| Introduction | ||
 Ladder as Working Platform | After watching a late night TV commercial about a folding extension ladder, Debbie thought she could use it to help her re-shingle her roof. The problem is she needs a flat surface to work on, but the roof has a slope of 38 degrees. However, this special folding extension ladder can pivot around the center joint which will allow it to be positioned on the roof as shown at the left. Debbie, who is a structural engineer, is concerned that the joint will experience high loads and may fail. What is known:
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 Location of Section a-a | Question | |
| What is the internal moment, shear force and axial force at point just to the left of the the central joint (section a-a)? | ||
| Approach | ||
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| It is important to understand, and be able to find, the internal moments, shear force and axial force at any point in a beam structure. These values can then be used to help design the beam. In this section, the internal loads (moment, shear and axial) will be calculated for a specific location. The next section will expand this concept, and develop diagrams showing the internals at all locations. Before the internal loads are determined, a few topics need to be reviewed, specifically, equilibrium equations, boundary conditions, and load types. | |||||
| Static Equilibrium Equations | |||||
 Typical Loading on Beam | Determining internal loads of beam structures requires a good understanding of static equilibrium equations and how to apply boundary conditions. For any static equilibrium problem, all forces and moments in all directions must be equal to zero. This is summarized in equation form as
ΣFx = 0 ΣFy = 0 ΣMz = 0 | ||||
| Boundary Conditions | |||||
 Typical Boundary Conditions | For typical 2D problems, there are three basic types of boundary conditions, fixed, pinned and slider (includes smooth surface). All three have different reaction forces that need to be used when the support is removed as illustrated in the diagram at the left. The Fixed condition has three reactions (2 force and 1 moment) since it cannot rotate, nor displace in either direction. The Pinned condition cannot displace but it can freely rotate. This requires only two reaction forces. The Slider condition allows the member to move in only one direction but it can freely rotate. This means there is only one reaction force perpendicular to the direction of the motion. A special case is when there is a smooth surface without friction. This allows the member to freely slide in the direction of the slope. The diagram uses triangular shapes to represent pinned connections. The triangle vertex is the pin location. When it is sliding, the triangle has two rollers to indicate free motion along the surface. Note that the rollers DO NOT indicate the member can lift off the surface, only sliding motion. | ||||
| Types of Loads | |||||
 Types of Loads | | Several types of loads can act on beams, such as concentrated loads, distributed loads, and couples. Concentrated loads are idealized from loads applied on a very small area. Distributed loads are spread along the axes of beams. For example, the weight of the beam can be assumed as a distributed force. Couples are moments applied on the beam. | |||
 Equivalent Load for a Distributed Load | Generally, distributed loads are converted into equivalent forces to make the solution process easier. For vertical loads, the equivalent force is The location of the equivalent load is found by
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 Example: Equivalent Loads for Uniform Distributed Load | |||||
| As illustrated in the example at the left, distributed loads can be split into separate distributed load. This makes finding the total equivalent force easier without doing the actual integration. The location of the equivalent force is through the centroid of the distributed load. Thus, if the distributed load is a basic shape, the centroid is easy to determine without integration. In fact, most basic shapes are listed in handbook tables (see Sections Appendix). | |||||
| Sign Conventions | |||||
 Sign Conventions for Bending Moments, Shear Forces and Axial Forces | The sign convention for shear forces and bending moments are not based on their directions along the coordinates axes. The positive bending moment tends to compress the upper part of the beam and elongate the lower part. The negative bending moment tends to elongate the upper part of the beam and compress the lower part of the beam. The positive shear force tends to rotate the material clockwise. The negative shear force tends to rotate the material counterclockwise. And the positive axial force tends to pull the material apart where as the negative force compresses the material. | ||||
| Internal Moments and Forces | |||||
 Typical Beam with Loads  Support Reactions | The first step in finding the internal loads (moment, shear force, and axial force) at a point is to determine the reactions at all supports. This is done by using the three equilibrium equations. To assist in this task, equivalent forces replace the distributed loads, as shown in the example at the left. Then, the basic equations give, ΣMA = 0 F(1.25) - By (1.5) = 0 [(7.2)(2.5)] (1.25) - By (1.5) = 0 By = 15 kN | ||||
 Free-Body-Diagram (FBD) of Ladder | A four foot folding ladder is being considered for a work platform on a sloping roof. The roof angle is 38 degrees and the ladder needs to be horizontal when placed on the roof as shown in the free-body-diagram (FBD) at the left. There are two loads on the ladder, the person working and the roofing materials. The person is a concentrated force of 150 lb and the materials is a distributed force ranging from 200 to 100 lb/ft. The location of both are indicated in the diagram. The total moment, shear force and axial force on the top member of the ladder needs to be determined at a location just to the right of the joint. | |
| Equivalent Loads | ||
 Equivalent Forces  Revised FBD of Ladder | The distributed load can be converted into two concentrated loads which will simplify the solution process. The two equivalent loads are, F1 = 0.5 (2 ft) (100 lb/ft) = 100 lb (triangle area) F2 = (2 ft) (100 lb/ft) = 200 lb (rectangle area) The location of each concentrated load is at the centroid of their area. For the triangle, it is (1/3)2 = 2/3 ft from the left edge. The revised FBD, reflecting the new equivalent forces, is shown at the left. Important angles and distances are also shown on the diagram. The vertical and horizontal distance between points A and B can be determined as (AB)x = 4 sin14 = 0.9677 ft (AB)y= 4 cos14 = 3.881 ft | |
| Reactions | ||
| Before the internal loads can be determined, the external reactions need to be found. In this case, there are three unknowns, so the three equilibrium equations for 2D can be applied to give, ΣMA = 0 6.304R - (2/3 + 0.9677)(100) - (1 + 0.9677)(200) - 150(3 + 0.9677) = 0 R = 182.8 lb ΣFy = 0 Ay - 200 - 150 - 100 + R cos38 = 0 Ay = 450 - 182.8 cos38 Ay = 306.0 lb ΣFx = 0 Ax - R sin38 = 0 Ax - 182.8 sin38 = 0 Ax = 112.5 lb | ||
| Section Cut at Specified Location | ||
 FBD of Left Section | The problem asks for the moment, M, shear, V, and axial load, A, just left of the middle joint, B. This can be done by cutting the ladder at that location and analyzing either the left or right section. The FBD of the left section is shown at the left. This section must be in static equilibrium, just like the full structure. Thus, the three equilibrium equations can be used to solve for the three unknowns, M, V, and A. ΣFx = 0 A + 112.5 = 0 A = -112.5 lb (compression) ΣFy = 0 -V + 306.0 = 0 V = 306.0 lb ΣMcut = 0 M + 112.5 (3.881) - 306.0 (0.9677) = 0 M = -140.5 lb-ft | |
| Alternate Solution Method - Check | ||
 FBD of Right Section | As mentioned previously, either side of the cut can be used to determine the moment, M, shear, V, and axial load, A. As a check, the right section will now be analyzed and compared to the previous solution. Of course, the two solutions should match but many times they do not due to calculation errors. The FBD of the right section is shown at the left. Notice, the loads M, V, and A are in the opposite direction as the previous analysis since the opposite section is used. Summing the forces and the moments give, ΣFx = 0 -A - 182.8 sin38 = 0 A = -112.5 lb (compression) ΣFy = 0 V - 100 - 200 - 150 + 182.8 cos38 = 0 V = 360.0 lb ΣMcut = 0 -M - 100(2/3) - 200(1) - 150(3) + 4(182.8) cos38 M = -140.5 lb-ft Notice, all three values match the previous calculations, as they should. | |
 Internal Moment of Ladder as a Function of Location | Change in Moment Across Ladder | |
| While not requested in this case, it is interesting to visualize the change in internal moment if the cut was moved to other locations. The graph at the left indicates the internal moment magnitude as a function of the cut location, s. The starting point is at A, then goes to B and then to C. Notice the maximum is not at joint B, but at the point where the person is standing. However, joint C does have a large moment and since it is a joint, it would probably fail before the straight section where the person is standing | ||
| Example | ||
 Two Member Structure | What is the internal moment at section a-a? | |
| Solution | ||
 Reaction Forces | Before a member can be cut, the reaction loads at the supports need to be determined. Since both members AB and CB are two force members, the direction of the reaction force at A and C are known. Considering the full structure, summation of forces at C gives ΣMC = 0 -15(10) + FAB cos45 = 0 FAB = 212.1 N | |
 Member Cut at a-a | Member AB can now be cut member at a-a. At the cut, both the shear and moment internal forces are unknowns. The vertical and horizontal components of the force FAB are FAB-x = FAB-y = FAB / cos45 = 150 N Vertical and horizontal distances are shown on diagram. Summing moments at the cut gives, ΣMcut = 0 M + 150(10)(1-cos25) - 150(10)(sin25) = 0 M = 493.4 N | |
