CITY AND GUILDS 9210 UNIT 135 MECHANICS OF SOLIDS Level 6 … · 2015. 6. 15. · Hardy Cross - the very same man who evolved the theory for solving pipe networks. The work can be

(c) D.J.Dunn www.freestudy.co.uk

1

CITY AND GUILDS 9210

UNIT 135

MECHANICS OF SOLIDS Level 6

TUTORIAL 5A - MOMENT DISTRIBUTION METHOD

Outcome 1 Calculate stresses, strain and deflections in a range of components under various load

conditions

The learner can:

1. Use Mohr’s Circle to determine:

a. stresses on inclined planes

b. combined bending torsion and axial loading.

2. Use and position on components strain gauge

rosettes.

3. Use calculations and or graphic means to determine:

a. shear force and bending moments in laterally

loaded beams

b. bending stress and shear stress distribution in

beams

c. deflection of beams

d. solution of statically indeterminate beams

e. centre of shear in beams.

4. Extend shear force, bending moment, bending

stress, shear stress and deflection analysis to:

a. beams of asymmetric cross section

b. composite beams

c. beams of elastic-perfectly plastic’ material.

5. Determine shear stress and twist of:

a. circular solid sections

b. thin walled cylinders

c. simple open sections.

6. Apply Euler critical loads to determine buckling for

a combination of:

a. free conditions

b. pinned conditions

c. built in end conditions.

7. Determine limiting stress condition.

8. Use analytical methods to determine stresses and

displacements in rings, cylinders and discs under

axi-symmetric loading:

a. internal/external pressure

b. shrink fits

c. rotation.

9. Apply Lame equations to problem

solving.

10. Employ Finite Element Analysis:

a. discretisation

b. types of elements

c. relationship between

i nodal forces

ii nodal displacements

iii stiffness matrix.

11. Represent examples of linear elements

using springs.

12. Obtain stiffness matrix using:

a. one-dimensional quadratic elements

b. displacement functions

c. shape functions

d. principle of virtual work.

13. Determine stresses from primary

unknown nodal displacements.

14. Explain the underlying assumptions and

approximate nature of the results of

Finite Element Method.

15. Analyse engineering materials behaviour

when loadings and service conditions:

a. involve

b. fatigue

c. yield criteria

d. fracture mechanics

e. creep

f. viscoelasticity.

16. Assess and select materials for

applications:

a. plastics

b. composites

c. ceramics

d. modern materials.

This is a stand alone tutorial on an alternative method for finding the bending moment in a beam.


2

MOMENT DISTRIBUTION METHOD

This is a ‘stand-alone’ tutorial for students studying structures. It is about a method of finding the bending

moment in beams that cannot easily be solved by other methods. The theory is due to the work of Professor

Hardy Cross - the very same man who evolved the theory for solving pipe networks. The work can be used

to solve the bending moment in frames but this is not covered here.

At the heart of this method is a way to

determine how a moment M applied at a

rigid joint (such as that shown) is

distributed to the members. When the

moment is applied the joint will rotate a

tiny angle θ and this is the same for all

the members at the joint and leads to the

solution.

The fraction of M that is distributed to any given member is KDM where KD is called the distribution factor.

The fraction depends on the rotational stiffness K of each member and this depends on:-

1. The length, material and cross section (E, I and L)

2. The way the other end of the member is fixed.

A fraction of M is also carried over to the opposite end of the member and this depends on the way the end

is supported. The moment carried over to the other end is KC KD M where KC is the carry over factor.

Usually we use Mx = KC My where My is the moment at the joint end and Mx the moment at the other end.

ROTATIONAL STIFFNESS K

Click here for the full derivation ROTATIONAL STIFFNESS

Relative Rotational Stiffness K for each member is the ratio θ

MK where M is the moment in the member

at the joint.

When the other end of the member is rigidly fixedL

4EIK

θ

M . (e.g. at end B of AB).

When the other end is simply supportedL

3EIK

θ

M . (e.g. at end B of CB).

At a free end (e.g. C on the diagram) the stiffness is zero as there is no moment M = 0.

At a rigid end (wall) (e.g. A on the diagram) the stiffness is infinity since θ = 0.


3

MOMENT DISTRIBUTION FACTOR

Click here for the derivation DISTRIBUTION FACTOR

The moment M applied at the joint is distributed to the members in a proportion that depends on the stiffness

of the member as seen from the joint. The moment distributed to each member at joint B is MBA, MBC and

MBD.

At the joint the sum of all the moments is zero so it follows MBA + MBC + MBD = M

The proportion of M distributed to a member is called the DISTRIBUTION FACTOR which we will

designate KD.

MBA = (KD) BAM MBC = (KD) BCM MBD = (KD) BDM

At a free end KD = 1 and at a rigidly fixed end KD = 0

CARRY OVER FACTOR

When a portion of M is distributed to a member, a moment is produced at the opposite end of the member.

The proportion of M at the opposite end is called the CARRY OVER FACTOR and we will define it as KC.

Moment carried over = KC x M e.g. at A the moment is MA = (KC)A x MB

The derivation shows that when the other end is rigidly fixed KC = ½ of the moment at the joint end. When

the other end is simply supported KC = 0 (no moment possible at a simple support or pin joint).


4

APPLICATION TO BEAMS

The moment distribution method may be used to solve difficult problems that cannot be solved by other

means because there are too many unknowns (Indeterminate beam or structure).

In this tutorial we will only study beams with joints such as the

one shown. The beam is rigidly held at the wall C and

simply supported at A and B. There will be three vertical

reactions due to the point load F and the uniform load w. These

are too many unknowns to solve by normal means.

We start be treating the beam as two parts AB and BC. It might be that the lengths AB and BC have a

different flexural stiffness EI and this can be accommodated in this method.

Next we clamp the end of each section rigidly so there is no deflection or rotation. This enables us to tackle

each section separately to begin.

When a joint is fixed there must be a fixing moment at each end MA, MB and MC.

It can be shown that for a single point load F 2

2

AL

aFbM and

2

2

BL

bFaM

And for a single uniformly distributed load w 12

wLM

2

A and 12

wLM

2

B

In this analysis, clockwise is positive and anticlockwise is negative which is consistent with the other

tutorials in the series where x is measured from the left and an upwards force causes sagging which is

always positive.

For any other combination of loading, the principle of superposition may be used. This principle allows us to

calculate the moments for each load separately and then add them together for the combined affect. Click

here for the derivation FIXING MOMENT

Since the joints are not in reality rigidly fixed, we now consider what happens if we release the fixing

clamps and restore the beam to its correct condition. The beam would take up its natural position. In this

example, the moment at A must be zero because it is a free end so a moment equal and opposite of MA must

be added and distributed at A. There will be a carry over to end B and we should take this into consideration

before balancing the joint B. Joint C is rigid so we do not need to add a balancing moment and there is

nothing to distribute. We calculate the imbalance at B and add an equal and opposite moment to balance it

and then distribute it. This would equalise the moment at B but a fraction of the moment is carried over to

the ends and this upsets the balance again so we have to redistribute the moment again and again until the

moment at a joint is balanced. A worked example will explain it better.


5

WORKED EXAMPLE No. 1

Find the bending moment at A, B and C for the beam shown. The EI values are the same for both

sections.

First treat the beam as two separate sections with rigid ends.

CALCULATE FIXING MOMENTS and STIFFNESS FACTORS.

MEMBER AB 3

2

23

2

2

A 10 x 37.52

1x 1 x 10 x 150

L

aFbM

3

2

23

2

2

B 10 x 37.52

1x 1 x 10 x 150

L

bFaM

Note that MB and MB are equal and opposite only because F is in the middle.

1.5EI2

3EI

L

3EIKBA

MEMBER CB 3232

B 10 x 11.2512

x315x10

12

wLM

32

C 10 x 11.2512

wLM EI333.1

3

4EI

L

4EIKBC

CALCULATE THE DISTRIBUTION FACTORS

At end B of BA KD = 0.5291.333EI1.5EI

EI5.1

K

KBA

4

At end B of BC KD = 0.47061.333EIEI5.1

1.333EI

K

KBC

Note 1 – 0.5294 is another way to get it.

Calculate the CARRY OVER FACTOR

When a balancing moment is applied at a point, ½ of the moment is carried over to the opposite end

unless the opposite end is free in which case it is zero. A rigid end does not carry over any moment to

the other end. If we released the joints that were fixed we have to add moments in the opposite direction

in order to balance them.

At the free end there should be no moment so we must add a moment of -37.5 kNm. At end C there is

no moment to be added and no carry over to B. We start by balancing point A.


6

Joint A Joint B Joint C

KD 0/1 0.529 0.471 0/1

Fixing Moment (X) 37.5 -37.5 11.25 -11.25

Balance A&C (Y) -37.5 -11.25 x 0

Carry over to B -37.5 x ½ -18.75 0 0

Totals (X+Y) 0 -56.25 11.25 -11.25

Imbalance at B -45

Balance B +45

Distribute using KD 23.805 21.195

Carry Over 0 23.805 x 0 21.195 x ½ 10.5975 Total 0 -32.445 32.445 -0.6525

At B we have achieved equal and opposite moments. The moment at A is zero, the moment at B is

32.445 kNm and at the wall (C) 0.653 kNm.

This problem is in fact solvable with Macaulay’s method and this gives similar answers. We did not

need extra correction cycles in this problem.


7


Find the moments at A, B and C for the beam below. The section AB has an EI value twice that of

section BC.

CALCULATE FIXING MOMENTS and STIFFNESS FACTORS.

MEMBER AB 3

2

23

2

2

A 10 x 88.8883

1x 2 x 10 x 200

L

aFbM

3

2

23

2

2

B 10 x 444.443

2x 1 x 10 x 200

L

bFaM

2EI3

6EI

L

3(2EI)KBA

MEMBER CB 3232

B 10 x 53.33312

4x 40x10

12

wLM

32

C 10 x 333.3512

wLM EI

4

4EI

L

4EIKBC

CALCULATE THE DISTRIBUTION FACTORS

At end B of BA KD = 0.6671EI2EI

2EI

K

KBA

At end B of BC KD = 0.3331EI2EI

1EI

K

KBC

Joint A Joint B Joint C

KD 0.667 0.333

Fixing Moment (X) 88.889 -44.444 53.333 -53.333

Balance A (Y) -88.889

Carry over A to B -88.889 x ½ -44.444

Totalise (X+Y) 0 -88.888 53.333 -53.333

Inbalance at B -35.555

Balance B +35.555

Distribute using KD 23.716 11.84

Carry Over 0 23.716 x 0 11.84 x ½ 5.92 Total 0 -65.173 65.173 -47.413

At B we have achieved equal and opposite moments. The moment at A is zero, the moment at B is

65.173 kNm and at the wall (C) -47.413 kNm. This problem is in fact solvable with Macaulay’s method

and this gives similar answers.


8

SELF ASSESSMENT EXERCISE No. 1

1. Solve the moments at A B and C for the beam shown. Section AB has an EI value three times that of

BC.

Answers 0, 34.4 kNm and 42.8 kNm

2. Repeat problem 1 if section BC has an EI value three times that of AB.

(Answers 0, 29.4 kNm and 45.3 kNm)

3. Solve the bending moment at the 3 supports for the beam shown.

Answers 1.81 kNm, 28.61 kNm and 0

Now we need to try harder problems with more than one joint. This requires more correction cycles.


9


Calculate the bending moment in the beam at A, B, C and D. The point loads at mid span. The middle

span (BC) has an EI value twice that of the other two spans.

CALCULATE FIXING MOMENTS and STIFFNESS FACTORS FOR B.

MEMBER AB 3

2

23

2

2

A 10 x 12.55

2.5x 2.5 x 10 x 20

L

aFbM

3

2

23

2

2

10 x 12.55

2.5x 2.5 x 10 x 20

L

bFaM


0.6EI5

3EI

L

3EIKBA

MEMBER CB 3232

B 10 x 10.41712

5x x105

12

wLM

32

C 10 x 10.41712

wLM

EI6.15

8EI

L

4(2EI)KBC

Note it is still rigidly fixed at C at this stage.

CALCULATE THE DISTRIBUTION FACTORS for B

At end B of BA KD = 0.2727EI6.1EI6.0

EI6.0

K

KBA

At end B of BC KD = 7273.0EI6.0EI6.1

1.6EI

K

KBC

CALCULATE FIXING MOMENTS and STIFFNESS FACTORS FOR C.

MEMBER BC

3232

B 10 x 10.41712

5x x105

12

wLM

32

C 10 x 10.41712

wLM

EI6.15

8EI

L

4(2EI)KCB

EI8.05

4EI

L

4EIKCD

Note both ends are rigid when C is released.


10

CALCULATE THE DISTRIBUTION FACTORS for C

At end C of BC KD = 0.6667EI8.0EI6.1

EI6.1

K

KCB

At end C of CD KD = 0.3333EI8.0EI6.1

EI8.0

K

KCB

MEMBER CD

3

2

23

2

2

C 10 x 12.55

2.5x 2.5 x 10 x 20

L

aFbM

3

2

23

2

2

D 10 x 12.55

2.5x 2.5 x 10 x 20

L

bFaM


The following table shows the process of balancing and correcting until we get very close to zero

difference at B and C. This shows that the moments at A, BC and D are 0, 11.57 kNm, 10.19 kNm and

13.66 kNm. The table was produced using a spread sheet.

A B C D KD 0.272727 0.727273 0.666667 0.333333

Fix Mom 14.7 -6.3 8.333333 -8.33333 12.5 -12.5

Balance A -14.7

Carry over

B ½ x 14.7 -7.35

Total B -13.65 8.333333

Difference -5.31667

Balance B 5.316667

Distribute B 1.45 3.866667

Carry Over 0 1.933333

Totals 0 -12.2 12.2 -6.4 12.5 12.5

Difference 0 6.1

Balance 0 -6.1

Distribute 0 0 -4.06667 -2.03333

Carry Over 0 -2.03333 ½ x -4.06667 ½ x-2.03333 -1.01667

Total 0 -12.2 10.16667 -10.4667 10.46667 -13.5167

Difference -2.03333 0

Balance 0 2.033333 0

Distribute 0.554545 1.478788 0 0

Carry Over ½ x 1.478 0.739394

Total 0 -11.6455 11.64545 -9.72727 10.46667 -13.5167

Difference 0 0.739394

Balance 0 0 -0.73939

Distribute 0 0 -0.49293 -0.24646

Carry Over 0 -0.24646 ½ x -0.49293 ½ x -0.24646 -0.12323

Total 0 -11.6455 11.39899 -10.2202 10.2202 -13.6399


Balance 0 0.246465 0

Distribute 0.067218 0.179247

Carry Over 0 ½ x 0.179247 0.089624

Total 0 -11.5782 11.57824 -10.1306 10.2202 -13.6399


Balance 0 0 -0.08962

Distribute 0 0 -0.05975 -0.02987

Carry Over 0 -0.02987 -0.01494

Total 0 -11.5782 11.54836 -10.1903 10.19033 -13.6548


Balance 0 0.029875 0

Distribute 0.008148 0.021727 0 0

Carry Over 0 0.010863 0

Total 0 -11.5701 11.57009 -10.1795 10.19033 -13.6548


Balance 0 0 -0.01086

Distribute 0 0 -0.00724 -0.00362

Carry Over 0 0 -0.00362 -0.00181

Total 0 -11.5701 11.56647 -10.1867 10.18671 -13.6566


11

SELF ASSESSMENT EXERCISE No. 2

Find the bending moments in the beams shown at A, B, C and D.

1.

Answers 0, 45.8 kN, 59 kN and 83 kN

Answers 0, 19.1 kN, 18 kN and 23.8 kN


12

FIXING MOMENTS

Fixing moments have been covered in the tutorial on deflection of beams under the heading ENCASTRÉ

BEAMS.

A fixing moment is the moment at the end of a beam

where it is rigidly fixed to prevent any deflection or

rotation. Finding them can be quite complicated

depending on the loading of the beam. For those

wishing to get to grips with the derivation the

following is given. Let the beam span be designated A

to B and the moments at the ends MA and MB. There

will also be reactions RA and RB.

We shall use Macaulay’s method to solve the slope and deflection and start with the derivation for a single

point load.

POINT LOAD

The bending moment at distance x from the left end is

AA2

2

MaxFxRdx

ydEIM

AxM

2

axF

2

xR

dx

dyEI Integrate A

22

A

The slope at both ends is zero so 0dx

dy

Put x = 0 and A is zero.

....(2)..........

2

xM

6

axF

6

xREIy

0 B that yields 0 x and 0 y putting then endsboth at zero is deflection theSince

B2

xM

6

axF

6

xREIy again Integrate .....(1)x.........M

2

axF

2

xR

dx

dyEI

2

A

33

A

2

A

33

AA

22

A

(The

constants of integration A and B are always zero for an encastré beam). Equations 1 and 2 give the slope and

deflection. Before they can be solved, the fixing moment must be found by using another boundary

condition. Remember the slope and deflection is zero at both ends of the beam so we have two more

boundary conditions to use.

A suitable condition is that y = 0 at x = L. From equation 2 this yields:-

2

LM

6

aLF

6

LREI(0)

2A

33

A

and note that L – a = b for simplicity

2

3A

A

2A

33

A3L

Fb

3

LRM

2

LM

6

Fb

6

LR0

We still have an unknown RA so using the other boundary condition that dy/dx = 0 when x = L then from

equation (2)

LM

2

bF

2

LR0

dx

dyEI A

22

A and substitute 2

3A

A3L

Fb

3

LRM

L

3L

Fb

3

LR

2

bF

2

LR0

2

3A

22

A

3L

Fb

3

LR

2

Fb

2

LR 32A

22A

3L

Fb

2

Fb

6

LR 322A

3

3

2

2

AL

2Fb

L

3FbR


13

Substitute into 2

3A

A3L

Fb

3

LRM

2

3

2

32

2

33

3

2

2

A3L

Fb

3L

2Fb

L

Fb

3L

Fb

3

LL

2Fb

L

3Fb

M

2

32

AL

Fb

L

FbM

2

2

2

22

AL

aFbbL

L

Fb

L

b1

L

FbM

and it can be shown that

2

2

BL

bFaM

UNIFORMLY DISTRIBUTED LOADS

A uniform load across the entire span will produce equal and opposite moments at either end of the beam.

In this case the reactions are RA = RB = wL/2

The bending moment at distance x from the left end is :

A

2

2

2

A

2

A2

2

M2

wx

2

wLx

dx

ydEI M

2

wxxR

dx

ydEIM

AxM6

wx

4

wLx

dx

dyEI Integrate A

32

....(2)..........2

xM

24

wx

12

wLxEIy

0 B that yields 0 x and 0 y putting then endsboth at zero is deflection theSince

B2

xM

24

wx

12

wLxEIy again Integrate .....(1)x.........M

6

wx

4

wLx

dx

dyEI

0 A that yields 0 x and 0dx

dy puttingit then endsboth at zero is slope theSince

2A

43

2A

43

A

32

As in the other case, A and B are zero but we must find the fixing moment by using the other boundary

condition of y = 0 when x = L

12

wLM

2

LM

24

wL0

2

LM

24

wL

12

wLEI(0)

2

A

2A

4

2A

44

If we worked out MB we would get 12

wLM

2

B


14

ROTATIONAL STIFFNESS and CARRY OVER

When a moment is applied to the end of a beam without any deflection, the end will rotate a small angle θ.

The ratio M/θ is the stiffness and this depends on the values of E, I and L. It also depends how the other end

of the beam is fixed.

FIXED END

Consider a section of beam with a moment M at the left end and fixed rigidly at the right end. The beam will

deflect as shown.

The left end must not deflect so an upwards force R must be applied also. The bending moment any distance

x from the end is:-

RxMdx

ydEIM

2

2

x

Integrate

A2

RxMx

dx

dyEI

2

but at x = L the slope is zero.

A2

RLML0

2

2

RLMLA

2

2

RLML

2

RxMx

dx

dyEI

22

Integrate again.

B2

xRLMLx

6

Rx

2

MxEIy

232

The deflection at x = 0 must also be zero hence B = 0

The deflection at x = L is zero hence:-

2

RLML

6

RL

2

ML0

32

32

3

RL

2

ML0

32

hence 2L

3MR

The gradient at the end is the rotation θ and at x = 0 we have:-

2

RLMLθ EI

2

substitute 2L

3MR

4

ML

4

3MLMLθ EI

The rotational stiffness is hence L

4EI

θ

M

CARRY OVER FACTOR

At end B the moment is MB = - MA + R L

2

MM

2

3MMM

AB

AB

The fraction ½ is called the CARRY OVER FACTOR


15

SIMPLY SUPPORTED END

Consider a section of beam with a moment M at the left end and fixed rigidly at the right end. The beam will

deflect as shown.

The reaction at the support must be equal and opposite to R. It follows that M = RL

xL

M-MRxM

dx

ydEIM

2

2

x

Integrate

A2L

Mx-Mx

dx

dyEI

2

Integrate again.

BAx6L

Mx-

2

MxEIy

32

The deflection at x = 0 must be zero hence B = 0

Ax6L

Mx-

2

MxEIy

32

The deflection at x = L is zero hence:-

AL6L

ML-

2

ML0

32

AL6

ML-

2

ML0

22

A6

ML-

2

ML0

3

ML-A

3

MLx

6L

Mx-

2

Mx

dx

dyEI

32

The gradient at the end is the rotation θ and at x = 0 we have:-

3

ML-EIθ

The rotational stiffness is hence L

3EI

θ

M

CARRY OVER FACTOR

The moment at the simply supported end is zero so the carry over factor is zero.


16

DISTRIBUTION FACTOR

The diagram shows three members AB, DB

and CB rigidly joined at B.

When a moment M is applied at B the joint

will twist an angler θ. The stiffness of each

member as seen from B is Kθ

M

The moment distributed to each member is

MBA, MBC and MBD. At the joint the sum of all the moments is zero so it follows MBA + MBC + MBD = M

BABA Kθ

M BC

BC Kθ

M BD

BD Kθ

M

θKθKθKM BDBCBA BDBCBA KKKθ

M Hence

K

Mθ

θ KM BABA and if we substitute

K

KM M BA

BA

Similarly

K

KM M BC

BC and

K

KM M BD

BD

The moment is distributed to each member by the distribution factor

K

KKD

MBA = (KD) BA M MBC = (KD) BC M MBD = (KD) BD M

At a free end, (e.g. C)

10K

K

K

K)K(

CB

CBCBD

At a rigidly fixed end (e.g. A)

0K

K

K

K)K(

AB

ABABD

Documents

CITY AND GUILDS 9210 UNIT 135 MECHANICS OF SOLIDS Level 6 … · 2015. 6. 15. · Hardy Cross - the very same man who evolved the theory for solving pipe networks. The work can be