CM 197
Mechanics of Materials
Chap 13: Shear Forces and Bending
Moments in Beams
Professor Joe Greene
CSU, CHICO
Reference: Statics and Strength of Materials, 2nd ed., Fa-Hwa Cheng, Glencoe/McGraw Hill,
Westerville, OH (1997)
CM 197
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Chap 13: Shear Forces and Bending
Moments
• Topics
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Introduction
Types of Beams
Types of Loading
Beam Reactions
Shear Force and Bending Moment in Beams
Shear Force and Bending Moment Diagrams
Relationships Among Load, Shear, and Moment
Sketching Shear and Moment Diagrams Using Relationships
Shear and Moment Formulas
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Introduction
• Beams are members that carry transverse loads and are subjected to
bending.
– Any structural member subjected to bending may be referred to as a beam.
• Shafts, girders, stringers, floor beams, joists, etc.
• Objective of chapter
– Determine the internal forces at various sections along a beam.
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Review Types of beams and supports.
Calculation of beam sections.
Calculation of internal forces (shear and moments) in the beam.
Produce graphical representations of shear and bending moments.
• Assumptions
– Beams are straight and of uniform cross sections that possess a vertical plane of
symmetry. Fig 13-1
– Horizontal positions
– Subjected to forces applied in the vertical plane of symmetry.
Plane of symmetry
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Types of Supports for Beams
• Supports
– Roller supports
• Roller or link support resists motion of the beam only
along the direction perpendicular to the plane of the
support (along axis)
– Reaction at roller support acts along the known
direction. Fig 13-2
– Have one unknown Reaction force, R
R
– Hinge supports
Rx
• Resists motion of the beam at the support in any
direction on the plane of loading.
• Have two unknown Reaction forces- Rx and Ry
Ry
– Fixed supports
• Beam is built into wall or column.
• End of beam is fixed and doesn’t move.
• Have two unknown Reaction forces, Rx and Ry, and
one unknown Moment, M
Rx
M
Ry
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Types of Beams
• Beams
– Simple Beam (Statically Determinate- 3 unknowns)
• Beam supported at its end with a hinge and a roller
– Overhanging Beam (Statically Determinate- 3 unknowns)
• Simple supported beam with an overhang from one or both ends
– Cantilever Beam (Statically Determinate- 3 unknowns)
• Beam that is fixed at one end and free at the other
– Propped cantilever beam (Statically Indeterminate >3 unknowns)
• Beam that is fixed at one end and simply supported at the other.
– Fixed Beam (Statically Indeterminate >3 unknowns)
• Beam with both ends are fixed to supports
– Continuous Beam (Statically Indeterminate >3 unknowns)
• Beam is supported on a hinge and two or more roller supports
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Types of Loading
• Loading
– Concentrated Loads
• Load is applied at a specific point on the beam and is considered a discrete
force acting at the point.
– Example, weight fastened to a beam by a cable that applies a
concentrated load.
– Uniform Loads
• Load is distributes over a part or the entire length of the beam.
– Load is force per unit length of beam (lb/ft or N/m)
– Replaced with equivalent force = uniform load times length of beam
» Example, weight of beam.
– Linearly Varying Loads
• Load is distributed with a uniform variation of intensity.
– Occurs on a vertical or inclined wall due to liquid pressure.
– Load changes along the length of the beam.
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Beam Reactions
• Beam Reactions
– Review of Chapter 3
– Procedure
• Write three independent equilibrium equations to solve for the three
unknowns.
• Three equations could be combinations of force-component or moment
equations, as long as they are independent of each other.
• This chapter assume no horizontal forces.
– Thus reaction force in horizontal direction is zero.
• Assume, sum of forces in x direction is 0, thus the horizontal reaction is 0.
– Example, 13-1
– Example, 13-2
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Shear Force & Bending Moment in Beams
• Shear Force and Bending Moment
– Developed in a beam to resist the external forces
and200tolb maintain
100 lb
equilibrium.
Rx
1 ft
3 ft
– Figure 13-7
R
R
y
10 ft
• Beam is 10 ft long with roller support on the right and hinge support on left.
• Beam is subjected to two concentrated loads.
• Results
– For cross section at 3 feet from the left, the forces and moments need to
be balanced.
– Balance the forces and moments at each section
» Normal forces: Rx = 0; Ry =150 pounds;
» Shear force: V = 150 – 100 = 50 lbs; Moment is acting clockwise
and = 3*150 – 100*2 = 250lb-ft;
» Internal resisting moment moment has to be equal and opposite of
moment and acting opposite direction. M = 250ft-lb
counterclockwise.
– For results for cross section from the right, the forces and moments are
calculated in the same manner.
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Beam Sign Conventions
• Positive Shear
• Shear force at a section is positive if the external forces on the beam produce a shear
effect that ends to cause the left side of the section to move up relative to the right
side. Fig 13-8a
• Positive Moment
• Bending moment at a section is positive if external forces on the beam produce a
bending effect that causes the beam to concave upward at the section. Fig 13-8b
– Positive internal shear force, V, at a given section of a beam viewed from both
directions is shown in Fig 13-8c.
– Positive internal moment M at a given section of a beam viewed from both
directions is shown in Fig 13-8d.
• Rule 1 for finding shear forces
• Internal shear force at any section of a beam is equal to the algebraic sum of the
external forces on either segment separated by the section.
– From the left of beam: upward force as positive
– From the right of the beam: downward force as positive
• Rule 2 for finding bending moments
• Internal bending moment at any section of a beam is equal to the algebraic sum of
the moments about a section due to external forces.
– From either side: moment due to upward force as positive
– Example, 13-1 and Example, 13-2
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Shear Force and Moment Diagrams
• Shear force and bending moment diagrams depict the variation of
shear force and bending moment along a beam.
– Help visualize the shear forces and bending moments along beam
– Construct diagram
• Ordinate (y-axis) values are from balance of forces and moments.
• Compute reaction forces and moment couples.
• Plot values of the shear just below the free body diagram on the y-axis with positive
values above the baseline and negative values below the baseline.
• Plot values of moments just below the shear diagram on same x-axis scale with
positive values above the baseline and negative values below the baseline.
• Important aspects of diagrams
– For concentrated loads
• The shear diagram has a horizontal line between the loads
• The moment diagram has a line with constant slope
– For distributed loads
• The shear diagram has a line with a constant slope.
• The moment diagram has a curved line between moment points.
– If the shear diagram crosses the horizontal line (at zero shear) then it has a
maximum or minimum moment at that point.
– Example 13-5
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Relationships among Load, Shear, and Moment
• Certain relationships exist among the loading diagram,
shear diagram and moment diagram.
• Assume incremental element of a beam.
– Relationships between Load and Shear
• For a distributed load, w, applied to a beam,
– The slope of the shear diagram (rate of change of the shear force per
unit length) at any section is equal to the load intensity at that section.
– The shear force at a section is equal to the shear force at the previous
section plus the total load between the two sections.
• Shear force at the section immediately to the right of a concentrated load is
equal to the shear force immediately to the left plus the load.
– Shear force has an abrupt change at the concentrated load, upward = +
– Relationship between shear and moment
• Moment at a section is equal to the moment at the previous section plus the
area under the shear diagram between the two sections.
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Using Relationships of Diagrams
• Loading Diagram or free body diagram
– Show all applied forces and reactions on beam plus dimensions
– Never replace a distributed load with an equivalent concentrated load
• Shear diagram procedure
– Draw shear diagram directly below the loading diagram.
• Horizontal line is drawn at proper location below loading diagram.
• Draw vertical lines from controlling sections for supports and loads.
– Start at left end and compute shear at controlling sections using Eqn 13-5
VB = VA + LoadB-A
• Note: Section with concentrated force is applied, shear force diagram has an abrupt
change at concentrated load location.
– Plot points on shear diagram using forces at each load.
• Connect points
• Note: Sections with distributed loads have a sloping line between them.
• Note: Slope of line is the intensity of the load. Concentrated loads have slope = 0
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Using Relationships of Diagrams
• Moment diagram procedure
– Draw moment diagram directly below the shear diagram.
• Horizontal line is drawn at proper location below loading diagram.
• Draw vertical lines from controlling sections for supports and loads
– Calculate areas from shear diagram for each section.
• Note: moments at free end or ends of a simple beam are always = 0.
– Start at the left end and compute the moments using Eqn 13-7,
MB = MA + AreaB-A
– Plot points on moment diagram using previous moments and areas
at each section
• Connect points with straight lines (for concentrated loads) or curves (for
distributed loads)
• Note: Maximum or minimum of moment occurs at the location where the
shear is = 0 or where the shear changes sign.
– Example, 13-6, 13-7, 13-8, 13-9
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Shear and Moment Formulas
• From previous section, shear force and bending moment
diagram can be plotted for some simple but typical loading
conditions.
– Table 13-1
• Procedure
– Match the loading conditions to one or more loading examples in
Table 13-1
• Get the loading solutions from the example and use it in problem for
– Reaction forces, Maximum shear, maximum moments, etc.
– Method of superposition
• Effect of each load is computed separately and the then combined effect is
added algebraically.
– Example, 13-10, 13-11
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