# 39-Solved problem 10-1-use three-moment equations 3/4.

## Solved problem 10-1- Moment Values By Three-Moment Equations For Continuous Beams-3/4.

### Brief description of the 0.90 rule for the elastic redistribution of the moment.

For part b of the requirement,  which is to Design the beam, using elastic analysis with the service loads and the 0.9 rule.

The 0.90 rule is a redistribution method of the bending moments in which he considers 90% only of the value of the negative moment.

Then add the average of the 10% value, the reduced value between the two supports from 0 to 10%, to the maximum positive moment for the end span.

For the case of a beam with a continuous edge, the average of 10% of the values is to be added to the maximum positive.

For our solved problem 10.1, for the elastic method, we will find that the author, Prof. Mccormac has given us the values of the negative and positive moment for each beam. The redistribution of the moment by using the 0.90 rule as explained.

## How to use the three-moment equations?

### The three-moment equations for the fixed joint A.

In the method of three moments equations, at the fixed support there is an imaginary beam length Lo with distance=0, so  L0=0.

From the next slide, review the equation of the three-moment equations. We apply the equation of three moments at Joint A, with no elastic reaction at the left span of Joint A, since no moment.

The second beam has a concentrated load we call it P1, its elastic reactions can be estimated after we draw the ding moment diagram, the maximum moment=P1*L1/4.

I will not write the value of P1 or W1, till I get the final relation through matrices, and then I will substitute these values.
The elastic reaction value is the area of half of the bending moment triangle, which is 0.50 *(P1*L/4)*L1/2=P1*L1^2/16, which is the half value of the triangle.

That was the value of the elastic reaction for the concentrated load.

But, we have a uniform load with intensity W1, which causes a bending moment = W1*L1^2/8. The elastic reaction value is the area of half of the parabola, which is=1/2*(2/3)*L2*(W1*L^2*(1/8)).

This value is the elastic reaction at the right side of joint A. The area of the half of the parabola=1/2*(2/3)*L1*(W1*L1^2/8)= W1*L1^32/24.

To solve the equation by using matrices, Consider the following:
1-Put moments MA, MB, MC, and MD, as a vertical column matrix (4×1).

2-While the row matrix is (1×4), from the three moments equation,  the matrix, can be arranged as, 2*L1*MA+L1*Mb+ 0*Mc+0*MD= -6*(P1*L1^2/16) -6*(W1*L1^3/24), at the right side of the equation.

We consider a column matrix 4×1, which is the load’s column (W1 W2 W3  W4), so (-6/24)=-1/4*L1^3.The row matrix 4×1 is(-1/4*L1^3 0  0  0).

For the concentrated load matrix, it has a column vector, 4×1, as (P1 P2 P3 P4), the row matrix 4×1 which is (-3/8*L1^2  0  0  0).

This row is to be multiplied by the vector matrix (W1 W2 W3 W4).

We have completed writing both sides of the matrix equation for the moment at joint A.

### The three-moment equations for the continuous joint B.

From the next slide, we will write the equation of the moment at joint B for the three-moment equations.
We have two beams at B, one is at the left side with length=L1,  loaded by a uniform load W1, and concentrated load P1.

While the second beam is at the right side of B and its length=L2, is loaded with a uniform load W2.

The moment at B is written as MA*L1+2*MB*(L1+L2)+MC*(L2) =-6*r-br-6*r-bL, where r-bL is the elastic reaction at B from the left part, while r-br is the elastic reaction at b from the right part.

The elastic reaction at b from the left, r-bL=elastic reaction from the uniform load W1+ elastic reaction from the concentrated load P1, the elastic reaction at B from the right, and the r-br=elastic reaction from the uniform load W2. -6*(r-br)=-6*(W2*L2^3/24).
-6*(r-bL)=-6*W1*L1^3/24+(-6)* (P1*L1^2/16).

We will create a matrix as follows, The moment vector-matrix is a vector matrix ( MA  MB  MC  MD).
From the three moments equations, we have a row matrix 1×4,(L1  2(L1+L2)  L2  0) to be multiplied by ( MA  MB MC  MD).

The Elastic reaction related to W1, W2, W3, W4, can be represented as a row matrix, 1x 4, as shown, (-L13^3/4 -L2^3/4 0 0).
While for the column matrix due to concentrated loads being 4×1, (P1 P2 P3 P4), we have a row-column 4×1 which is (-3/8*L1^2 0 0 0 ).

### The three-moment equations for the continuous joint C.

From the next slide, the moment at joint C can be represented by the two beams intersecting at C.

The second beam is loaded by W2, with span =L2, while the third beam has a load of W3 with span =L3, Te three moments equations to be written as MB*L2+2*MC*(L2+L3)+MD*0= -r-cL-6*r-cr= -1/4*W2*L2^3-1/4*W3*L3^3. We have a row matrix 1×4 which is(0 L2+2*(L2+L3) 0) and a vertical matrix, MA MB MC MD), For the row matrix 4×1,(0 -L2^3/4 -L3^3/4 0) and a column matrix (W1 W2 W3 W4).

We do not have any elastic reaction values from the concentrated loads, which can be represented by the row matrix 1×4, (0 0 0 0) and a column matrix (P1  P2  P3  P4).

### The three-moment equations for the end joint D.

From the next slide, three-moment equations.

We have left with The moment at joint D. The elastic reaction at D=0, the moment at D=0, since joint D supports. The row matrix 4×1,(0 0 0 0) and a column matrix (W1 W2 W3 W4) Plus the row matrix 4×1,(0 0 0 0) and a column matrix (P1 P2 P3 P4). This matrix is written to suit the condition when the last support is a hinge.

### The final form matrix equations for the continuous beams.

We have a matrix 4×4 multiplied by a column vector (MA MB MC MD) plus that matrix 4×4 has the data related to the span lengths. While at the right-hand side, we have the coefficient of elastic reactions 4×4, multiplied by the column vector (W1 W2 W3 W4) plus the coefficient of elastic reactions 4×4, P3 P4).

In the next post, we will find out the final values of moments.

This is the pdf file used in the illustration of this post.