ENG1020 Engineering Structures

Lecture 8 Beam bending

Department of Civil EngineeringMonash University

Lecturer: Dr. Bill Wong

Beam analysis

Purpose of beam analysis: calculate the maximummoment and shear force (internal forces) in the beam and check against its capacity for safety

Beam design

Where is the most stressed part of the hangar? Will it fail under the loading?

Beam analysis

What is the maximum moment in the following beam ?

Distribution of internal shear force

Distribution of internal bending moment

Beam failure

Sunbury

Internal forces in a beam

Internal forces at any point along a beam - found by drawing a free body diagram which cuts through that point

Purpose of using free body diagram: By trial and error (or other more scientific ways), we can obtain the set of maximum forces when calculating the forces at different locations of X.

X

A B

Internal forces

The forces at X can be found by drawing a free body diagram through X. Internal forces (2 sets, equal & opposite) at cut face X:

X

A B

At cut face

Sign convention: Positive (+ve) forcesM = bending momentV = Shear forceN = Axial force

N

V

M

Example

Beam with UDL: First, calculate reactions

Forces at a distance X from A Total equivalent load P = 10X kN

at a distance of X/2 from A

w = 10 kN/m

L = 10mX

A B

w = 10 kN/m

X

A

Ay = 50 kN

V

N

MP

C

Example

Take moment about C, Mc = 010X (X/2) - 50X + M = 0M = -5X2 + 50X (1)

Fy = 050 - 10X - V = 0V = 50 - 10X (2)

Fx = 0N = 0

Plotting equations (1) and (2) for X = 0 to 10 m

w = 10 kN/m

X

A

Ay = 50 kN

V

N

MP

C

Diagrams

Bending moment diagram (BMD)

Shear force diagram (SFD)

Maximum bending moment = 125 kNm Maximum shear force = 50 kN Maximum axial force = 0

X = 5 m,M = 125 kNmX = 2.5 m

M = 93.75 kNm

-ve space

+ve space

50 kN

50 kN

In general, Max. M = wL2/8Max. V = wL/2

Short-cuts to BMD and SFD

Shear force diagram (SFD):Follow the directions of the loads (including reactions) from LEFT to RIGHT and plot to scale

Bending moment diagram (BMD):Calculate M as sum of areas under the shear force diagram, and join the points by straight lines - for segments without UDL curves - for segments with UDL

(see notes for more details of the shapes of the diagrams)

Example

From equilibrium Ax = 0, Ay = By = P/2 Shear force diagram Bending moment diagram

L/2

A B

P

L/2C

P/2

-P/2

M = PL/4

Area = PL/4

Area = -PL/4

Relationship between Load, Shear Force and Bending Moment

Consider a small length x of a beam element:

Fy = 0V – wx – (V + V) = 0

w = V/ x M(right hand face) = 0

-M - V x + wxx/2) + (M + M) = 0 V = M/ x

N

VM

x V+V

N+N

M+Mw/length

This term is too small and can be ignored

Maximum moment

From above, Slope of the shear force diagram = UDL; ie, if UDL = 0,

the shear force is a horizontal line (zero slope) Slope of the moment diagram is the shear When V = 0, the moment is a maximum (M/ x = 0) Since V = M/ x

M = That is, bending moment M at x2 is the area between x1 and x2 under the shear force diagram

2x1x Vdx

V

dxx1 x2

Example

Determine the magnitude and location of the maximum bending of the beam

Ans. Max. moment = 140.625 kNm (where?) Max. shear force = 75 kN

w = 20 kN/m

4 m

A B

30 kN

4m

Example

Determine the magnitude and location of the maximum bending of the beam

w1 = 20 kN/m

6 m

A B

w2 = 10 kN/m

4m

Example

Bent beam: Plot the bending moment and shear force diagrams

10 kN

6 kN

20 kN

2m4m4m

5m

BC

D

E