Calculation of Reactions, Internal Shears and Moments Using Influence Lines

1. Calculation of Reactions, Internal Shears and Moments Using Influence Lines • Objectives of the materials covered: Given an influence line and the optimal load positioning, students should be able to compute the resulting magnitude of reactions, shears, and moments caused by any combination of loads placed on the structure.

2. Calculation of reactions due to concentrated loads • Once the influence line has been developed, and the critical placement of dead and live loads has been determined, the influence line can then be used to quickly determine the value of the resulting reaction, shear, or moment. • For example, to determine the maximum possible value for the left reaction on a simple beam, the influence line is first generated and the loads placed such that they cause the greatest possible influence. This was accomplished in the previous section, and is repeated here:

3. Calculation of reactions due to concentrated loads • It is now possible to calculate the maximum left reaction that the truck wheels could cause by simply multiplying the magnitudes of the wheel loadings times the corresponding heights of the influence line at the locations of these loads: Reaction at A = 20,000# x 1.0 + 10,000# x 0.75 = 27,500# • Note that by using a dimensionless unit load, the resulting influence line also has no units, and the answer obtained by multiplying the height of the influence line times the applied loads gives the proper units for the reaction.

4. Calculation of reactions due to concentrated loads • The influence line could also be expressed by the following equation, where y is the height of the influence line at a distance x from the left reaction: y = 1.0 – x/L • Thus if you wanted to know the magnitude of the left reaction corresponding to the application of a load at x = 0.75L, it could be found from Left Reaction = Load*y = Load*(1.0 – x/L) = Load*(1.0 – 0.75L/L) = 0.25*Load • If you wanted the magnitude of the left reaction due to the application of a 5000N force at x = 0.875L, it could be found as Left Reaction = 5000N * (1.0 – x/L) = 5000N * (1.0 – 0.875L/L) = 625N • Note that this is why we used a dimensionless unit load, so that when a real load with units of kips or Newtons is applied to the beam, the result would take on the units of the applied loads.

5. Calculation of reactions due to uniform loads • Results for uniform live loads can also be determined as follows. Assume that a differentially small load dP were applied to the beam of Figure 1, at a location x.

6. Calculation of reactions due to uniform loads • As demonstrated earlier, such a load would cause a differential left reaction of magnitude dReaction = dP*(1-x/L) • Thus the total reaction caused by a continuous uniform load across the beam could be found by integrating this incremental load across the beam, wherever the uniform load is applied: R = dP(1-x/L) = w(1-x/L)dx = w(1-x/L)dx R = w(height of the influence line)dx = w*(area under the influence line)

7. Calculation of reactions due to uniform loads • Thus the maximum possible reaction on a simply supported beam due to a uniformly distributed load of any length can be determined using the influence line of Figure 1. First, the influence line tells us where to place the uniform load for maximum positive effect, namely across the entire length of the beam where the influence line is positive.

8. Calculation of reactions due to uniform loads Next, the value of the reaction can be computed by multiplying the magnitude of the uniform load w by the area under the influence line over which the uniform load is applied: Ra = Height of uniform load * Area under the influence line triangle Ra = w kips/ft * (L ft * (1.0+0)/2) = wL/2 kips Now admittedly this result is rather obvious, and most people would not create an influence line for such a simple case. However, we will later show that determining where to place these loads for maximum effect can be quite confusing unless influence lines are available.

9. Calculation of maximum positive shear at a point in a beam • Previously we showed the generation of an influence line for shear at the quarter point in a simply supported overhanging beam, with critical placement of a set of concentrated and uniform loads, which will be repeated here. Note that for maximum positive shear, the uniform live load is placed only in the ranges where the influence line is positive.

10. Calculation of maximum positive shear at a point in a beam • Assuming a uniform load of 2,000 pounds/foot, and L = 10 feet, we can calculate the maximum positive shear at the quarter point by: V ¼ positive = Concentrated Load*height + Concentrated Load*height + Uniform Load*Area under diagram + Uniform Load*Area under diagram V ¼ positive = 20k * 0.75 + 10k * (2/3)*0.75 + 2k/ft * 0.25 * (10ft/4)/2 + 2k/ft * 0.75 * (3*10ft/4)/2 = 26.25 kips

11. Calculation of maximum negative shear at a point in a beam • Previously we showed the generation of an influence line for shear at the quarter point in a simply supported overhanging beam, with critical placement of a set of concentrated and uniform loads, which will be repeated here:

12. Calculation of maximum negative shear at a point in a beam • Assuming a uniform load of 2,000 pounds/foot, and L = 10 feet, we can calculate the maximum negative shear at the quarter point by: V ¼ negative = Concentrated Load*height + Concentrated Load*height + Uniform Load*Area under diagram + Uniform Load*Area under diagram V ¼ negative = 20k *(- 0.50) + 10k * (1/2)*(-0.50) + 2k/ft *(-0.25) * (10ft/4)/2 + 2k/ft * 0.50 * (10ft/2)/2 = 15.625 kips

13. Calculation of maximum positive moment at a point in a beam • Previously we showed the generation of an influence line for moment at the quarter point in a simply supported overhanging beam, with critical placement of a set of concentrated and uniform loads, which will be repeated here. Note again that loads are placed only in regions where the influence line is positive, the large wheel load is placed over the largest positive value, and the smaller wheel load is placed to the right, or left, of the large wheel load, depending on which side results in the highest value.

14. Calculation of maximum positive moment at a point in a beam • Previously we showed the generation of an influence line for moment at the quarter point in a simply supported overhanging beam, with critical placement of a set of concentrated and uniform loads, which will be repeated here. Note again that loads are placed only in regions where the influence line is positive, the large wheel load is placed over the largest positive value, and the smaller wheel load is placed to the right, or left, of the large wheel load, depending on which side results in the highest value.

15. Calculation of maximum positive moment at a point in a beam • Assuming a uniform live load of 2,000 pounds/foot, and L = 10 feet, we can calculate the maximum positive moment at the quarter point by: M ¼ positive = Concentrated Load*height + Concentrated Load*height + Uniform Load*Area under influence line + Uniform Load*Area under influence line = 20kip * 0.1875*10ft + 10kip * 0.1875*10ft*2/3 + 2kip/ft * (0.1875*10ft*(10ft/2)) = 406.25 kip ft

16. Calculation of maximum negative moment at a point in a beam • Previously we showed the generation of an influence line for moment at the quarter point in a simply supported overhanging beam, with critical placement of a set of concentrated and uniform loads, which will be repeated here:

17. Calculation of maximum negative moment at a point in a beam • Assuming a uniform load of 2,000 pounds/foot, and L = 10 feet, we can calculate the maximum negative moment at the quarter point by: M ¼ negative = Concentrated Load*height + Concentrated Load*height + Uniform Load*Area under influence line + Uniform Load*Area under influence line = 20kip * -0.375*10ft + 10kip * (-0.375*10ft)/2 + 2kip/ft * (-0.375*10ft * (10ft/2)/2) + 2kip/ft * (-0.03125*10ft)*(10ft/8)/2 = 131.64