| Introduction | ||
 Truck Bed Failure due to Loading (© 2003, Kurt Gramoll) | "Well, this is a new one for me" said the highway patrolman when he was called to investigate the accident. "I have never seen a semi snap in two due to the load. I wonder why it failed near the front and not in the middle?" His partner added, "I am sure the investigation team will let us know. I wonder if the truck was overloaded?" What is known:
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| Question | ||
 Assumed Location of Trailer Supports | Does the failure location agree with a basic beam analysis? What is the value of the maximum bending moment and does it exceed the design moment of 200 kip-ft? | |
| Approach | ||
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| Introduction | ||
| It is important to know how the shear forces and bending moments vary along the length of a beam that is being designed. Graphs are used to describe the change of shear forces and moments. These graphs are called shear and moment diagrams. Employing these diagrams, the maximum and minimum shear and moment are easily identified and located. | ||
| General Method to Develop Shear and Moment Diagrams | ||
 Basic Example to Construct a Shear and Moment Diagram | Constructing shear and moment diagrams is similar to finding the shear and moment at a particular point on a beam structure. However, instead of using an exact location, the location is a variable distance 'x'. This allows the shear and moment to be a function of the distance, x. In theory, this appears to be simple, but the equations can be complex, especially with distributed loads that are also a function of the location, x. Also, if there are multiple loads or supports, more than one function must be developed, i.e. one shear and moment function for each section or span of the beam. The general steps for shear and moment diagrams are as follows:
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| Sign Convention | ||
 Positive Internal Moment and Shear | When constructing shear and moment diagrams, the sign convention is important so viewers will know what direction the beam is bending or shearing. Generally, it is assumed that a positive moment causes a beam to bend downward as shown in the diagram. A positive shear will skew the beam with the left side going up and the right going down, as shown. Each textbook can have different conventions, so it is important that you check what directions are positive or negative. | |
 Multiple Beam Sections for Different Loading Configurations | Notice, all internal moments and shears need to cancel out if the cut section is reassembled. This means the direction is opposite on the right and left faces of the cut. | |
| Multiple Beam Sections (or Segments) | ||
| One confusing aspect of developing moment and shear diagrams is the need to have separate moment and shear functions for each beam segment. This is because a single function cannot model the moment (or shear) change over a load or support (it is a discontinuous function). Each new beam section will have its own moment and shear equations as a function of the location, x. The diagram at the left gives various beams and typical loadings. After each section is cut, then a FBD is drawn for either side of the beam, just like a single section beam discussed above. Then the M and V is determined for that cut and plotted. | ||
| Example: Multiple Beam Sections | ||
 | Moment and shear diagrams are best understood by examining an example. The simple support beam at the left has a single point load between the supports. The first step is to draw a FBD of the whole beam and solve for the reactions. | |
 | ΣMA = 0 (10 ft) RB - (6 ft)(120 lb) = 0 RB = 72 lb ΣFy = 0 RA - 120 lb + 72 lb = 0 RA = 48 lb | |
 | Determine Beam Section Section (1), to the left of the applied load, will have an expression for the shear force and bending moment that will differ from the section (2), to the right of the applied load. Therefore, the sections must be evaluated separately and each will have their own moment and shear equations. | |
 | Section (1) (0 ≤ x ≤ 6 ft) First, cut section (1) a distance x from the left side and form a FBD, as shown. Remember, the discarded right beam section needs to be replaced by unknown an internal shear force and moment, V1 an M1. The left beam section must stay in static equilibrium. ΣFy = 0 48 lb - V1 = 0 V1 = 48 lb ΣMcut = 0 M1 - x (48 lb) = 0 M1 = 48x ft-lb | |
  | Section (2) (6≤ x ≤ 10 ft) Now the next and last section can be cut. The left or right section can be discarded. For consistency with section (1), the left side will be examined. Again, form a FBD, as shown. The internal loads are labeled, V2 an M2 to distinguish them from section (1) shear and moment equations. Again, applying the static equilibrium equations gives, ΣFy = 0 48 lb - 120 lb - V2 = 0 V2 = -72 lb ΣMcut = 0 M2 - x (48 lb) + (x - 6)(120 lb) = 0 M1 = 720 - 72x ft-lb Plot Shear and Moment Diagrams The functions for V and M for both beam sections can be plotted to give the shear and moment over the length of the beam. The plots are given at the left. The location for maximum and minimum shear force and bending moment are easily found and evaluated. | |
| Complex Distributed Load Example | ||
| Distributed loading is one of the most complex loading when constructing shear and moment diagrams. This causes higher order polynomial equations for the shear and moment equations. Recall, distributed loads can be converted to equivalent forces which are easier to work with. Also, complex, non-uniform distributed loads can be split into simpler distributed loads and treated separately. | ||
 | An example is the best way to illustrate how to work with non-uniform distributed loads. Take a simple cantilever beam with a linear varying distributed load as shown at the left. This example has only one beam segment, so only one cut will be needed. | |
 | Cut the beam some distance x from the left. If the right part of the cut beam is used, then the support reactions at A do not need to be determined (this is a unique situation). The distributed load intensity, wcut, is a simple linear relationship between A and B, or wcut = 200 + (x/20)(500-200) = 200 + 15x N/cm | |
 | The distributed load can be split into two parts, a rectangular and triangular shape. The equivalent loads, F1 and F2 of each shape are calculated as F1 = (20 - x) (200 + 15x) = 4,000 + 100x - 15x2 F2 = 0.5 (20 - x) [500- (200 + 15x)] = 3,000 - 300x + 7.5x2 The internal moment and shear, M and V, can now be determined using the equilibrium equations, ΣFy = 0 V = (4,000 + 100x - 15x2) + (3,000 - 300x + 7.5x2) V = 7,000 - 200x - 7.5x2 N | |
 | ΣMcut = 0 M = -[(1/2) (20-x)] (4,000 + 100x - 15x2) + [(2/3) (20-x)] (3,000 - 300x + 7.5x2) M = -80,000 + 7000x - 100x2 - 2.5x3 N-cm Both the shear and moment can now be plotted as a function of position, x, to give the moment/shear diagrams. Since the distributed load was changing linearly, the shear is a quadratic equation and the moment is a cubic. | |
| Moment and Shear Relationship | ||
| From the examples given in this section, it is possible to see an interesting relationship between the shear and the moment. The shear at a given location is the slope of the moment function. In mathematical terms, this is written as, V = dM/dx This relationship may be helpful in determining both the moment and shear diagram without calculating the actual function. Derivation of this relationship is done in the Integration of Load Equation. | ||
 Example Graphic  Beam Model with Loading | Introduction | |
| The trailer is 53 ft long and carries 79,500 lb. Since the load is assumed to be evenly distributed across the trailer bed, the load intensity, w, is w = (79,500 lb)/(53 ft) = 1,500 lb/ft = 1.5 kip/ft The trailer is support by two sets of wheels which can be modeled as simple pin joints as shown at the left. Joint A is at the left edge and joint B is at the center of the back wheels. Both of the support joints are approximate. | ||
| Free Body Diagram with Reactions | ||
 Free Body Diagram | The basic trailer can be modeled with a free body diagram to assist in finding the reaction forces. There will be a reaction at joint A and B. These can be found by using the static equilibrium equations, ΣMA = 0 39 RB - 26.5 (79,500) = 0 RB = 54,020 lb = 54.02 kip The reaction at A can be found by summing forces, ΣFy = 0 RA + 54.02 - 79.5 = 0 RA = 25.48 kip | |
 Beam Sections | Sections | |
| Now that the reaction forces are known, the beam can be cut into sections. This beam has one continuous load and two supports. Since one of the supports, the right one, is not at the beam edge, this will cause a discontinuity in the moment and shear diagrams. The beam will need to be analyzed at two places, once in each of the sections shown in the diagram. | ||
| Section 1) | ||
 Section 1 Cut | The first section is cut and the left part of the cut beam is used as shown in the diagram. The unknown internal shear V1 and moment M1 are applied at the cut edge to keep the beam section in equilibrium. These two loads can be determined from the equilibrium equations, giving ΣFy = 0 25.48 - 1.5x - V1 = 0 V1 = 25.48 - 1.5x 0 ≤ x ≤ 39 ft ΣMcut = 0 M1 + (x/2) (1.5x) - 25.48x = 0 M1 = 25.48x - 0.75x2 0 ≤ x ≤ 39 ft | |
| Section 2) | ||
 Section 2 Cut | The second cut is done between joint B and the right edge of the beam. This time, the right beam section will be analyzed. Again, the unknown internal shear V2 and moment M2 are applied at the cut edge. Note, that the direction of the internal loads are opposite from section 1 cut. This is because the opposite beam section (the right side) is used. This follows the sign convention specified in the previous section. Applying the equilibrium equations gives, ΣFy = 0 V2 - 1.5 (53 - x) = 0 V2 = 79.5 - 1.5x 39 ≤ x ≤ 53 ft ΣMcut = 0 -M2 - [(53 - x)/2] [1.5 (53 - x)] = 0 M2 = -2,107 + 79.5x - 0.75x2 39 ≤ x ≤ 53 ft | |
| Plot Shear and Moment Equations | ||
 Shear and Moment Diagrams | Each of the two segments have different functions for the shear and moment. These can be plotted over each section to give a complete shear and moment diagram. These are shown at the left. With many shear and moment diagrams, a maximum is needed. In this problem, the first beam section increases and then decreases. To find the location of the maximum, equate the first derivative with respect to the locations, x, to zero. This gives, dM1/dx = 0 25.48 - 1.5x = 0 x = 16.99 ft The maximum moment at 16.99 ft is 216.4 kip-ft which is a large moment. The moment is positive so the beam will be bent downward. This moment does exceed the maximum allowed of 200 kip-ft, so failure would be expected. | |
 Truck Failed Near the Location of the Maximum Moment | Discussion | |
| It is interesting to note that the maximum moment from the analysis agrees well with the location where the trailer failed | ||
| Example 1 | ||
 Example Graphic | A simple supported beam needs to support two loads, a point force of 500 lb and a distributed load of 50 lb/ft as shown. Plot the shear and moment over the length of the beam. What is the maximum shear and moment? In this example, there is a point load and a distributed load. This will require the beam to be sectioned into three segments. Each segment will have a separate moment and shear equation. | |
| Solution 1 | ||
 Free-Body Diagram | The first step in analyzing any beam is to determine the reaction forces at the supports. Using the free body diagram at the left, the moments can be summed at point A to give, ΣMA = 0 12 RD - 9 (300) - 3 (500) = 0 RD = 350 lb The reaction at A can be found by summing forces, ΣFy = 0 RA - 500 - 300 + 350 = 0 RA = 450 lb | |
 Beam Sections  Section 1 with Sign Convention | The beam needs to be split into three sections. Each section starts and ends at a load or support. After the beam is organized into sections, then each section can be cut and the internal shear and moment determined. Section 1 The first cut is between points A and B. The location of the cut is not set, and thus labeled as x. For any location x between x = 0 and x = 3 ft, the shear and moment are given by V = 450 0 ≤ x ≤ 3 ft M = 450x 0 ≤ x ≤ 3 ft | |
 Section 2 | Section 2 The second cut is done between points B and C. This beam section includes the 500 lb point force. The section is kept in equilibrium by internal loads, V2 and M2. These two loads can be determined from the equilibrium equations, giving ΣFy = 0 450 - 500 - V2 = 0 V2 = -50 lb 3 ≤ x ≤ 6 ft ΣMcut = 0 M2 + 500 (x - 3) - 450 x = 0 M2 = 1,500 - 50x 3 ≤ x ≤ 6 ft | |
 Section 3 | Section 3 The third cut is done between points C and D (left edge). This beam section is the most complex due to the distributed that is cut. Only that part of the distributed load that is on the section is used. The width of the distributed load on the remaining beam section is (x - 6) ft. Just like the other two sections, the unknown internal loads, V3 and M3, can be determined from the equilibrium equations, giving ΣFy = 0 450 - 500 - (x - 6) 50 - V3 = 0 V3 = 250 - 50x lb 6 ≤ x ≤ 9 ft | |
