Control of Stresses in a Multi-Span Bridge

Ruo Jia

1. Objection and theory The aim of this project is using MATLAB as a tool to optimize the Britannia Bridge design. The new design should be more efficient than Edwin Clark`s work, as the peak bending moment was minimised, which leads to a bigger safety factor against yield. The specification of the design is that the bridge is formed of four spans, length of each span are shown in Figure 1. It should be built to support a uniformly self-weight of 99.6kN/m, a box structure with Young’s modulus of 210Gpa and second moment of area of 12.07m^4.

Figure1: Free body diagram of the Britannia Bridge

Compare with a simply supported design at each end by the piers, building the bridge as one continuous beam supported by all five piers introduces sagging moments at the middle of each span. The ideal situation is achieved when the maximum sagging equals the maximum hogging moment in Figure 2.

Figure2: Ideal moment diagram

Controlled settlement method is used to achieve the ideal situation. This method provides a hogging moment over the pier and connects the spans continuously. The way to do it is that, every time the newly span installs, angled the span a little upwards and then lower it down after they joined. With the right controlled settlement distances being used, the optimum moment distribution could be achieved. Clark has attempted these calculations by hand but unfortunately he made a few errors so the design is not optimum.This report indicates how to carry out controlled settlement using MATLAB.The first several scripts were written to understand the basic idea then the main problem is solved by optimization.

2. Simple bending model-via Macaulay Brackets Specifications for 1) and 2): The bridge length L=1, the load per unit length W/L=1, the bending stiffness=1

1) Simply supports beamThe first scenario is for a uniformly distributed load on a beam simply supported on each end as shown in figure3. This section need to be done both by hand and MATLAB.

Figure3. Uniformly distributed load on a beam simply supported on each end

Firstly, by cutting into the beam and looking left (the length of the cutting is x), the bending moment equation can be done:

M=12

x2 - 12

x

The first term donates self weight of the span and the second term donates upward reaction on the

first end. Using the equation M=EId2 yd x2

, the deflection distribution along the beam can be calculated by

a double integration with respect to x:

d2 yd x2

=12

x2 - 12

x

dydx

=16

x3 -14

x2+c1

y ¿ 124

x4 - 112

x3+c1x+c2

Calculate by hand, as no displacement at each end, two constant can be found easily (x=0, y=0;x=L,

y=0): c1= 124

, c2=0. The bending moment and displacement diagram can be draw.

Instead of substitution of boundary conditions into the displacement function manually, optimization can be performed in MATLAB to find these unknowns. Optimization is done by two steps. Firstly, write an optimization function (OB) with unknowns which need to be minimised. Then, a function `fminsearh` must be introduced to find the minimum of the optimization function (OB) defined earlier.To perform fminsearch, initial estimate of every variable set earlier need to be settled. Then, from the initial estimate, fminsearch will start iterating through values and find the minimum it could reach at last. The accuracy of fminseach can be improved by reduce the step size, reduce the tolerance etc.

%bending moment, slope and deflection function

M=@(x)w*x.^2/2-w*L*x/2;

slope=@(x,c1)(-1/EI)*(w*x.^3/6-w*L*x.^2/4+c1);

deflection=@(x,c1,c2)(-1/EI)*(w*x.^4/24-w*L*x.^3/12+c1*x+c2);

%OB defines a boundary condition function. Values of independent terms will be

minimised by fminsearch. Square each term to ensure the whole function will be

minimised instead of cancelled by each other.

OB=@(c1,c2)(deflection(0,c1,c2)).^2+(deflection(L,c1,c2)).^2;

[C]=fminsearch(@(c)OB(c(1),c(2)),[0 2]);

C1=C(1);

C2=C(2);

The output of fminsearch is C, consist the optimum value of C1=0.0417 and C2= -1.840e-05.Almost

same with the calculation by hand. It cannot reach the exact value since fminsearch find the point discontinuously with tolerance, but it is close enough. So the bending moment distribution and deflected shape are the same for both method and shown in figure 4.

Figure 4. Bending moment distribution and deflected shape in first scenario

2) Simply supported beam with a central rigid prop.The second scenario is for a uniformly distributed load on a simply supported beam with a central rigid prop. The main difference made in this situation is that the middle support gives another unknown factor, the force in the middle. This force adds a step of discontinuity to moment equation. If

solve it by hand, x< L2

and x> L2

will have different expressions. Since the unknown force in the middle

does not to take into account when write moment expression x< L2

. After x= L2

, f*x has to take into

account. Therefore, the Macaulay bracket is introduced in this situation. This bracket introduces a

step function. Bracket x- L2

means when x< L2

, the term equals 0; when x> L2

, the term equals 1. In

MATLAB, Macaulay bracket can be expressed as Heaviside. Integration of a term multiplies Heaviside will be the same as the result of integration of that term in the region of Heaviside. The integration always has to add a constant (c1, c2) to adjust the expression shape when applies the boundary condition.

%the whole beam length is L, L=1

%bending moment, slope and deflections expressions

%f is the supporting force in the middle of the beam, force at each end is (w*L-f)/2

since symmetric; c1 and c2 are constant as previous

%there are 3 unknowns f, c1, c2 in the equations

M=@(x,f)w*x.^2/2-(w*L-f)/2*w*L*x-f*w*L*(x-L/2).*heaviside(x-L/2); slope=@(x,f,c1)-

1/EI*(w*x.^3/6-(w*L-f)/2*w*L*x.^2/2-f*w*L/2*(x-L/2).^2.*heaviside(x-L/2)+c1);

deflection=@

(x,f,c1,c2)-1/EI*(w*x.^4/24-(w*L-f)/2*w*L*x.^3/6-f*w*L*(x-L/2).^3/6.*heaviside(x-

L/2)+c1*x+c2);

As three unknowns from deflection expression, in correspond, the optimization function (OB) need 3

terms to limit the boundary condition as well. The deflections at x=0, x= L2

and x=L all equal to zero.

Every term squared with the same reason as before, the situation that they cancel out each other should not exist.

%find the value of c1, c2, f by optimization, the square of each error component should

be zero

OB=@(c1,c2,f)(deflection(0,f,c1,c2)).^2+(deflection(L/2,f,c1,c2)).^2+

(deflection(L,f,c1,c2)).^2;

[C]=fminsearch(@(c)OB(c(1),c(2),c(3)),[2 2 1]);

The process is same as before, but three unknowns need to be searched for in the fminsearch function. The results are: C1=0.0026; C2=2.1160e-07; F=0.6250. The bending moment distribution and deflected shape are the same for both method and shown in figure 5. The results look reasonable as the moment distribution is symmetrical with zero moment at both ends. The middle moment is a positive value, which means hogging in the middle and sagging for both sides away from it. The deflection distribution is symmetric as well with every point deflect downwards and at ends and middle point, deflections are zero.

Figure5.Bending moment distribution and deflected shape in second scenario

3) The Britannia BridgeThis question extends the second scenario into Britannia Bridge which has 4 supporting points. The specifications and structure has been specified in the first part. Bending moment equation could easily get same as the method in second part, integrate twice and the deflection expression should be :

% f1=fc, f2=fb=fd, c1 is the integration constant from moment integration, c2 is the

integration constant from slope integration

deflection = @(x,f1,f2,c1,c2) (-1/EI)*(w*(x.^4)/24 -(6*w*L-2*f2-f1)*(x.^3)/12 - f2*((x-

L).^3)/6.*heaviside(x-L)-

f1*((x-3*L).^3)/6.*heaviside(x-3*L)-f2*((x-5*L).^3)/6.*heaviside(x-5*L) + c1*x + c2);

As this scenario is more complicated with more unknowns, two ways of increase the accuracy of optimization has been used. Firstly, as four unknowns from deflection expression, in correspond, the optimization function (OB) need 4 terms to limit the boundary condition as well. But it turns out the accuracy is not high enough. Therefore, every limitation of deflection on the supporting points is added. There is an extra condition that the moment at the x=6L is zero.

%finds the value of f1, f2, c1 and c2 by optimization with 6 boundary conditions

OB=@(f1,f2,c1,c2)(deflection(0,f1,f2,c1,c2)).^2+(deflection(L,f1,f2,c1,c2)).^2+

(deflection(3*L,f1,f2,c1,c2)).^2+(deflection(5*L,f1,f2,c1,c2)).^2+

(deflection(6*L,f1,f2,c1,c2)).^2++(M(6*L,f1,f2)/(10e5)).^2;

%Set TolFun, MaxFunEvals and TolX in optimset reaches higher accuracy

options = optimset('TolFun',1e-10,'MaxFunEvals',10e5,'TolX',1e-8);

[C]=fminsearch(@(c)OB(c(1),c(2),c(3),c(4)),[10e5 10e5 0 0],options);

Secondly, increase the accuracy of the fminsearch by optimset. Optimset can chang the default values of fminsearch. Tolfun defines termination tolerance on the function value, maxFunEvals defines maximum number of function evaluations allowed and TolX defines termination tolerance on x. With higher accuracy number for these settings, the result turns reasonable. The initial value of c1 and c2 are in the order of 10e4, as they represent the value of f1 and f2.The total weight of the beam is 4.18e8, therefore a reasonable order of f1 and f2 should be e8 or e7. The results are F1=1.4815e+07N, F2=1.1765e+07N, C1=0.0034, C2=0.0022m, and the BM diagram and Deflection are as shown in Figure6.

Figure6.Bending moment distribution and deflected shape in third scenario

The BM and deflection are both in a reasonable shape. In BM diagram, M (6L) =1.048e-09, almost zero, M (0)=0. In defection diagram, y(0)=-7.543e-14, y(L)=-5.771e-06, y(3L)=-3.864e-07, y(5L)=-5.771e-06, y(6L)= 2.692e-12. These boundary conditions all minimised to zero at the extent lower than the order of e-06, which is accurate enough for construction.

3. Bending moment optimization 4) Optimum bending moment of central supported beamThe fourth script is based on the structure of the second scenario, a uniformly distributed load on a simply supported beam length2L with a central rigid prop. The moment equation is same as before, as

the moment equation is M= x2

2−(1− F2 )x−f ( x−1 )

The aim for this part is to optimize the BM distribution reaching the ideal case, which means make max hogging moment, equals max sagging moment. By adjusting the middle supporting force, the lowest possible maximum absolute moment in the beam can be achieved. N-norm is the method used to estimate the maximum bending moment without searching. Calculate value of norm by the script below.% each row represents a set of norm with same N, which means first row is for N=2,

second row is N =4, etc. Every column represents a set of norm with same input of same

force, which means the first column is the value of norm for force=1/1001, N=2, 4, 8,

16, INF. Therefore, a 5*1001 normvector matrix is formed to be filled by all the values

of norm and the minimum will be found later.

row=1

for N=[2,4,8,16,inf]

column=1;

F and x both in a range from 0 to2, as they should be the same size, 1001 points selected. Order of

1000 is for higher accuracy, so more pointed will be near the max and the biggest value is closer to exact value.1001 is used as the maximum norm peak expected at the location of supporting point. For higher accuracy, select the point at the peak instead of select two points near it is preferred.

F=linspace(0,2,1001);

for force=F

x=linspace(0,2,1001);

moment=M(x,force);

normvector(row,column)=norm(moment,N);

%column increment, a new norm calculated with new value of x

column=column+1;

end

%row increment, a new row is added when change to a new N

row=row+1

end

Compare the Norm as a function of F in Figure7. The figure shows that a smooth parabola of N=2, tends to a pair of intersecting lines. The calculation of 2-Norm is to make square of each term in BM equation then root square the sum. The 2-norm line is higher than the rest is because it adds up every term’s effect while inf-norm only takes the maximum term into account. Consider the physical meaning of the curve, if the central reaction is zero, the bridge will be as same as the first scenario, which means totally sagging; if only the central force hold the beam then it’s completely hogging. Therefore, as two lines intersect at the minimum point, it suggests that the left line represents sag and the right line represents hog. The minimum point is where the maximum hogging moment equals the maximum sagging moment. Also, the optimum has been reached at this point.

Figure7. N-norm as a function of F in fourth scenario Figure8.Optimum bending moment diagram

After MATLAB calculated every value of Norm, a find function can help to find the minimum value of norm with F, and F=1.1720. The optimum BM diagram is in Figure8. It shows the max sagging moment is 0.0857 while the max hogging moment is 0.086, almost the same.5) Optimum bending moment of Britannia Bridge

The fifth scenario goes through a similar route as the fourth scenario, with unity EI and w, total length of the beam 3 with 5 supporting forces. Since the structure is symmetrical, and total weight of the 5 forces is fixed as 3, there are 2 unknown forces need to be varied reaching optimum situation. In fourth scenario, only single supporting force varies and a 5*1001 matrix formed. Therefore, a 5*1001*1001 matrix of norm needed as 2 unknowns vary with 2, 4, 8, 16, INF norm.

In the 5*1001*1001 matrix, each row represents a set of norm with same N, which means first row is for N=2, second row is N =4, etc. Every column represents a set of norm with same input of same f1(supporting force at A and E), which means the first column is the value of norm for force=1/1001,

N=2, 4, 8, 16, INF. Every depth represents the same f2(supporting force at B and D) in the same way. Therefore, a 5*1001*1001 normvector matrix is formed by all the values of norm and the minimum will be found later.

The coding is similar, just when filling the normvector, another for loop is nested to get the matrix three dimensions. By varying the value of f1 and f2, a minimum of max bending moment can be found using find function in MATLAB and min function twice.The 3D plot of INF-Norm and optimum BM diagram are shown below:

Figure 9. 3Dplot of INF-Norm Figure10. Detail Graph

The same phenomenon as scenario four, the INF-norm 3D plot is sharper edged than 2-Norm and the value of norm is smaller. From the detailed graph, it could be seen that the surface is not smooth with sharp edges. The optimum situation is reached in Figure11, with end supporting force f1 (Fa Fe) 0.875 and f2 (Fb Fd) 0.125, the optimum moment is 0.0625(both max hogging and max sagging).

Figure 11. Optimum BM graph of Britannia Bridge

4. Solve real problem of Britannia Bridge6) Settlement distance for Britannia BridgeThe first part of Scenario 6 is similar to Scenario 5; only it requests to use Britannia Bridge specifications. Copy the 5th Scenario script with new scaled constants, the moment diagram is shown with lower accuracy. This problem arises because the scale is different. In scenario5, to find the optimum forces supporting at A and B, two force vectors created in the domain from 0 to3, 1001 points are picked with equal space out. The 1*1001 force vector is 0, 0.003, 0.006, 0.009, 0.0120…2.997, 3.000 the gap between two consecutive forces is 0.003.By calculating 1001*1001 value of norm, the minimum norm could be found. But in scenario 6, two 1*1001 force vectors have values: 1.0e+07 *(0, 0.0042, 0.0084, 0.0125, 0.0167 …4.1790, 4.1832) the gap between two consecutive forces is 4.2e+04. The gap increases 14e+06 times using the Britannia Bridge scale. As the exact value of max BM must be a point between two values calculate near it, MATLAB can only tell the exact value of these two points value. As the gap getting bigger, the accuracy is lower. The same thing happens to the

x division. To higher the accuracy, fminsearch function is used after the ‘for’ loop found the optimum f1 and f2: [i,j]=find(n== min(min(normvector)));%find the position of the minimum value by for

loop

F1=f1(i); %find F1=FA+FE at this minimum

F2=f2(j); %find F2=FB=FD at this minimum

OBf = @(f1,f2)norm(M(x,f1,f2),inf)

options = optimset('TolFun',1e-20,'MaxFunEvals',10e20,'TolX',1e-20);

[B] =fminsearch(@(b)OBf(b(1),b(2)),[F1,F2],options);

Fopt1=B(1);

Fopt2=B(2);

The optimum function is used to find the minimum value of norm, and fminsearch is introduced use F1 and F2 as initial value. The optimum force is found combined for loop and optimization. Fopt1=1.7430e+06, Fopt2=1.2201e+07.

for theideal case ,|M (70)|¿|M (140)|¿|M (210)|¿|M (280)|¿1.22e+08 , shown in Figure12.

Figure12. Optimum BM diagram in scenario 6

The second part of question is to find settlement distance. This need plug Fopt1 and Fopt2 in the first step into the deflection equation. Optimization equation limits both ends deflections are zero. Use fminsearch with tolerance set 'TolX',1e-10,'MaxFunEvals',1e30. The deflections at B and D are0.118m and deflection at C is 0.275m as third plot in Figure 12.Unfortunately, it is impractical to construct the bridge like the deflection graph. It needs lower point C 275mm and B,D 118mm after settle the bridge horizontal. This is also very hard to apply as lower the supporting point displacement of 420m beam is difficult to hold and control.

7) The controlled settlement distances to obtain the optimum BM distribution After the optimum moment distribution had been found in scenario 6, the last part is to obtain this distribution in practice. It can be constructed by parts. Firstly, place BC as simple support, then attach CD (D is raised with a slight angle). Lower D after two spans rivets together, a hogging moment introduces at point C. Do the same way to add span DE and AB, finally an optimum BM diagram will appear as the first graph in figure12.

For reaching the optimum BM distribution, the distance that each span is to be raised have to be

calculated. This can be achieved by starting with the optimum distribution, then work backwards to reach a series of simply supported spans like Figure 13.

Figure 13. BM diagram of a series of simply supportes spansThe general idea is Mtotal = Mideal +Mpoint load. The point loaded beam has the identical dimensions and properties as the Britannia Bridge without self weight. Superimpose two moment equation can get the total moment equation which at the end of elimination, Mtotal is shown as Figure13. When a span is eliminated from the whole beam, the central moment of connection is zero. With this limitation, the controlled settlement at the free end can be found. The script below shows how to eliminate AB as an example.%remove AB

To remove span AB, point A is released from boundary condition, which means the defection and moment are not zero at A. As moment at B is zero, Fa1 is calculated by Fa1*L + Mideal = 0. Therefore, Fa1=Mideal(B)/-L= -122000000/LFa1=-122000000/L; % The moment, slope, deflection of the point load function can be settled with unknown

force Fb1, Fc1, Fd1

M1=@(x,Fb1,Fc1,Fd1)Fa1.*x+Fb1.*(x-L).*heaviside(x-L)+Fc1.*(x-3.*L).*heaviside(x-3.*L)

+Fd1.*(x-5.*L).*heaviside(x-5.*L);

slope1=@(x,Fb1,Fc1,Fd1,c1)(-1/EI).*((Fa1/2).*x.^2+(Fb1/2).*(x-L).^2.*heaviside(x-L)+

(Fc1/2).*(x-3.*L).^2.*heaviside(x-3.*L)+(Fd1/2).*(x-5.*L).^2.*heaviside(x-5.*L)+c1);

deflection1=@(x,Fb1,Fc1,Fd1,c1,c2)(-1/EI).*((Fa1/6).*x.^3+(Fb1/6).*(x-

L).^3.*heaviside(x-L)+(Fc1/6).*(x-3.*L).^3.*heaviside(x-3.*L)+(Fd1/6).*(x-

5.*L).^3.*heaviside(x-5.*L)+c1.*x+c2);

Use the optimization function with fminsearch to find the optimum value of unknown force and constant. With these constant known, the BM distribution and Deflection can be found. The initial value for the fminsearch is found by a trial run of the script, read the value of Fb1,Fc1,Fd1,c1,c2, then use them as initial value reaching higher accuracy value. %zero deflections at point B,C,D,E and zero moment at point EOB=@(Fb1,Fc1,Fd1,c1,c2)(deflection1(L,Fb1,Fc1,Fd1,c1,c2)).^2+

(deflection1(3.*L,Fb1,Fc1,Fd1,c1,c2)).^2+(deflection1(5.*L,Fb1,Fc1,Fd1,c1,c2)).^2+

(deflection1(6.*L,Fb1,Fc1,Fd1,c1,c2)).^2+((M1(6.*L,Fb1,Fc1,Fd1)/10.^9).^2);

[c]=fminsearch(@(c)OB(c(1),c(2),c(3),c(4),c(5)),

[3e6,5e6,4e5,1e10,1e10],optimset('TolX',1e-20,'MaxFunEvals',1e30));

Therefore, the deflection at A (the controlled settlement at A) can be found by reading the value at x=0 on Figure 14. Point A need to be raised by 0.2144m.

Figure 14 The controlled settlement of A

Then the new Mtotal can be found by adding the point loaded moment distribution to the optimum moment distribution, as the red line on figure 15. This distribution takes the place of the optimum distribution when span DE is removed.%MT is Mtotal after AB is removed,M is the optimum BM.

MT=@(x,Fb1,Fc1,Fd1,f1,f2) M1(x,Fb1,Fc1,Fd1)+M(x,f1,f2);

When remove DE, the same procedure as before, but the difference is the new x is not from left to right but the other way around, this is achieved by introducing a new variable x2=420-x. The new point force at E is worked out by the same procedure, find the total moment at D and Fe2=-Md/L. The results are in Figure15, the settlement distance is 0.1966 at E.

.

Figure15.The controlled settlement at EA similar procedure for eliminating CD, and the settlement distance is 0.9544 at D.

Figure16.The controlled settlement at DAs a result, a series of simply supported span BM diagram could be reached. In reality, the bridge is constructed from the last step of coding.5.ConculsionThe optimum BM distribution is reached from MATLAB analysis. The results are higher accuracy than Clark’s work. The new settlement parameters can reach the max hogging BM=max sagging BM=1.22e+08Nm, increasing the safety factor from 2.4 to 4.8. Further analysis to optimize the design further can be done, by taking into account the weight of vehicles pass by, wind effect and corrosion effect etc.