Beam Reactions and Diagrams

Beams are structural elements with various engineering applications like roofs, bridges, mechanical assemblies, etc.  In general, a beam is slender, straight, rigid, built from isotropic materials, and most important, subjected to loads perpendicular to their longitudinal axis.  If instead of perpendicular loads the same structural member would be subjected to longitudinal loads it would be called column or post.  If the same member would be subjected to a torque, it would be called and treated as a shaft.  Therefore, when identifying mechanical or structural components, consideration of the manner of loading is very important.

Note that when it comes to orientation, beams can be horizontal, vertical or any inclination in between (like submerged plates analyzed in fluid mechanics)… provided the loading is perpendicular to their major axis.

Beam supports:

Support Type	Looks like	Symbol	Reactions
Roller, also called simple movable sliding			Vertical reaction only Allows horizontal movement Allows rotation
Pinned, also called hinged			Vertical reaction Horizontal reaction Allows rotation
Fixed			Vertical reaction Horizontal reaction Moment reaction

Beam Loads^[1]:

Loads	Symbol	Examples	Covered
Point, also called concentrated		vehicle wheels columns person on diving board	Yes
Uniform Distributed		beam weight snow load on roof truss	Yes
Variable Distributed		hydrostatic load on submerged surface aggregate pile beam of variable cross-section	Yes
Concentrated Moments		machine components	No

Beam types:

Types	Diagram	Examples	Covered
Simple beams, or simply supported		short span bridge	Yes
Overhanging beams		patio support beams diving board	Yes
Cantilever beams		building entrance covers overhanging street signs	Yes
Compound beams		machine components	No
Continuous beams		long span bridge pipe supports	No

When solving for reactions, the following steps are recommended:

1. Draw the beam free body diagram
2. Replace the uniform distributed load (if any) with the equivalent point load
3. Solve ΣM_A = 0 (sum of moments about support A). This will give you R_B (reaction at support B).
4. Solve ΣM_B = 0. This will give you R_A.
5. Using R_A and R_B found at steps 3 and 4 check if ΣV = 0 (sum of all vertical forces) is satisfied.
   1. Note that steps 4 and 5 can be reversed.
   2. For a cantilever beam use ΣV = 0 to find the vertical reaction at the wall and ΣM_wall = 0 to find the moment reaction at the wall. There is no other equation to validate your results.

The shear force at any section of a beam may be found by summing all the vertical forces to the left or to the right of the section under consideration.

Similarly, the bending moment at any section of a beam may be found by adding the moments from the left or from the right of the section considered. The moment’s pivot point is the location under consideration.

By convention, internal shearing forces acting downward are considered positive. They counteract upward external forces. Therefore, when representing the shear forces you can draw them in the direction of external forces. This is visually easier than following the sign convention.

Clockwise moments, conventionally, are considered negative while counter-clockwise moments are considered positive.  When representing the bending moment variation, consult the following table showing qualitative bending moment curves dependent on the shape of the shear force graphs.

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When drawing the shear force and bending moment diagrams, while the sign convention is important, consistency is crucial.  For instance, consider a simple beam loaded with a point load applied on a UD load. Starting the diagrams at support A, looking towards the page, will generate the following:

Now, flip the beam horizontally 180º (or change the observation point, looking at the beam from the opposite side) and draw the diagrams, starting from the same point A. The diagrams will appear as follows:

Note that, while the shear force diagrams appeared to be mirrored images (flipped horizontally), the bending moment diagram is not affected.  Additionally, the most important result of this analysis, illustrates that maximum shear force and bending moment magnitudes will always be the same.

Shear Forces Diagrams:

• At the ends of a simply supported beam the shear force is zero.
• At the wall of a cantilever beam the shear force equals the vertical reaction at the wall. At the beam’s free end the shear force is zero.
• On any beam segment where no loads are applied, the shear force remains constant (horizontal line).
• A point load or reaction on a shear force diagram generates an abrupt change in the graph, in the direction of the applied load.
• A uniform distributed load acting on a beam is represented by a straight line shear force with a negative or positive slope, equal to the load per unit length.

Bending Moments Diagram:

• At the ends of a simply supported beam the bending moments are zero.
• At the wall of a cantilever beam, the bending moment equals the moment reaction. At the free end, the bending moment is zero.
• At the location where the shear force crosses the zero axis the corresponding bending moment has a maximum value.
• The shape of the bending moment curve between two points on the beam is as shown in the above two tables.
• The change in bending moment between two points on the beam equals the area under the shear force diagram between the same two points.

The above guidelines will assist you in generating the beam diagrams;  they also serve as a check.

When solving beam diagrams in class and at home you may check your answers by using this free online beam calculator:   SkyCiv Cloud Engineering Software

Problem 1: State the maximum shear force and bending moment values.

Problem 2: State the maximum shear force and bending moment values.

Problem 3: A 24 meters long beam is simply supported at 3 meters from each end.  The beam carries a point load of 18 kN at the left end and 22 kN at the right end of the beam.  The beam weighs 400 kg/m.  Sketch the beam diagrams and determine the location on the beam where the bending moment is zero.

Problem 4: A simple overhanging beam 112 ft long overhangs the left support by 14 ft.  The beam carries a concentrated load of 90 kips 12 ft from the right end and a uniform distributed load of 12 kips/ft over a 40 ft section from the left end.  Sketch the beam diagrams and determine the shear force and the bending moment at a section 50 ft from the left end.

Problem 5: Suggest an improvement to this chapter.