Shear Force and Bending Moment Diagrams Dr Jonathan Leaver

Structural Joints

Source: Wei Loo

Structural Joints (cont.)

Source: Wei Loo

Determining Reactions on a Beam An object that is not accelerating (e.g. stationary or moving at constant

velocity) must have zero net force acting on it. e.g. Fx = 0, Fx = 0, Fx = 0 If an object is not rotating it must have zero net moment acting on it. Consider the following beam:

Loads (downwards) are forces imposed on the beam Reactions (upwards) result from the imposed loads.

To determine the reactions. Follow these steps:

1. Draw free body diagram (i.e. replace supports with forces and moments as applicable)

2. Replace any UDLs with a single force acting at the midpoint of the UDL.

3. Find the right hand reaction by taking moments about the point of action of the left hand reaction.

4. Find the left hand reaction by adding all the forces together for the x and y directions and putting these equal to zero.

Determining Reactions on a Beam (cont.)

Calculate reactions

Sum moment about A: 6kN*3m + 10kN*5m RD*10m = 0

therefore RD = 6.8 kN

Sum forces: RA 6kN -10kN +6.8 = 0

therefore RD = 9.2 kN

Example 1 Find the reactions.

1) Determine reactions MA = 0 (Clockwise +ve) 2kN*1m - RB*3m + (2kN/m*1m)*2.5m + 4kN*2m = 0 RB = 5kN F = 0 gives RA 2kN + 5kN 4kN 2kN/m*1m = 0 RA = 3 kN

Example 2

Examples

(a) (a)

(b)

(c)

Find the reactions.

Solutions

(a) (a)

(b)

(c)

Reactions: 2.857 kN (left support), 7.143 kN . No moments. 50 kN (left support), 150 kN. No moments. 150 kN, -410 kN-m

45o

Example

Find the reactions RA and RB.

RAX = 4.24 kN, RAY = -0.621 kN, RB = 7.86 kN

RA RB

Solution

Source: Wei Loo

Example Find the reactions and moment at A.

3 kN/m

1.5 m 2 m

4 kN-m

A

1 m

1) Draw free body diagram

2) MA = 0 MA +3 kN/m*2m*(1.5+1) 4 kN-m = 0 Notice how the distance of the 4 kN-m moment from A does not enter the equation. Hence the influence of the moment is independent of location. MA = - 11 kN-m 3) FAY = 0 RAY 3kN/m*2m = 0 RAY = 6 kN 4) FAx = 0 RAX = 0

3 kN/m

1.5 m 2 m

4 kN-m

MA

1 m

RAY RAx

Solution

Calculate the reactions at the supports.

Answers: RA = 75 kN RB= 60 kN

Problem

For equilibrium: Sum of forces in x and y directions = 0 sum of moments =0

DE4101 Students can ignore slides past this point.

ENGG5038 students should study all slides.

1. Draw Free Body Diagram

Draw free body diagrams (i.e. replace supports with forces and moments as applicable)

2. Find Reactions

Take moments about one of the supports to determine one reaction.

Sum forces in x and y directions to determine the other reaction.

Draw deflected shape

3. Find Shear Forces

Determine SFs by taking cuts just before and just after each applied load and reaction.

Draw the SF diagram remembering +ve shear convention (see diagram).

Finding Reactions Shear and Bending Moments

+ve shear +ve shear

1. Find Bending Moments

a) Theoretical method

Determine BMs by taking cuts at each applied load. In civil engineering a positive BM is usually drawn on the tension side of the beam. (As long as you get the shape of the BM correct no marks will be deducted for drawing on the compression side of the beam as mechanical engineers use this method)

b) Simplified method

BM is the area under the shear force diagram. (SF is the gradient of the BM diagram.)

a. SF = 0 when the BM is a maximum or minimum.

b. BM=0 at points of contraflexure.

Finding Reactions Shear and Bending Moments

Bending Shear and Deflection

1) Determine shear forces

(assume SF at cut is +ve upwards)

From A

Cut just left of B: SFB- +4.4 = -4.4kN

Cut just right of B: SFB+ +4.4kN 12kN = +7.6 kN

Cut just right of C: SFc + 4.4kN 12kN +17kN = -9.4kN

Cut just right of D: SFc + 4.4kN 12kN +17kN -8kN = -1.4kN

Cut just right of E: SFc + 4.4kN 12kN +17kN -8kN 12 kN= +10.6kN

2) Determine moments

(assume moment at cut is +ve clockwise)

From Left:

Cut at B: MB + 4.4kN*2 = 0 MB = -8.8kN-m

Cut at C: Mc + 4.4kN*5m - 12kN*3m = 0 Mc = +14 kN-m

Cut at D: MD + 4.4kN*8m - 12kN*6m +17 kN*3m = 0MD= -14.2kN-m

Cut at E: ME + 4.4kN*9.5m - 12kN*7.5m +17 kN*4.5m -8kN*1.5m = 0

ME = -16.3kN-m

At F: MF = 0

Bending Shear and Deflection (Simplified Method) 1) Determine shear forces

(assume SF at cut is +ve upwards)

From A

Cut just left of B: SFB- +4.4 = -4.4kN

Cut just right of B: SFB+ +4.4kN 12kN = +7.6 kN

Cut just right of C: SFc + 4.4kN 12kN +17kN = -9.4kN

Cut just right of D: SFc + 4.4kN 12kN +17kN -8kN = -1.4kN

Cut just right of E: SFc + 4.4kN 12kN +17kN -8kN 12 kN= +10.6kN

2) Determine moments

Moments are the area under the SF diagram.

From Left:

At B: MB = -4.4kN*2 = 0 MB = -8.8kN-m

At C: Mc = -8.8 kN-m + 3m * 7.6 kN Mc =+14 kN-m

At D: MD = 14 kN-m + 3m * (-9.4) kN = MD = -14.2kN-m

At E: ME = -14.2 kN-m + 1.5m*(-1.4) kN ME = -16.3kN-m

At F: MF = 0

Common Results of Beam Analysis

Note: W(N) = w(N/m)*L(m) in right side figure

Ref: Steel Designers Manual CSRDC 1972

Exercises

Exercises

Exercises

Exercises

Solution

Source: Wei Loo

Point SF (Sum forces to the left) kN

BM (Sum BM to left) kN-m

A+ = 9.2 = 0 B+ = 9.2-6

= 3.2 = 9.2*3 = 27.6

C+ = 9.2-6 -10 = -6.8

=9.2*5-6*2 = 34

Calculate reactions

Sum moment about A: 6kN*3m + 10kN*5m RD*10m = 0

therefore RD = 6.8 kN

Sum forces: RA 6kN -10kN +6.8 = 0

therefore RD = 9.2 kN

Example

Exercises

Source: Wei Loo

i. Draw the deflected shape.

ii. Find reactions. Take moments at one support and then sum forces to find other reaction.

iii. Find BMs. Take cuts at bend and where a force acts. Draw the BM on the compression side of the beam.

iv. Find SFs. Take cuts from an end and remember +ve shear rule. The shear force is the slope of the BM diagram

Solution

Source: Wei Loo

Find SF

Start from an end say C:

Take a cut at D: SFD 5 = 0; SFD = 5kN. Either +5kN or 5kN is acceptable.

The shear force at D from D-C becomes an axial force in member DE . Hence there is no shear force in DE but the axial force is 5kN. You do not need to show the axial force.

A

F

E D

C

B

H

G

Solving for Shear Forces

Solving for Shear Forces

Take a cut just above A: SFA- 5kN = 0 SFA- = 5kN

Take a cut just to the right of A:

Summing all the forces in the vertical direction on the structure to the left

SFA+ + 4.375 = 0 SFA+ = -4.375kN

5 kN

SFA

2m

E

A

Solving for Shear Forces

Source: Wei Loo

Solving for Bending Moments

Source: Wei Loo

Solving for Bending Moments

Note: The BM here is drawn on the tension side of the beam. Drawing on the compression side means that the shear force represents the slope of the BM with the correct sign. (Either method is correct)

Exercises (i) Draw the deflected shape.

(ii) Find the reactions

(iii) Draw SF and BM diagrams

A

D

C B

E

Solution: Deflected shape

3 m

2 m

1 m -3.203,-0.01,0-3.203,-2.579,0

0,0,0 0,0,0y

x

1) Determine reactions MA = 0 (Clockwise +ve) 2kN*1m - RB*3m + (2kN/m*1m)*2.5m + 4kN*2m = 0 RB = 5kN F = 0 gives RA 2kN + 5kN 4kN 2kN/m*1m = 0 RA = 3 kN 2) Determine moments From Left: Cut at C: Mc + 3kN*1m = 0 Mc = -3 kN-m Cut at B: MB + 3kN*3 2kN*2m = 0 MB = -5kN-m At D since the distance to forces is the same as for B then MD = MB = -5kN-m At E: ME = 0 3) Determine shear forces From A Cut just left of C: SF = 3kN Cut just right of C: SF = 3kN 2kN = 1 kN Cut just left of B: SF = same as above = 1kN From E Cut just right of E: SF = - 1kN Cut just left of D: SF = -4kN Just right of E SF = - 4 kN, At D SF = 4 + (2 kN/m*1m) = - 6 kN SF in D-B = 0 (but the axial force = 6 kN)

Solution

Solution

3 m

2 m

1 m -3.203,-0.01,0-3.203,-2.579,0

0,0,0 0,0,0y

x

Exercises

Find the shear force and bending moment diagrams.

Calculate the magnitude of the SF and BM at point K.

Source: Materials and Structures: Whitlow 1991

Solutions

Source: Materials and Structures: Whitlow 1991

Shear Force and Bending Moment Diagrams Dr Jonathan LeaverStructural JointsStructural Joints (cont.)Determining Reactions on a BeamSlide Number 5Slide Number 6Slide Number 7Slide Number 8Slide Number 9Slide Number 10Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22Slide Number 23Slide Number 24Slide Number 25Slide Number 26Slide Number 27Slide Number 28Slide Number 29Slide Number 30Slide Number 31Slide Number 32Slide Number 33Slide Number 34Slide Number 35Slide Number 36Slide Number 37Slide Number 38Slide Number 39