Bridge Truss AnalysisDarren Gabos

Brendan O’Leary

Intro to Comput for Engr & Sci – ENGR 108Nicholas Carlson

11-2-2013

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Introduction A client has requested for our corporation’s (GO Engineering) aid in order to evaluate a

truss bridge design. To do this, we must confirm that each member of the bridge can

support the given loads (see Figure 1), according to the load restrictions. A member

cannot exceed a tension load of 40 klb (kilo-pounds) or a compression load of 20 klb.

The bridge will fail if one of those values is surpassed.

Given Diagram:

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Theory There are 2 angles that are unknown in the diagram. To find these, we used the Law of

Sines, which states:

o sin Aa

=sin Bb

Before finding the member forces, we must first consider the reaction forces at each end

joint (joints 1 and 8). Joint 1 is a pin, meaning it is restricted in the x and y direction but

has a moment. Therefore, it has both a reaction force in the x direction (R1x) and in the y

direction (R1y). Joint 8 is a roller, meaning it is only restricted in the y direction.

Therefore, it has a reaction force in the y direction (R8). To find these forces, we must use

the equilibrium equations and the moment equation:

o ΣF y=0 (sum of forces in y direction is zero)

o ΣF x=0 (sum of forces in x direction is zero)

o ΣM 1=0 (sum of moments about joint 1 is zero)

M=F∗d (moment equals force times perpendicular distance)

To calculate the loads on each member, we must look at each individual joint and use

equilibrium equations:ΣF x=0 ,Σ F y=0

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ProcedureFirst, we identified variables and found missing angles and reactions.

Next, the force at each end point was calculated. The qualities (roller vs. pin) of each were taken into consideration:

Listing variables in MATLAB.

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The remaining members were then evaluated to find the forces applied to them. 2 & 3 as an example:

Finally, we included the forces of each member into a matrix to clearly display the results:

Calculating Forces in MATLAB

More Force Calculations

Making a Matrix in MATLAB

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Figures Figure 1: Truss Bridge

Sketch of Bridge(Modeled in Adobe AutoCAD 2013)

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ResultsForce Subject Force (klb) Direction and Type of Force

Member 1 14.6 CompressionMember2 24.3 TensionMember 3 10.0 CompressionMember 4 1.20 CompressionMember 5 24.3 TensionMember 6 24.8 CompressionMember 7 4.01 TensionMember 8 20.7 TensionMember 9 29.2 Compression

Member 10 20.7 TensionMember 11 8.00 TensionMember 12 11.0 TensionMember 13 10.0 TensionReaction 1x 14.0 LeftReaction 1y 10.3 UpReaction 8 20.7 Up

Conclusion

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In conclusion, the bridge fails due to member 6 and 9. After evaluating each member of

the bridge by assembling a system of linear equations and using givens to calculate

characteristics of the joints and members of the bridge, our results show multiple force

magnitudes exceeding the given load restrictions for a member. Members 6 and 9 both

exceeded 20 klb of compression, resulting in a failure of the entire bridge. Therefore, we cannot

approve of this bridge design for our client.

AppendixMATLAB Code%% Finding Reaction Forces

% Because joint 1 is a pin, there is both an x and y reaction force. Joint% 8 only has a vertical reaction because it is a roller% All answers are in units of kilo-poundsF_1 = 14F_2 = 8F_3 = 13F_4 = 10R_1x = -14% M_1 = 0 (sum of moments about joint 1 is zero)R_8 = (8*12+13*24+10*36+14*16)/48% F_y = 0 (sum of forces in y direction is zero)R_1y = F_2 + F_3 + F_4 - R_8%% Finding unknown angles% By using Law of Sinestheta_1 = asind((16*sind(45))/(sqrt(16+144)))theta_2 = 180-45-theta_1%% Calculating other forces at joint 1% F_y = 0 (sum of forces in y direction is zero)M_1 = -R_1y/sind(45)% F_x = 0 (sum of forces in x direction is zero)M_2 = -R_1x + -M_1*cosd(45)%% Calculating other forces at joint 8% F_y = 0 (sum of forces in y direction is zero)M_9 = -R_8/sind(45)% F_x = 0 (sum of forces in x direction is zero)M_10 = -M_9*cosd(45)

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%% Calculating forces at joint 7% F_y = 0 (sum of forces in y direction is zero)M_13 = F_4% F_x = 0 (sum of forces in x direction is zero)M_8 = M_10%% Calculating forces at joint 3% F_y = 0 (sum of forces in y direction is zero)M_11 = F_2% F_x = 0 (sum of forces in x direction is zero)M_5 = M_2%% Calculating forces at joint 2% F_x = 0 (sum of forces in x direction is zero)% M_3 = (M_1*cosd(45)-M_4*cosd(45))/cosd(18.43)% F_y = 0 (sum of forces in y direction is zero)% 0 = -M_1*sind(45)-M_11-M_4*sind(45)+M_3*sind(18.43)% After substituting for M_3M_4 = (-M_1*sind(45)-M_11+M_1*cosd(45)*tand(18.43))/(sind(45)+cosd(45)*tand(18.43))% After substituting M_4 into the equation for M_3M_3 = (M_1*cosd(45)-M_4*cosd(45))/cosd(18.43)%% Calculating forces at joint 4% F_x = 0 (sum of forces in x direction is zero)M_6 = (M_3*cosd(18.43)-14)/cosd(18.43)% F_y = 0 (sum of forces in y direction is zero)M_12 = -M_3*sind(18.43)-M_6*sind(18.43)%% Calculating the remaining force at joint 6% F_x = 0 (sum of forces in x direction is zero)M_7 = (-M_6*cosd(18.43)+M_9*cosd(45))/cosd(45)%% Placing values into a matrix%Type of force (compression or tension): a 1 means the member is undergoing%a tension force, a 2 means the member is undergoing a compression force.%Also, the magnitudes (absolute values) of the forces will be taken shown%in the matrixMatrix_values = [abs(M_1),2;abs(M_2),1;abs(M_3),2;abs(M_4),2;abs(M_5),1;abs(M_6),2;abs(M_7),1;abs(M_8),1;abs(M_9),2;abs(M_10),1;abs(M_11),1;abs(M_12),1;abs(M_13),1]

MATLAB OutputF_1 =

14

F_2 =

8

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F_3 =

13

F_4 =

10

R_1x =

-14

R_8 =

20.6667

R_1y =

10.3333

theta_1 =

63.4349

theta_2 =

71.5651

M_1 =

-14.6135

M_2 =

24.3333

M_9 =

-29.2271

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M_10 =

20.6667

M_13 =

10

M_8 =

20.6667

M_11 =

8

M_5 =

24.3333

M_4 =

-1.1775

M_3 =

-10.0143

M_6 =

-24.7712

M_12 =

10.9973

M_7 =

4.0079

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Matrix_values =

14.6135 2.0000 24.3333 1.0000 10.0143 2.0000 1.1775 2.0000 24.3333 1.0000 24.7712 2.0000 4.0079 1.0000 20.6667 1.0000 29.2271 2.0000 20.6667 1.0000 8.0000 1.0000 10.9973 1.0000 10.0000 1.0000

Handwritten Calculations