a-a (note that there is no axial force in this beam):

Reactive forces are developed at each support and equal and opposite internal shear force (V) and bending moment (M) are present to keep each beam segment in equilibrium.

The above senses for V and M are the “usual” Civil Engineering convention for positive shear and positive bending moment.

It is good practice to show the sign convention next to shear and moment diagrams to make the user clear of the convention for positive shear and moment.

a

a

a

aM

a

aM

V V

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Some common icons for showing sign convention

Differential and integral relationships between V, M, and w:

𝑑𝑉𝑑𝑥 = 𝑤 slope of tangent to V diagram at a point

= (distributed load intensity at that point)

𝑑𝑀𝑑𝑥 = 𝑉 slope of tangent to M diagram at a point = value of V at that point

VB – VA = (Area of distributed load between points A and B)

MB – MA = Area of V diagram between points A and B

Example Problem

Construct the Shear and Bending Moment Diagram for the following beam:

A

B

E

4 ft 2 ft

10 k2 k/ft

2 ft 2 ft

20 k-ft

C D

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Find support reactions at A and E

F.B.D. of entire beam cut loose from its supports

Need to replace distributed load as an equivalent point load (equivalent point load = area of distribution placed at the centroid of the distribution)

Apply equations of equilibrium to find unknown reactions

Moment Equilibrium

𝑀! = 0

Counterclockwise moments about point A positive

− 8 𝑘)(2 𝑓𝑡 + − 10 𝑘 )(6 𝑓𝑡 − 20 𝑘-𝑓𝑡 + 𝐸! )(10 𝑓𝑡 = 0 𝑬𝒚 = 𝟗.𝟔 𝒌 (Ey is positive so Ey acts upward as assumed)

Ax

B E

4 ft 2 ft

10 k2 k/ft

Ay Ey

A

2 ft 2 ft

C D

20 k-ft

Ax

B

E

2 ft 2 ft

10 k

(2 k/ft)*(4 ft) = 8 k

Ay Ey

A

2 ft

20 k-ft

C D

2 ft 2 ft

+

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Force Equilibrium

+↑ 𝐹! = 0

Upward forces positive

𝐴! − 8 𝑘 − 10 𝑘 + 𝐸! = 0

𝑨𝒚 = 𝟖.𝟒 𝒌 (Ay is positive so Ay acts upward as assumed)

Force Equilibrium

+→ 𝐹! = 0

forces positive to the right

𝐴! = 0

𝑨𝒙 = 𝟎 (no axial force in beam)

F.B.D of beam with known support reactions shown

Find internal shear and moment from A to B

Make a cut at section a-a, an arbitrary point between points A and B F.B.D of beam to left of cut, show unknown internal forces V and M in their positive sense

B E

4 ft 2 ft

10 k2 k/ft

8.4 k 9.6 k

A

a

a

2 ft 2 ft

20 k-ft

DC

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Replace distributed load as an equivalent point load

Moment Equilibrium of segment

𝑀! = 0

Counterclockwise moments about point a are positive

− 8.4)(𝑥 + 2𝑥)(𝑥/2 + 𝑀 = 0 𝑴 = 𝟖.𝟒𝒙− 𝒙𝟐 (units in k and ft)

Force Equilibrium

+↑ 𝐹! = 0

Upward forces positive

8.4 − 2𝑥 − 𝑉 = 0

2 k/ft

A

8.4 k

M

Va

a

x

A

8.4 k x/2

M

Va

a(2 k/ft)*(x) = 2x

x/2

+

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𝑽 = 𝟖.𝟒− 𝟐𝒙 (units in k and ft)

Plot V and M between points A and B

At point A Evaluate at x = 0

𝑉! = 8.4− 2)(0 = 𝟖.𝟒 𝒌 𝑀! = 8.4)(0 − 0! = 𝟎 𝒌− 𝒇𝒕

At point B Evaluate at x = 4 ft

𝑉! = 8.4− 2)(4 = 𝟎.𝟒 𝒌 𝑀! = 8.4)(4 − 4! = 𝟏𝟕.𝟔 𝒌− 𝒇𝒕

V is linear in x between A and B

M is parabolic in x between A and B

Find internal shear just to the left of point E Make a cut just to the left of point E (point E-)

F.B.D of beam

Force Equilibrium

+↑ 𝐹! = 0

Upward forces positive

9.6 𝑘 + 𝑉!! = 0

𝑽𝑬! = −𝟗.𝟔 k

E

9.6 k

VE-

ME- ~ 0

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Fill in rest of shear diagram

Shear between B and C, C and D, D and E will be constant since no distributed load is found in these regions; Since VB = 0.4 k shear between B and C will be equal to 0.4 k; Note that a shear discontinuity will be expected at C due to the 10 k point load; Since VE = -9.6 k shear between D and E will be equal to -9.6 k; Point moment at D will not affect shear so shear between C and D will be -9.6 k; Shear discontinuity at C is equal to point load of 10k (0.4 k to -9.6 k) O.K.

Find Moment from B to C

MC – MB = Area of V diagram between points B and C

Area of shear diagram between B and C = (0.4 k) (2 ft) = 0.8 k-ft

𝑀! −𝑀! = 0.80 𝑘 − 𝑓𝑡 𝑀! = 17.6 + 0.8 = 𝟏𝟖.𝟒 𝒌− 𝒇𝒕

M is linear in between B and C

Find Moment from C to D

Note that a moment discontinuity will be expected at D due to the 20 k-ft point moment; Area of shear diagram between C and D = (-9.6 k) (2 ft) = -19.2 k-ft

At point D- (just to the left of point D) 𝑀!! −𝑀! = −19.2 𝑘 − 𝑓𝑡 𝑀!! = 18.4− 19.2 = −𝟎.𝟖 𝒌− 𝒇𝒕

M is linear in between C and D

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Find Moment from D to E

Area of shear diagram between D and E = (-9.6 k) (2 ft) = -19.2 k-ft ME = 0 at the roller support

At point D+ (just to the right of point D) 𝑀! −𝑀!! = −19.2 𝑘 − 𝑓𝑡 𝑀!! = − −𝑀! − 19.2 = − 0− 19.2 = 𝟏𝟗.𝟐 𝒌− 𝒇𝒕

M is linear in between D and E

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Plot V and M diagrams under the F.B.D of the beam

Note again differential and integral relationships between V, M, and w:

𝑑𝑉𝑑𝑥 = 𝑤 slope of tangent to V diagram at a point

= (distributed load intensity at that point)

𝑑𝑀𝑑𝑥 = 𝑉 slope of tangent to M diagram at a point = value of V at that point

VB – VA = (Area of distributed load between points A and B)

MB – MA = Area of V diagram between points A and B

M

B C

4 ft 2 ft

10 k2 k/ft

8.4 k 9.6 k

A

x

V

- 9.6 k

8.4 k

0

17.6 k-ft

0

E

20 k-ft

D

0.4 k

18.4 k-ft 19.2 k-ft

-0.8 k-ft

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