Structural Theory 2 (Part 25)1

THEORY2 Structural Theory 2 Chapter 5

Moment Distribution Method;

General;- the moment distribution method of analyzing continuous beams and frames

involves little more labor than the approximate methods but yields accuracy equivalent to that obtained from the infinitely more laborious “exact” methods. The inlays of statically indeterminate structures previously discussed frequently involved the solution of inconvenient simultaneous equations. Simultaneous equations are not necessary in solutions by moment distribution, except in a few rare situations for complicated frames. The Cross method, introduced by the late Prof. Hardy Cross in 1924, involves successive cycles of computation, each cycle drawing closer to the “exact” answer. The calculations may be stopped after two or three cycles, giving a very good approximate analysis or they may be carried on to whatever degree of accuracy is desired. When these advantages are considered in the light of the fact that the accuracy obtained by the lengthy “classical” method is often questionable, the true worth of this quick and practical method is understood.

Introduction;- the beauty of moment distribution lies in its simplicity of theory and

application. The following discussion pertains to structures having members of constant cross section throughout their respective length (that is, prismatic members). It is assumed that there is no joint translation where two or more members frame together, but there can be some joint rotation (that is, the members may rotate as a group but may not move with respect to each other). Finally, axial deformation of members is neglected.

Considering the structure on Figure 1a, joints A and C are seen to be fixed. Joint B, however, is not fixed, and the loads on the structure will cause it to rotate slightly, as represented by the elastic curve on Figure 1b. If an imaginary clamp is placed at B, fixing it so that it cannot be displaced, the structure will, under load, take the shape of Figure 1c. For this situation, with all ends fixed, the fixed-end moments can be calculated with very little difficulty by the usual expressions.

If the clamp at B is removed, the joint will rotate slightly, twisting the ends of the members meeting there and causing a redistribution of the moments in the member ends. The changes in the moments or twists at B ends of members AB and BC cause some effect at their other ends. When a moment is applied to one end of a member, the other end being fixed, there is some effect or “carryover” to the fixed end.

After the fixed-end moments are computed, the problem to be handled may be briefly stated as consisting of the calculation of;

- the moments caused in the B ends of the members by the rotation of B

- the magnitude of the moments carried over to the other ends of the members

- the addition or subtraction of these latter moments to the original fixed-end moments.

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CBA

Figure 1a Figure 1b

C

B

A CBA

Figure 1c


These steps can be simply written as being the fixed-end moments plus the moments due to the rotation of joint B as;

M = +

Basic Relations; - there are two questions that must be answered in order to apply the moment-

distribution method to actual structures. They are;

- what is the moment developed or carried over to a fixed end of a member when the other end is subjected to a certain moment?

- when a joint is unclamped and rotates, what is the distribution of the unbalanced moment to the members meeting at the joint, or how much resisting moment is supplied by each member?

Carryover Moments;- to determine the carryover moment, the unloaded beam of constant cross section

in Figure 2a is considered. If a moment is applied to the left end of the

beam, a moment will be developed at the right end. The left end is at a

joint that has been released and the moment causes it to rotate an amount . There will, however, be no deflection or translation of the left end with respect to the right end.

The second moment-area theorem may be used to determine the magnitude of . The deflection of the tangent to the elastic curve of the beam at the left

end with respect to the tangent at the right end (which remains horizontal) is equal to the moment of the area of the M/EI diagram taken about the left end and is equal to zero. By drawing the M/EI diagram in Figure 2b and dividing it into triangles to facilitate the area computations, the following expression may be written and solved for ;

=

0 = +

= -

“A moment applied at one end of a prismatic member, the other end being fixed, will cause a moment half as large and of opposite sign at the fixed end with the carryover factor equal to –½”. The minus sign refers to strength-of-

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1MB

A

L

a

EI

M2EI

M1

b

EI

M1

EI2

M1

c

Figure 2


materials sign convention: A distributed moment on one end causing tension in bottom fibers must be carried over so that it will cause tension in the top fibers of the other end. A study of Figure 2 shows that carrying over with a +½ value with the moment-distribution sign convention automatically takes care of the situation, and it is unnecessary to change the signs with each carryover.

Distribution Factors;- usually a group of members framed together at a joint have different stiffness.

When a joint is unclamped and begins to rotate under the unbalanced moment, the resistance to rotation varies from member to member. The problem is to determine how much of the unbalanced moment will be taken up by each of the members. It seems reasonable to assume the unbalance will be resisted in direct relation to the respective resistance to end rotation of each member.

The beam and M/EI diagram of Figure 2b are redrawn in Figure 2c, with the proper relationship between and , and an expression is written for the

amount of rotation caused by moment .Using the first moment-area theorem, the angle may be represented by the

area of the M/EI diagram between A and B, the tangent at B remaining horizontal.

= =

Assuming that all members consist of the same material, having the same modulus of elasticity values, the only variables in the foregoing equation affecting the amount of end rotation are the L and I values. The amount of rotation occurring at the right end of a member obviously varies directly as the L/I value for the member. The larger the rotation of the member, the less moment it will carry. The moment resisted varies inversely as the amount of rotation or directly as the L/I value. This latter value is referred to as the stiffness factor, K.

K =

To determine the unbalanced moment taken by each of the members at a joint, the stiffness factors at the joint are totaled and each member is assumed to carry a proportion of the unbalanced moment equal to its K value divided by the sum of all the K values at the joint. These proportions of the total unbalanced moment carried by each of the members are the distribution factors.

= =

Fixed-End Moments;- when all of the joints of a structure are clamped to prevent any joint

rotation, the external loads produce certain moments at the ends of the members to which they are applied. These moments are referred to as fixed-end moments.

Unbalanced Moments;- initially the joints in a structure are considered to be clamped. When a joint

is released, it rotates if the sum of the fixed-end moments at the joints is not zero. The difference between zero and the actual sum of the end moments is the unbalanced moment.

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Carryover Moments;- the distributed moments in the ends of the members cause moments in the other

ends, which are assumed fixed, and these are the carryover moments.

Distributed Moments;- after the clamp at a joint is released, the unbalanced moment causes the joint

to rotate. The rotation twists the ends of the members at the joint and changes their moments. In other words, rotation of the joint is resisted by the members and resisting moments are built up in the members as they are twisted. Rotation continues until equilibrium is reached at which time the sum of the moments at the joint is equal to zero. The moments developed in the members resisting rotation are the distributed moments.

Sign Convention;- the moments at the end of a member are assumed to be negative when they tend to

rotate the member end clockwise about the joint (the resisting moment of the joint would be counterclockwise). The continuous beam, with all joints assumed to be clamped, has clockwise (or -) moments on the left end of each span and counterclockwise (or +) moments on the right end of each span.

EX. Analyze the continuous beam shown below.

SPAN AB BC CD

JOINT A B B C C D

K 8 12 12

DF 0 0.40 0.60 0.50 0.50 1.00

FEM -75.00 +75.00 -53.33 +53.33 -15.00 +15.00

BAL -8.67 -13.00 -19.17 -19.17 -15.00

CO -4.34 -9.59 -6.50 -7.50 -9.59

BAL +3.84 +5.75 +7.00 +7.00 +9.59

CO +1.92 +3.50 +2.88 +4.80 +3.50

BAL -1.40 -2.10 -3.84 -3.84 -3.50

CO -0.70 -1.92 -1.05 -1.75 -1.92

BAL +0.77 +1.15 +1.40 +1.40 +1.92

CO +0.39 +0.70 +0.58 +0.96 +0.70

BAL -0.28 -0.42 -0.77 -0.77 -0.70

FM -77.73 +69.26 -69.26 +33.86 -33.87 0.00

VL +50.00 +50.00 +40.00 +40.00 +15.00 +15.00

VM +1.41 -1.41 +4.43 -4.43 +5.65 -5.65

VT +51.41 +48.59 +44.43 +35.57 +20.65 +9.35

R +51.41 +93.02 +56.22 +9.35

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10 kN/m

CBA

D

100 kN5 kN/m

3.0 m 8.0 m 6.0 m

2I 4I 3I

3.0 m


relative stiffness factors,

= = = ; = = 8

= = = ; = = 12

= = = ; = = 12

distribution factor,

= 0

= = 0.40

= = 0.60

= = 0.50

= = 0.50

= 1.0

fixed-end moments,

= - = - = -75.0 kN-m

= + = + = +75.0 kN-m

= - = - = -53.33 kN-m

= + = + = +53.33 kN-m

= - = - = -15.0 kN-m

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= + = + = +15.0 kN-m

EX. Analyze the structure shown by moment distribution method.

SPAN AB BC CD

JOINT A B B C C

K 4 6 -----

DF 1.00 0.40 0.60 1.00 -----

FEM -37.50 +37.50 -37.50 +52.50 -40.00

BAL +37.50 -12.50

CO +18.75 -6.25

BAL -5.00 -7.50

CO -2.50 -3.75

BAL +2.50 +3.75

CO +1.25 +1.88

BAL -1.25 -1.88

FM 0.00 +51.25 -51.25 +40.00 -40.00

VL +25.00 +25.00 +25.00 +35.00 +20.00

VM -8.54 +8.54 +1.41 -1.41 0.00

VT +16.46 +33.54 +26.41 +33.59 +20.00

R +16.46 +59.95 +53.59


= = = ; = = 4

= = = ; = = 6


= 1.00

= = 0.40

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A B C D

50 kN 40 kN20 kN 20 kN

I

3 m 4 m3 m 2 m 2 m 2 m

2I


= = 0.60

= 1.00

fixed-end moments,

= - = - = -37.5 kN-m

= + = + = +37.5 kN-m

= - -

= - - = -37.5 kN-m

= + +

= + + = +52.5 kN-m

EX. Determine all moments for the beam shown, which is assumed to have the following support settlements: B = 20 mm and C = 40 mm.Given: E = 200 GPa I = 4.0 x


= = ; = = 1

= = ; = = 1

= = ; = = 1

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50 kN50 kN 40 kN

B CA D4 m 3 m4 m4 m4 m 5 m


SPAN AB BC CD

JOINT A B B C C D

K 1 1 1

DF 1.00 0.50 0.50 0.50 0.50 1.00

FEM +100.00 +200.00 +100.00 +200.00 -346.88 -271.88

BAL -100.00 -150.00 -150.00 +73.44 +73.44 +271.88

CO -75.00 -50.00 +36.72 -75.00 +135.94 +36.72

BAL +75.00 +6.64 +6.64 -30.47 -30.47 -36.72

CO +3.32 +37.50 -15.24 +3.32 -18.36 -15.24

BAL -3.32 -11.13 -11.13 +7.52 +7.52 +15.24

CO -5.57 -1.66 +3.76 -5.57 +7.62 +3.76

BAL +5.57 -1.05 -1.05 -1.03 -1.03 -3.76

CO -0.53 +2.79 -0.52 -0.53 -1.88 -0.52

BAL +0.53 -1.14 -1.14 +1.21 +1.21 +0.52

CO -0.57 +0.27 +0.61 -0.57 +0.26 +0.61

BAL +0.57 -0.44 -0.44 +0.16 +0.16 -0.61

CO -0.22 +0.29 +0.08 -0.22 -0.31 +0.08

BAL +0.22 -0.19 -0.19 +0.27 +0.27 -0.08

CO -0.10 +0.11 +0.14 -0.10 -0.04 +0.14

BAL +0.10 -0.13 -0.13 +0.07 +0.07 -0.14

CO -0.07 +0.05 +0.04 -0.07 -0.07 +0.04

BAL +0.07 -0.05 -0.05 +0.07 +0.07 -0.04

FM 0.00 +31.86 -31.87 +172.50 -172.48 0.00

VL +25.00 +25.00 +25.00 +25.00 +25.00 +15.00

VM -3.98 +3.98 -17.58 +17.58 +21.56 -21.56

VT +21.02 +28.98 +7.42 +42.58 +46.56 -6.56

R +21.02 +36.40 +89.14 -6.56


= 1.00

= = 0.50

= = 0.50

= = 0.50

= = 0.50

= 1.00fixed-end moments,

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= - +

= - + = +100.0 kN-m

= + +

= + + = +200.0 kN-m

= - +

= - + = +100.0 kN-m

= + +

= + + = +200.0 kN-m

= - -

= - - = -346.88

kN-m

= + -

= + - = -271.88

kN-m

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Structural Theory 2 (Part 25)1