Bending

BADI 1

John Errington MSc

Loaded Beams

• When a beam is supported at both ends and loaded it will bend.

• The amount by which it bends depends on – the load, and the way it is distributed– the distance between supports– the elastic modulus for the material, and – the distribution of material in the beam

Loaded beam

Here we have a solid square section beam, centrally loaded with a force F and unsupported distance 2x

The beam bends into an arc, so the top is shorter than the bottom. This means the top is under compression, and the bottom under tension.

The material in the middle (the neutral axis) contributes less to the strength of the beam than the material in the top and bottom quarters of the beam.

F

2x

For our design we need to know how much the beam will bend, and the maximum stress in the beam to determine whether it will break.

Figure 1. A side view of a simply supported beam (top) bending under a distributed lateral load (bottom).

Figure 2. The internal forces and the cross-sectional stress distribution of a beam in bending.

Note the position of the neutral axis

Joists and floorboards

Floor loading• The timber floor of a property obviously weighs something. This weight consists

of the timber joists, the plasterboard ceiling underneath it, and the floorboards. All of these are known as "the dead load" and the joists themselves must be able to support this dead load without sagging. For a normal property, with chipboard or timber floors and plasterboard ceilings, this dead load is generally taken to be no more than 0.50 kiloNewtons per square metre. (kN/sq.m)

• The weight we place upon a floor by way of bathroom suites, beds, wardrobes etc, is known as the "imposed load". It is again accepted that, for normal household requirements, the imposed load will not exceed 1.5kN/sq.m.

• The building regulations tables A1 and A2 list the size of joist necessary to support this weight, over a maximum span. Table A2 uses timbers known as "SC4" which are high strength timbers containing very few, if any, knots. They are not common in modern day construction unless specified and we will deal with the more generally used timbers, dealt with in table A1. These timbers are known as SC3 and will have C16 stamped on them.

This table is for a dead load of more than 0.25 but not more than 0.50 and allows for an imposed loading of no more than 1.5 kN/sq.m.

This is an abridged representation of  Building regulations Table A1

Spacing (distance apart) of joists. Spacing (distance apart) of joists.

400mm

450mm

600mm

400mm

450mm

600mm

Size of joist Size of joist

38 x 97 mm 1.72 m 1.56 m 1.21 m 50 x 170 mm 3.61 m 3.47 m 3.08 m

38 x 122 mm 2.37 m 2.22 m 1.76 m 50 x 195 mm 4.13 m 3.97 m 3.5 m

38 x 140 mm 2.72 m 2.59 m 2.17 m 50 x 220 mm 4.64 m 4.47 m 3.91 m

38 x 147 mm 2.85 m 2.71 m 2.33 m

38 x 170 mm 3.28 m 3.1 m 2.69 m 63 x 97 mm 2.19 m 2.08 m 1.82 m

38 x 184 mm 3.53 m 3.33 m 2.9 m 63 x 122 mm 2.81 m 2.7 m 2.45 m

38 x 195 mm 3.72 m 3.52 m 3.06 m 63 x 147 mm 3.37 m 3.24 m 2.95 m

38 x 220 mm 4.16 m 3.93 m 3.42 m 63 x 170 mm 3.89 m 3.74 m 3.4 m

38 x 235 mm 4.43 m 4.18 m 3.64 m 63 x 195 mm 4.44 m 4.28 m 3.9 m

63 x 220 mm 4.91 m 4.77 m 4.37 m

47 x 97 mm 1.92 m 1.82 m 1.46 m

47 x 122 mm 2.55 m 2.45 m 2.09 m 75 x 122 mm 2.97 m 2.86 m 2.6 m

47 x 147 mm 3.06 m 2.95 m 2.61 m 75 x 147 mm 3.56 m 3.43 m 3.13 m

47 x 170 mm 3.53 m 3.4 m 2.99 m 75 x 170 mm 4.11 m 3.96 m 3.61 m

47 x 195 mm 4.04 m 3.89 m 3.39 m 75 x 195 mm 4.68 m 4.52 m 4.13 m

47 x 220 mm 4.55 m 4.35 m 3.79 m 75 x 220 mm 5.11 m 4.97 m 4.64 m

Steel and concrete

• Steel or concrete are often used to support loads, in large buildings and for bridges.

• Steel is good under tension and compression

• Concrete is excellent under compression, and is lighter than steel, but has low tensile strength

Steel sections

Rounds Hexagons Squares Flats Equal Leg Angles Unequal Leg Angles

Channels "H" Beams "I" Beams Tee Bars Diamond Floor Plate Rectangular Tubing

Seamless and Welded Pipe Seamless and Welded Tubing Half Rounds

Bending moment for a point load

A uniform beam 5m long is supported at its ends and has a load of 6kN at 3m from one end.

1. What is the reaction in the supports?

2. What is the bending moment?

6kN

5m

3m

Calculate reaction in supports

Taking moments about A

we find a CW moment of 3m * 6kN

and a CCW moment of -5m * Rb

As the system is in equilibrium

CW = CCW so

3m * 6kN = -5m * Rb

Rb = - 18/5 kN = -3.6kN

Also Ra + Rb = - 6kN

Ra = - 6kN – (-3.6kN) = -2.4kN

Ra = - 2.4kN Rb = - 3.6kN

6kN

5m

3m

A B

Ra Rb

CW = clockwise

CCW = counter clockwise

Bending moment for beam with point loadTo find the bending moment at a point p, x metres from a we take moments about P.

To the left of P:

Reaction Ra causes a CW moment of 2.4x

To the right of P:

The load causes a CW moment of 6 * (3-x)

Reaction Rb causes a CCW moment of 3.6*(5-x)

Total moment = (18 -6x) – (18 -3.6x) = -2.4x

i.e. 2.4x CCW

So the bending moment for this loaded beam at a distance x from A is Mx = 2.4x.

6kN

5m

3m

AB

Ra Rb

3.6kN2.4kN

p

X m

Bending moment and shear force for point load

Plotting BM vs x we get this graph, showing BM is a maximum of 7.2 at x =3.

We can also plot the shear force acting on the beam. Imagine the beam cut at point p and work out the force acting on

the left side of the cut.

6kN

5m

3m

AB

Ra

3.6kN2.4kN

p

X m

Bending moment and shear force for point load

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

distance from A

Bending moment

Shear force

Bending moment for a distributed load

A beam 3m long is supported at its ends and has a distributed load of 2 tonnes/m.

1. What is the reaction in the supports?

2. What is the bending moment?

3m

2 tonnes/mA B

Calculate reaction in supports

Distributed load of 2 tonnes / m = 2000kg * 9.81 = 19620N / m

The total weight of the load is

19620N * 3m = 58860N

By symmetry this is evenly split between the ends, so

Ra = Rb = 588600/2 = 29430N

Ra Rb3m

2 tonnes/mA B

Bending moment for beam with distributed load

To find the bending moment at a point p, x metres from a we take moments about P.To the left of P:

Reaction Ra causes a CW moment of 29430x

The load causes a CCW moment of 19620 * x *x/2

= 9810 x2

Bending moment Mx= 29430x - 9810 x2

We can show (by plotting, or by differentiating this expression and looking for stationary turning points, then differentiating again) that the maximum bending moment M! for this loaded beam occurs at the mid point, where

M! = 29430* 1.5 - 9810 *1.5*1.5

M! = 44145 – 22072 = 22073 Nm

Ra Rb3m

19620N/mA B

29430N

p

x

CW = clockwise

CCW = counter clockwise

29430N

Bending moment and shear force for distributed load

Here you can see that the bending moment is a maximum at the centre, while the shear force is a maximum at the ends.

Bending moment and shear force for distributed load

-30000

-20000

-10000

0

10000

20000

30000

0 0.5 1 1.5 2 2.5 3

Distance from A

BM

SF

TO BE CONTINUED

• Enough for one week.

• Next week we will use this information to evaluate how much stress we will get in a loaded beam, and how much it will sag.

Calculate maximum stress in beam and radius of curvature of neutral axis

Suppose our beam from the previous example was a steel box girder with dimensions as shown here. • How much would it bend? • Would it break?

To answer these questions we need to calculate the radius of curvature , and the maximum stress in the beam.

HOW?

0.2m

0.1m

Thickness t = 0.02m

/ y = M / I = E / RWhere: is the strain at y metres from the neutral axis

M is the bending moment

I is the Moment of Inertia for the beam

E is Young’s Modulus of elasticity for the material of the beam, and

R is the radius of curvature at the neutral axis

Maximum stressThe stress is a maximum at the outer faces (top and bottom) of the beam, at a distance D/2 from the neutral axis, where D is the depth of the beam.

Using / y = M / I = E / R ! = M! D / 2 I

Remember we are using ! to signify a maximum value

For our beam M! is 22073 Nm and

I = 0.1 * 0.23 /12 – 0.06 *0.163 / 12

I = 6.667*10-5 -2.048*10-5 I = 4.619*10-5 m4

(Formula for calculation of I shown later)

! = 22073 * 0.2 / 2 * 4.619*10-5

! = 47.8 * 106 Pa (1Pa = 1N/m2)

Steel has a tensile strength of 1.5*109 N/m2 so no problem

Radius of curvatureFor our steel beam M! is 22073 Nm (calculated)

E = 2.1 * 1011 N/m2 (tables)

I = 4.619*10-5 m4 calculated)

Using / y = M / I = E / R R = I E / M

The radius of curvature is given by

R = (4.619*10-5 )* (2.1 * 1011) / 22073

R = 439m

So how much will our beam sag?

How far does the beam sag?The amount by which the beam sags, x is found using Pythagoras theorem.

R2 = a2 + (L/2)2

a2 = R2 - (L/2)2

a = sqrt(R2 - (L/2)2)

x = R - sqrt(R2 - (L/2)2)

Substituting values for R=439m and L=3m

x = 439 - sqrt(4392 - (3/2)2)

x = 439 - sqrt(192721 - 2.25)

x = 439 – 438.9974 = 2.56 *10-3 m

The beam sags by 2.56mm in the centre

x

R

L/2

a

Moment of Inertia Defined

• The moment of inertia measures the resistance to a change in rotation.– Change in rotation from torque– Moment of inertia I = m * r2 for a point mass

• The total moment of inertia I is due to the sum of masses at a distance from the axis of rotation.

N

iiirmI

1

2

Moment of Inertia

More importantly for our purposes the moment of inertia describes the way in which the mass of the body is distributed.

As we have already seen we can use this parameter to evaluate how a regular isotropic beam will respond to a load.

Mass at a Radius

• Extended objects can be treated as a sum of small masses.

• A straight rod (M) is a set of identical masses m.

• The total moment of inertia is

• Each mass element contributes

• The sum becomes an integral

axis

length L

distance r to r+r

2)( rmI

rrLMI

rLMm2)/(

)/(

23

0

2

)3/1()3/)(/(

)/(

MLLLMI

drrLMIL

Rigid Body Rotation

• The moments of inertia for many shapes can found by integration.

– Ring or hollow cylinder: I = MR2

– Solid cylinder: I = (1/2) MR2

– Hollow sphere: I = (2/3) MR2

– Solid sphere: I = (2/5) MR2

Point and Ring

• The point mass, ring and hollow cylinder all have the same moment of inertia.I = MR2

• All the mass is equally far away from the axis.

• The rod and rectangular plate also have the same moment of inertia.I = (1/3) MR2

• The distribution of mass from the axis is the same.

M

R M R

axislength R length R

MM

Moment of Inertia of common shapes

y is the distance of the NA from the top of the shape. ** indicates raised to power e.g. D**3 = D cubed

b

h

b2

b1

h2 h1

Rectangle Box

3

12

1hbI rect

322

311 12

1

12

1hbhbIbox

Moment of Inertia of a Cross Section

A box girder of dimension 0.2m * 0.1m and 0.02m thickness has a csa of 0.012m

This amount of material would make a solid bar of 0.12m * 0.1m

I(bar) = 1.44 *10-5 m4 I(box) = 6.66 *10-5 – 1.36 *10-5 = 5.3 *10-5 m4

Other shapes

• We can calculate the M of I for other shapes just by adding or subtracting, as we did for the box girder. See also next slide.

• We can evaluate the M of I at a different point within the shape, using the parallel axis theorem

• We can evaluate the M of I about an axis in another plane, using the perpendicular axis theorem

Comparison of moments of inertia for different shapesall having the same cross-sectional area

a: Ia =bh**3 / 12 = 1*(1*1*1)/12 = 1/12

Ia = 1/12

b: Ib = I(b outer) - I(a) = (1.414 * 1.414**3) /12 – 1/12

= 4/12 – 1/12 = 3/12

Ib = 3/12

c: Ic = I -I -I +I

Ic = { (2*(2**3)) - (1*(2**3)) - (2*(1**3)) + (1*(1**3)) } / 12

Ic = { (2*(8)) - (1*(8)) - (2*(1)) + (1*(1)) } / 12

Ic = { 16 - 8 - 2 + 1} / 12 = 7 / 12

Ic = 7/12

a

1m

1m

b

1.41m

c

2m

Parallel Axis Theorem

• Some objects don’t rotate about an end, but rather about the axis at the center of mass.

• The moment of inertia depends on the distance between axes.

• The moment of inertia for a rod about its center of mass:

2MhII CM

axis

M

h = R/2

2

22

22

)12/1(

)4/1()3/1(

)2/()3/1(

MRI

MRMRI

RMIMR

CM

CM

CM

Perpendicular Axis Theorem

For flat objects the rotational moment of inertia of the axes in the plane is related to the moment of inertia perpendicular to the plane.

yxz III

M

Ix = (1/12) Mb2

Iy = (1/12) Ma2

a

b

Iz = (1/12) M(a2 + b2)

Top view

Side view

To Increase the Moment of Inertia

• Increase the size:– But as you increase the size, you increase the weight

and cost

• Use a denser material– But it may not have the properties you need and it will

be heavier

• Change the cross-sectional shape:– A hollow cross-section is stronger for the amount of

material used

SummaryWe can now work out the stress in a loaded beam, such as a table top or beam bridge. We can find out whether the structure will be strong enough, and how much it will deform under load.

Question.A bridge is made of reinforced concrete using pre-stressed reinforcing rods. Should the rods be placed at the top, middle or bottom of the section?WHY?