Introduction: Introduction to Finding Forces in Bridge
Engineers use a method of static vector mechanics to identify the forces acting on bridges and other trusses. In this problem we are given a truss system with certain given information (length of sides, and two forces acting on certain points) and asked to find the forces in the members CD, CF, and FG (the supports between points C and D, C and F, F and G) and whether they are in compression (forces pushing together) or tension (forces pulling apart).
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Step 1: Free Body Diagram
The first step is to draw a free body diagram of the entire system (bridge). A free body diagram is a sketch of the bridge that includes all the given information and the information you are looking for. This includes the lengths of sides, points where beams intersect, vector forces acting on those points (broken down into the forces acting in the X coordinate system and Y coordinate system), and angles between beams.
The picture given has a force unknown at point E in the Y-direction facing up, force given at points G and F in the Y-direction facing down, forces unknown at point A in the X and Y-direction pointing to the right and up, length given for each beam, and unknown angle theta between each beam.
Step 2: Finding Theta
The next step is to find the unknown angle, theta. To do this draw a new free body diagram of triangle ABG. Now draw a line from point B to the mid-point (labeled point B') of line AG. Draw a new free body diagram of triangle ABB'. Since this new triangle is a Right Triangle and two of the side lengths are known, use the law of cosines to find the value of theta.
Adjacent side: side between theta angle and 90 degree angle
Hypotenuse side: side opposite the 90 degree angle
cos(theta) = (Adjacent side length/Hypotenuse side length)
simplified---> theta = cos^(-1)(Adjacent/Hypotenuse)
Step 3: Summing Moment About Point a to Find Ey
Now look back to your original free body diagram to find the moment about point A.
Select point A and all force vectors that do not directly pass through point A. To sum a moment about point A, multiply the magnitude of each of force vectors by the perpendicular distance from that vector to point A. Add those values together and set equal to zero. When adding together imagine you are walking in a counter-clockwise circle around point A, the forces that would be acting against you when you are walking parallel to them (the two forces given at point G and F) need to have a negative sign put in front of them.
Your equation should look like:
(Ey)(3 x length of beam) - (force at point F)(2 x length of beam) - (force at point G)(length of beam) = 0
Plug in known values and solve for force Ey.
Step 4: Draw Sectional Free Body Diagram
Now in order to solve for the interior forces acting on each beam, you must use a sectional free body diagram. To make this diagram imagine you cut the beams CD, CF, and GF in half. Now draw a new free body diagram using triangle FDE and the parts of the remaining beams. Draw three for vectors pointed into the ends of the three beams that were cut off. Label them Fcd (for the force acting on beam CD), Fcf (for the force acting on beam CF), and Fgf (for the force acting on beam GF).
Step 5: Take Moment About Point F
In order to take the moment you must first find the vertical distance between point D and the midpoint of beam FE. Draw a line between point D and the midpoint of FE. Label this midpoint, point D'. Draw a right triangle using points D, E, and D'. Since this is a right triangle and two of the side lengths are known, you can use pythagorean theorem to solve for the remaining side. Use the equation below:
a^2 + b^2 = c^2 Where c=Hypotenuse (given beam length), a=line ED' (given beam length divided by 2), and b is unknown length
Solve for b.
Now to solve for the moment about point F use the same process as solving for moment about point A.
The equation used should be:
(Ey)(length of beam) - (Fcd)(b) = 0
Solve for Fcd. Since this is a positive value the force is a compression force.
Step 6: Sum of the Forces in the Y-Direction
For step 6 use the same sectional free body diagram as in step 4, but with only forces acting in the y-direction.
To find the force acting on beam CF sum up all the forces acting in the Y-direction.
To find the force of Fcf acting in the y-direction use the equation:
Fcf (y-direction) = Fcf * sin(theta)
Now sum all the forces acting in the y-direction and set equal to zero. If any for is pointing down put a negative sign in front of it. Your equation should look like:
Ey - (given force at point F) - (Fcf * sin(theta)) = 0
Solve for Fcf. Since this is a negative value the force acting on beam CF is in tension.
Step 7: Sum of the Forces in the X-Direction
For step 7 use the same free body diagram as in step 4, but with only forces acting in the x-direction.
To find the force acting on beam GF sum up all the forces acting in the X-direction.
To find the force of Fcf acting in the x-direction use the equation:
Fcf (x-direction) = Fcf * cos(theta)
Now sum all the forces acting in the x-direction and set equal to zero. If any force is pointing left put a negative sign in front of it. Your equation should look like:
Fgf + Fcd + (Fcf * cos(theta)) = 0
Solve for Fgf. Since this is a negative value the force acting on beam GF is in tension.
You have now found all the forces in the required beams.