moment distribution - frames

7/28/2019 Moment Distribution - Frames

1/7

School of the Built Environment

Structural Analysis Notes

Moment Distribution Frames with Sway

These notes are designed to complement the lecture material, not replace it. They should

serve as a reminder of what is already in your mind and are not intended as a self-teaching

aid.

Moment distribution can be applied to rigidly-jointed frames as well as continuous beams.

We will look here at single-bay portal frames, though there is no reason why you should not

be able to extend your work to multi-bay frames. Frames with more than one storey are more

complex, and you will need to refer to a textbook if you wish to look at those.

Think back to your work on continuous beams and consider the problem below:

Figure 1: Continuous beam with varying EI value.

Note that the side spans of the beam have a higher value of EI than the central span. This may

seem odd for a continuous beam, but will become clear when we turn it into a frame.

The FEMs and DFs are calculated as usual. Refer back to the notes on Moment Distribution

if you need to.

MBC = -26.7 kNm MCB = 53.3 kNm

DFBA = DFBC =

DFCD = DFCB =

The distribution is tabulated in figure 2:

4m

EI 2EI2EI

DCB

8m8m 12m

A

30kN


2/7

-26.7 53.3

-6.7 -13.3 -40.0 -20.012.5 25.0 8.4 4.2

-0.5 -1.0 -3.2 -1.6

0.2 0.4 0.1

12.7 25.4 -25.4 43.2 -43.2 -21.6

Figure 2:Distribution and final moments at joints.

If you are unsure of the distribution then write it our stage by stage so that you can see where

each value has come from. I started at joint C.

Now look at the portal frame in figure 3. It has dimensions which are, in some senses,

identical to the beam in figure 1, except the two outer spans have been wrapped around to

form legs.

Figure 3:Simple portal frame.

As far as moment distribution is concerned, there is no difference between the frame and the

beam. In both cases, joints B and C are rigid joints between members, the ends of which must

rotate by the same amount when released during distribution. There are, however, important

physical differences because in the frame, sideways movement of these joints has important

effects, so a simple distribution is not sufficient. Look first at the initial simple distribution,

laid out around the frame. This is numerically identical to the beam distribution.

C

2EI8m2EI

3/4 1/40 01/4 3/4

A CB D

30kN

4mEI

D

B

12m

A


3/7

Figure 4: Distribution laid out around the frame.

Before proceeding, satisfy yourself that you understand the way in which the figures are laid

out. If you prefer to tabulate your calculations, then write them out in a table.

The moments at the end of the distribution in figure 4 look as if they are in equilibrium

moments either side of each joint are equal and opposite. However, if we look at the

moments acting on the legs of the frame and calculate the shear forces which must exist, then

we see that the frame as a whole is not in equilibrium:

Leg AB: HAB 8 = 25.4 + 12.7

i.e. HAB = 4.8 kN

Leg CD: HCD 8 = 43.2 + 21.6

i.e. HCD = 8.1 kN

(note: HAB and HCD are in opposite directions)

Figure 5: Shear forces on frame legs, calculated from end moments.

(-)43.2

HCDHAB

3/4

1/4

0 0

1/4

3/4

A

CB

D

-25.4

0.1

0.5

8.4

-6.7

-26.7

25.0

0.4

25.4

12.50.2

12.7

43.2

-1.0

4.2

-13.3

53.3

-40.0

-3.2

-43.2

-20.0-1.6

-21.6

HAB HCD

25.4

12.7 (-)21.6

8m


4/7

The shear force calculations show that the frame is not in horizontal equilibrium, since a force

of 8.1 4.8 = 3.3 kN would have to be applied to the top of the frame to keep it in

equilibrium:

Figure 6:Frame with prop force.

The prop force, which is imaginary, is the force which would be required to prevent the topof the frame from moving sideways, in this case to the left. This movement would actually

take place in a real frame loaded like this, since there are no restraints preventing it, other than

the stiffness of the legs themselves. This sideways movement is known as sway and must be

taken into account when analysing portal frames.

To remove the prop force, we simply have to add an equal and opposite sway force:

Figure 7: Sway force.

We need to calculate the effect of this prop force on the reactions at A and D and on thebending moments in the frame, so that we can add these in to the no sway distribution

which we have already completed. However, it is different in nature to the forces we have

looked at so far, as it is applied to a joint rather than to a member. We currently have no

method to calculate FEMs for this force, so we have to look instead at the sway displacement

which will take place.

3.3 kN

prop force

8.1 kN4.8 kN

3.3 kN

sway force


5/7

Figure 8:Sway displacement of legs.

Refer back to your class notes or to a textbook if you would like the derivation of this

relationship. Note that there is no value for the sway force in the equation, so we must

assume a value for the FEMs and then factor the resulting moments accordingly. This

proportion depends solely upon the properties (length, stiffness and type of support) of the

legs. So in the case of our symmetrical frame, all four FEMs will be the same we will

choose 20kNm as a starting point. The value (or its sign) does not matter, as the consequent

factoring will take account of this. Note that the same DFs are used.

Figure 8:Sway distribution.

A

B

MBA

MAB

L

MBA = MAB = 6EI / L2

3/4

1/4

0 0

1/4

3/4

A

CB

D

-6.6

0.6

-2.2

-5.0

20.0

-15.0

1.6

6.6

20.0

-7.5

0.8

13.3

-6.7

-0.1

0.3

-4.4

-2.5

20.0

-13.1

-0.26.7

20.0

-6.5

13.5


6/7

Notice the small rounding errors it should be clear from inspection that the legs should be

identical, so we will use 6.7 kNm at the top and 13.4 kNm at the bottom:

HS 8 = 13.4 + 6.6

i.e. HS = 2.5

Total prop force (2 legs)

= 2 2.5 = 5 kN (leftwards at the top)

Figure 9: Sway force calculation.

We need to apply a sway force of 3.3 kN leftwards to the top of the frame, so we must factor

this down:

Sway factor = required sway force = 3.3 / 5 = 0.66

force from sway distribution

The moments from the sway distribution are multiplied by the sway factor before adding them

to the moments from the no-sway distribution:

+ =

No-sway 0.66 sway Final moments

Figure 10: Combination of no-sway and sway moments.

Take particular note of the direction of the moment arrows, and hence the sign of the

moments.

38.84.44.4

HS

HS

6.6

13.4

8m

43.225.4

12.7 21.6 8.8 8.8

29.8

21.5 12.8


7/7

All that remains is to check that the frame is now in horizontal equilibrium:

Leg AB: (29.8 + 21.5) / 8 = 6.4 kN (rightwards at the base)

Leg CD: (38.8 + 12.8) / 8 = 6.5 kN (leftwards at the base)

The results reveal a small rounding error which is actually only 0.04 kNm, so the result isacceptable.

This basic method can be applied to any portal frames with or without sway. You should be

able to see that the only frames which do not sway are those which are themselves

symmetrical and also carry symmetrical loading.

Sideways forces (e.g. wind loads) are dealt with in exactly the same way as above, except that

the total horizontal reaction force will be equal and opposite to the applied horizontal force.

Some additional examples of the way sway FEMs are derived may be helpful:

Leg AB: 6 2EI / 82 3/16

2EI EI Leg CD: 6 EI / 82 3/32

Ratio AB : CD 3/16 : 3/32 i.e. 32 : 16

Use 32 kNm for AB and 16 kNm for CD

(or you could use 20 and 10 same ration of 2:1)

Leg AB: 6 2EI / 82 3/16

2EI EI Leg CD: 3 EI / 82 3/64 (note pinned end)

Ratio AB : CD 3/16 : 3/64 i.e. 64 : 16 or 4:1

Use, e.g., 40 and 10.

Leg AB: 6 2EI / 82 3/16

2EI EI Leg CD: 6 EI / 122 1/24

Ratio AB : CD 3/16 : 1/24 i.e. 3/16 : 3/72

i.e. 72:16 Use 72 for AB and 16 for CD

Figure 11:Examples of sway FEMs.

C

D

C

8m

A

A

C

D

B

8m

12mA

D

B

8m

12m

B

12m

12m

moment distribution - frames

Documents