COURSE NO-416( PRE-STRESSED CONCRETE LAB)

Presented By Sharmeen MahmudId no:10.01.03.078

Sec: BYear & sem : 4-2

Solving Statically Indeterminate Structure: Moment Coefficient Method.

Introduction

In the moment distribution method, every joint of the structure to be analysed is fixed so as to develop the fixed-end moments. Then each fixed joint is sequentially released and the fixed-end moments (which by the time of release are not in equilibrium) are distributed to adjacent members until equilibrium is achieved. The moment distribution method in mathematical terms can be demonstrated as the process of solving a set of simultaneous equations by means of iteration.

The moment distribution method falls into the category of displacement method of structural analysis.

Indeterminate structure

The structures which cannot be solved by the equilibrium equation are known as indeterminate structure.

Moment distribution method

The moment distribution method  is a structural analysis method for statically indeterminate beams and frames developed by Hardy Cross. It was published in 1930 in an ASCE journal. The method only accounts for flexural effects and ignores axial and shear effects. From the 1930s until computers began to be widely used in the design and analysis of structures, the moment distribution method was the most widely practiced method.

Some Basic Definition

Flexural stiffness

The flexural stiffness (EI/L) of a member is represented as the product of the modulus of elasticity (E) and the second moment of area (I) divided by the length (L) of the member. What is needed in the moment distribution method is not the exact value but the ratio of flexural stiffness of all members.

Fixed end moments

Fixed end moments are the moments produced at member ends by external loads when the joints are fixed.

Distribution factors

When a joint is released and begins to rotate under the unbalanced moment, resisting forces develop at each member framed together at the joint. Although the total resistance is equal to the unbalanced moment, the magnitudes of resisting forces developed at each member differ by the members' flexural stiffness. Distribution factors can be defined as the proportions of the unbalanced moments carried by each of the members.

Carryover factors

When a joint is released, balancing moment occurs to counterbalance the unbalanced moment which is initially the same as the fixed-end moment. This balancing moment is then carried over to the member's other end. The ratio of the carried-over moment at the other end to the fixed-end moment of the initial end is the carryover factor.

Steps

The steps involved in moment distribution method are described below :

1.Calculate the fixed end moments for all the spans of the beams by considering all the joints as fixed.

2. Calculate the stiffness coefficients for all the members. The stiffness coefficient ( kAB ) for a member AB is calculated as follows:

kAB = 4EI/L if the far end B is fixed .

kAB = 3EI/L if the far end B is pinned or on a roller.

E is modulus of elasticity, I is moment of inertia of section and L is span AB. While calculating the stiffness coefficients all the intermediate joints are taken as fixed but the joints at end of beam are kept as they are.

3.Calculate the distribution factors based on the stiffness coefficient of the member.

Distribution factor can be defined as the ratio of stiffness coefficient of a member to the sum of the stiffness coefficient of all the members meeting at that joint. If BA, BC and BD are connected at joint B, then the Distribution Factor (D) can be easily calculated as follows;

DBA= kBA/(kBA+kBC+kBD)

DBC=kBC/(kBA+kBC+kBD)

DBD=kBD/(kBA+kBC+kBD)

Distribution factor for a pinned support or roller at the of the beam is taken as 1 whereas for a fixed support at the end of beam the distribution factor are taken as zero.

4. Balance all the joints by applying the balancing moments in the proportions of distribution factors.

5.Carry over half of the balancing moments to the opposite ends of the span. If the opposite end is pinned there should be no carry over moment to that end (as in the case of pinned support at the ends of the beams ).

6. Continue these cycles of balancing and carry over till the joints reach equilibrium state when the unbalanced moment is negligible (based on desired accuracy).

7. Take the sum of all the moments (fixed end moment, balancing moment, carry-over moment) at each end.

Sign convention

Once a sign convention has been chosen, it has to be maintained for the whole structure. The traditional engineer's sign convention is not used in the calculations of the moment distribution method although the results can be expressed in the conventional way. In the BMD case, the left side moment is clockwise direction and other is anticlockwise direction so the bending is positive.

Framed Structures

Framed structures with or without sidesway can be analysed using the moment distribution method.

Sides way Frame

Problem

15 kN/m 10 kN/m150 kN

8 m 6 m 8 m

A B CI I I

3 m

D

SOLUTION OF PROBLEMS -

Solve the previously given problem by the moment distribution method

Fixed end moments

mkNwl

MM

mkNwl

MM

mkNwl

MM

DCCD

CBBC

BAAB

.333.5312

)8)(10(

12

.5.1128

)6)(150(

8

.8012

)8)(15(

12

22

22

Stiffness Factors

EIEI

K

EIEIEI

K

EIEI

L

EIKK

EIEI

L

EIKK

DC

CD

CBBC

BAAB

5.08

4

5.08

4

8

4

667.06

))(4(4

5.08

))(4(4

Distribution Factors

00.1

4284.0500.0667.0

500.0

5716.0500.0667.0

667.0

5716.0667.05.0

667.0

4284.0667.05.0

5.0

0.0)(5.0

5.0

DC

DC

DC

CDCB

CD

CD

CDCB

CB

CB

BCBA

BC

BC

BCBA

BA

BA

wallBA

BA

AB

K

KDF

EIEI

EI

KK

KDF

EIEI

EI

KK

KDF

EIEI

EI

KK

KDF

EIEI

EI

KK

KDF

stiffnesswall

EI

KK

KDF

Joint A B C D

Member AB BA BC CB CD DC

Distribution Factors 0 0.4284 0.5716 0.5716 0.4284 1

Computed end moments -80 80 -112.5 112.5 -53.33 53.33Cycle 1

Distribution 13.923 18.577 -33.82 -25.35 -53.33

Carry-over moments 6.962 -16.91 9.289 -26.67 -12.35Cycle 2

Distribution 7.244 9.662 9.935 7.446 12.35

Carry-over moments 3.622 4.968 4.831 6.175 3.723Cycle 3

Distribution -2.128 -2.84 -6.129 -4.715 -3.723

Carry-over moments -1.064 -3.146 -1.42 -1.862 -2.358Cycle 4

Distribution 1.348 1.798 1.876 1.406 2.358

Carry-over moments 0.674 0.938 0.9 1.179 0.703Cycle 5

Distribution -0.402 -0.536 -1.187 -0.891 -0.703

Summed up -69.81 99.985 -99.99 96.613 -96.61 0moments

Moment Distribution Table

Computation of Shear Forces

Simply-supported 60 60 75 75 40 40

reaction

End reaction

due to left hand FEM 8.726 -8.726 16.665 -16.67 12.079 -12.08

End reaction

due to right hand FEM -12.5 12.498 -16.1 16.102 0 0

Summed-up 56.228 63.772 75.563 74.437 53.077 27.923 moments

8 m 3 m 3 m 8 mI I I

15 kN/m 10 kN/m150 kN

AB C

D

Shear Force and Bending Moment Diagrams

56.23

3.74 m

75.563

63.77

52.077

74.437

27.923

2.792 m

-69.80698.297

35.08

126.704

-96.613

31.693

Mmax=+38.985 kN.mMax=+ 35.59 kN.m

3.74 m84.92

-99.985

48.307

2.792 m

S. F. D.

B. M. D

Thank You