Beams, Shear Force & Bending Beams, Shear Force & Bending Moment DiagramsMoment Diagrams

A A BEAMBEAM is a long, slender, structural is a long, slender, structural member designed to support member designed to support transverse loadings.transverse loadings.

A transverse loading is applied A transverse loading is applied perpendicular to the axis of the perpendicular to the axis of the beam.beam.

Beams classified by their supports.Beams classified by their supports.

Beams, Shear Force & Bending Beams, Shear Force & Bending Moment DiagramsMoment Diagrams

• Shall approach this through examining Shall approach this through examining BEAMs.BEAMs.

• There are a number of steps that you There are a number of steps that you MUST go through to get the solution.MUST go through to get the solution.

• Make sure you follow this every time – Make sure you follow this every time – makes solution a lot easier!makes solution a lot easier!

Bending of BeamsBending of BeamsMany objects in everyday life can be analysed as beams.

If we ignore mass of the beam, the forces on the beam are as shown:

Support reaction forces.

Load Force

We shall normally ignore the beam mass in these lectures.

The design and analysis of any structural member requires knowledge of the internal loadings acting within it, not only when it is in place and subjected to service loads, but also when it is being hoisted.

In this lecture we will discuss how engineers determine these loadings.

Bending of BeamsBending of Beams

Support reaction forces.

Load Force

In the figure, the force shown downwards is acting on the beam.

It is a point load, acting at a single point on the beam.

Beam loading:Beam loading:Point loadingPoint loading

Uniformly Distributed Loading Uniformly Distributed Loading (UDL)(UDL) UDL e.g. UDL – 20N/m,

3kN/m

Combined Combined loadingloading

Beams can have a range of different forms of section.

Example Example Determine the reactions of a beam of length 4.5 m which is supported at its ends and subject to a point load of 9 kN a distance of 1.5 m from the left-hand end. Neglect the weight of the beam.

The reactions at the supports can be found by taking moments about LHS:

MA = 0 = RB 4.5 – 9 1.5

RB = (9 1.5)/4.5 = 3kN

Fy = 0 = RB + RA – 9

RA = 6kN

9

RBRA

SHEAR FORCESHEAR FORCE

• The algebric sum of the vertical The algebric sum of the vertical forces on either side of the section of forces on either side of the section of a loaded beam is called Shearing a loaded beam is called Shearing ForceForce

Bending MomentBending Moment

• The algebric sum of the moments of The algebric sum of the moments of the forces on either side of the the forces on either side of the section of a loaded beam is called section of a loaded beam is called Bending Moment.Bending Moment.

Fy = 0 = 360 – 120 – V1

V1 = 240 kN

Mat 1m = 0 = -540 + (360 1) - (120 0.5) – M1

M1 = - 540 + 360 - 60 = - 240 kNm

To determine the shear force at 1m from the LHS:

To determine the bending moment at 1m from the LHS:

• If we wanted to determine the SF and BM at If we wanted to determine the SF and BM at any point along the beam, would need to go any point along the beam, would need to go through this process each time.through this process each time.

• You will agree this tedious. You will agree this tedious. • We need an alternative method to enable us We need an alternative method to enable us

to determine SF and BM at any point along to determine SF and BM at any point along the beam.the beam.

Shear force and bending moment Shear force and bending moment diagrams.diagrams.

• Shear force diagrams and bending moment Shear force diagrams and bending moment diagrams are graphs used to show the diagrams are graphs used to show the variations of the shear forces and bending variations of the shear forces and bending moments along the length of a beam. moments along the length of a beam.

A steel beam 8m long is pin-jointed at the left-hand end and A steel beam 8m long is pin-jointed at the left-hand end and simply supported 4m from the right-hand end. The beam is simply supported 4m from the right-hand end. The beam is loaded as shown. For the beam:loaded as shown. For the beam:

(a)(a) Determine the reactions at A and C.Determine the reactions at A and C.

(b)(b) Derive equations for the shear force and bending Derive equations for the shear force and bending moment as a function of distance ‘x’ (horizontal moment as a function of distance ‘x’ (horizontal displacement from the left-hand end).displacement from the left-hand end).

(c)(c) Draw the shear force and bending moment diagrams Draw the shear force and bending moment diagrams for the beam, and determine the position of for the beam, and determine the position of counteraflexure (should one exist).counteraflexure (should one exist).

2 2 1 3

15kN 2kN/m

EDCBA

(a) Draw the FBD.

15kN

EDCBA

2 3 = 6kN

1

Rax

Ray Rc

By inspection Rax = 0

Have TWO unknowns: Ray and Rc. Therefore, require TWO equations:

Fy = 0 & M = 0

Fy = 0 = Ray – 15 + Rc – 6

Ray = 21 - Rc (1)

MA = 0 = (15 2) – (Rc 4) + (6 6.5)

30 - 4Rc + 39 = 0

Rc = 69/4 = 17.25kN

2

Ray = 21 - 17.25 = 3.75 kN

Sub for Rc into equation (1)

(b) To derive the equations, split the beam into a number of spans.

Start from the LHS and whenever you come across a force OR bending moment, you get a span for which an equation has to be derived.

15kN

EDCBA

17.253.75

Span AB: 0 x 2

Span BC: 2 x 4

Span CD: 4 x 5

Span DE: 5 x 8

X=0

X=2

X=4

X=5 X=8

M = -x2 + 16x -64

M/x = -2x + 16 = Vde

CHECK:

(c) To plot the shear force and bending moment diagrams is now a straight forward process of putting in the limits positions of ‘x’ in each of the equations and plotting the values on a graph.

Point of counterflexure is a point where the bending moment graph crosses the axis.

There are situations where there is NO point of counterflexure, i.e. the bending moment graph DOES NOT cross the axis.

In this question we do have point of counterflexure.

This occurs over span BC, .i.e. when 2 x 4

At the point of counterflexure, the bending moment is EQUAL TO ZERO.

Mbc = -11.25x + 30 = 0

x = 30/11.25 = 2.67m

The point of counterflexure is when x = 2.67m