Chapter 1 – Introduction to structural design: Reactions

Example 1.1 Find reactions for simply supported beam

Problem definition. Find the three reactions for a simply supported beam supporting a distributed load of 100 kips/ft over a span of 20 ft. Simply supported means that the beam is supported by a hinge and a roller, and is therefore determinate.

Solution overview. Draw load diagram with unknown forces and/or moments replacing the reaction (constraint) symbols; use the three equations of equilibrium to find these unknown reactions.

Problem solution

1. Redraw load diagram (Figure 1.15a) by replacing constraint symbols with unknown forces, H_A, R_A, and R_B, and by showing a resultant for all distributed loads (Figure 1.15b).

2. The solution to the horizontal reaction at point A is trivial, since no horizontal loads are present: ΣF_x = H_A = 0. In this equation, we use a sign convention, where positive corresponds to forces pointing to the right and negative to forces pointing to the left.

3. The order in which the remaining equations are solved is important: moment equilibrium is considered before vertical equilibrium in order to reduce the number of unknown variables in the vertical equilibrium equation. Moments can be taken about any point in the plane; however, unless you wish to solve the two remaining equations simultaneously, it is suggested that the point be chosen strategically to eliminate all but one of the unknown variables. Each moment is the product of a force times a distance called the moment arm; this moment arm is measured from the point about which moments are taken to the "line of action" of the force and is measured perpendicular to the line of action of the force.

Where the moment arm equals zero, the moment being considered is also zero, and the force "drops out" of the equation. For this reason, it is most convenient to select a point about which to take moments that is aligned with the line of action of either of the two unknown vertical reactions so that one of those unknown forces drops out of the equation of equilibrium. The sign of each moment is based on an arbitrary sign convention, with positive used when the moment causes a clockwise rotation of the beam considered as a free-body diagram and negative when a counterclockwise rotation results (the opposite convention could be chosen as well). In the equation that follows, each product of two numbers represents a force times a distance so that, taken together, they represent the sum of all moments acting on the beam. Any force whose moment arm is zero is left out.

ΣM_B = R_A(20) – 2000(10) = 0

Solving for the vertical reaction at point A, we get: R_A = 1000 kips.

4. Finally, we use the third equation of equilibrium to find the last unknown reaction. Another sign convention is necessary for vertical equilibrium equations: we arbitrarily choose positive to represent an upward-acting force and negative to represent a downward-acting force.

ΣF_y = R_A + R_B – 2000 = 0

or, substituting R_A = 1000 kips:

1000 + R_B – 2000 = 0

Solving for the vertical reaction at point B, we get R_B = 1000 kips. The two vertical reactions in this example are equal and could have been found by simply dividing the total load in half, as we did when considering tributary areas. Doing this, however, is only appropriate when the structure's geometry and loads are symmetrical.

Example 1.2 Find reactions for three-hinged arch

Problem definition. Find the reactions for the three-hinged arch shown in Figure 1.17a.

Solution overview. Draw load diagram with unknown forces and/or moments replacing the reaction (constraint) symbols; use the three equations of equilibrium, plus one additional equation found by considering the equilibrium of another free-body diagram, to find the four unknown reactions.

Problem solution

1. The three-hinged arch shown in this example appears to have too many unknown variables (four unknowns versus only three equations of equilibrium); however, the internal hinge at point C prevents the structure from behaving as a rigid body, and a fourth equation can be developed out of this condition. The initial three equations of equilibrium can be written as follows: a. ΣM_B = R_A(60) – 20(30) = 0, from which R_A = 10 kips. b. ΣF_y = R_A + R_B – 20 = 0; then, substituting R_A = 10 kips from the moment equilibrium equation solved in step a, we get 10 + R_B – 20 = 0, from which R_B = 10 kips. c. ΣF_x = H_A – H_B = 0.

Sign conventions are as described in Example 1.1. This last equation of horizontal equilibrium (step c) contains two unknown variables and cannot be solved at this point. To find H_A, it is necessary to first cut a new FBD at the internal hinge (point C) in order to examine the equilibrium of the resulting partial structure shown in Figure 1.18.

2. With respect to this FBD, we show unknown internal forces H_C and V_C at the cut, but we show no bending moment at that point since none can exist at a hinge. This condition of zero moment is what allows us to write an equation that can be solved for the unknown, H_A:

ΣM_C = 10(30) – H_A(20) = 0

from which H_A = 15 kips.

Then, going back to the "horizontal" equilibrium equation shown in step c that was written for the entire structure (not just the cut FBD), we get:

ΣF_x = H_A – H_B = 15 – H_B = 0

from which H_B = 15 kips.

Example 1.3 Find reactions for a cable

Problem definition. Find the reactions for the flexible cable structure shown in Figure 1.20a. The actual shape of the cable is unknown: all that is specified is the maximum distance of the cable below the level of the supports (reactions): the cable's sag.

Solution overview. Draw load diagram with unknown forces and/or moments replacing the reaction (constraint) symbols; use the three equations of equilibrium, plus one additional equation found by considering the equilibrium of another free-body diagram, to find the four unknown reactions.

Problem solution

1. The cable shown in this example appears to have too many unknown variables (four unknowns versus only three equations of equilibrium); however, the cable's flexibility prevents it from behaving as a rigid body, and a fourth equation can be developed out of this condition. The three equations of equilibrium can be written as follows:

a. ΣM_B = R_A(80) – 10(65) – 20(40) = 0, from which R_A = 18.125 kips.

b. ΣF_y = R_A + R_B – 10 – 20 = 0; then, substituting R_A = 18.125 kips from the moment equilibrium equation solved in step a, we get 18.125 + R_B – 10 – 20 = 0, from which R_B = 11.875 kips.

c. ΣF_x = –H_A + H_B = 0.

Sign conventions are as described in Example 1.1. This last equation of horizontal equilibrium (step c) contains two unknown variables and cannot be solved at this point. By analogy to the three-hinged arch, we would expect to cut an FBD and develop a fourth equation. Like the internal hinge in the arch, the entire cable, being flexible, is incapable of resisting any bending moments. But unlike the arch, the cable's geometry is not predetermined; it is conditioned by the particular loads placed upon it. Before cutting the FBD, we need to figure out where the maximum specified sag of 10 ft occurs: without this information, we would be writing a moment equilibrium equation of an FBD in which the moment arm of the horizontal reaction, H_A, was unknown.

2. We find the location of the sag point by looking at internal vertical forces within the cable. When the direction of these internal vertical forces changes, the cable has reached its lowest point (Figure 1.21). Checking first at point C, we see that the internal vertical force does not change direction on either side of the external load of 10 kips (comparing Figure 1.22a and Figure 1.22b), so the sag point cannot be at point C.

However, when we check point D, we see that the direction of the internal vertical force does change, as shown in Figure 1.23. Thus, point D is the sag point of the cable (i.e., the low point), specified as being 10 ft below the support elevation.

We can also find this sag point by constructing a diagram of cumulative vertical loads, beginning on the left side of the cable (Figure 1.24). The sag point then occurs where the "cumulative force line" crosses the baseline.

3. Having determined the sag point, we cut an FBD at that point (Figure 1.25a) and proceed as in the example of the three-hinged arch, taking moments about the sag point: ΣM_D = 18.125(40) – 10(25) – 10(H_A) = 0, from which H_A = 47.5 kips.

Once the location of the sag point is known, a more accurate sketch of the cable shape can be made, as shown in Figure 1.25b.

Then, going back to the "horizontal" equilibrium equation shown in step 1c that was written for the entire structure (not just the cut FBD), we get ΣF_x = –H_A + H_B = –47.5 + H_B = 0, from which H_B = 47.5 kips. In this last equation, the value of H_A is written with a minus sign since it acts toward the left (and our sign convention has positive going to the right).