Shear Force & Bending Moment Diagrams
When designing a beam, there is the need to know how the bending moments vary throughout the length of the beam, particularly the maximum and minimum values of these quantities .
1- Finding Internal Reactions
• Pick left side of the cut:
– Find the sum of all the vertical forces to the left of the cut, including V. Solve for shear, V.
– Find the sum of all the horizontal forces to the left of the cut, including N. Solve for axial force, N. It’s usually, but not always, 0.
– Sum the moments of all the forces to the left of the cut about the point of the cut. Include M. Solve for bending moment, M
– Find the sum of all the vertical forces to the left of the cut, including V. Solve for shear, V.
– Find the sum of all the horizontal forces to the left of the cut, including N. Solve for axial force, N. It’s usually, but not always, 0.
– Sum the moments of all the forces to the left of the cut about the point of the cut. Include M. Solve for bending moment, M
• Pick the right side of the cut:
Same as above, except to the right of the cut .
Same as above, except to the right of the cut .
2- Beam Shear
2-Beam Shear
• Why?
• necessary to know the maximum value of the shear .
– necessary to locate where the shear changes from positive to negative .
• where the shear passes through zero.
• Use of shear diagrams give a graphical representation of vertical shear throughout
• Why?
• necessary to know the maximum value of the shear .
– necessary to locate where the shear changes from positive to negative .
• where the shear passes through zero.
• Use of shear diagrams give a graphical representation of vertical shear throughout
the length of a beam .
3-Bending Moment
• Bending moment: tendency of a beam to bend due to forces acting on it .
• Magnitude (M) = sum of moments of forces on either side of the section can be determined at any section along the length of the beam .
• shear is dependent on the loads and reactions
– when a reaction occurs; the shear “jumps” by the amount of the reaction .
– when a concentrated load occurs; the shear “jumps” by the amount of the load Moment is dependent upon the shear diagram .
– the area under the shear diagram = change in the moment (i.e. Ashear diagram = ΔM)
straight lines on shear diagrams create sloping lines on moment diagrams sloping lines on shear diagrams create curves on moment diagrams .
– when a reaction occurs; the shear “jumps” by the amount of the reaction .
– when a concentrated load occurs; the shear “jumps” by the amount of the load Moment is dependent upon the shear diagram .
– the area under the shear diagram = change in the moment (i.e. Ashear diagram = ΔM)
straight lines on shear diagrams create sloping lines on moment diagrams sloping lines on shear diagrams create curves on moment diagrams .
- positive shear = increasing slope
negative shear = decreasing slope
negative shear = decreasing slope
In beam design, only need to know:
– reactions
– max. shear
– max. bending moment
– reactions
– max. shear
– max. bending moment
Now let’s solve some problems
Example 1 :
Example 2 :
Example 3 :
Example 4 :
Now let’s draw the moment Diagrams also ..
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| press on image for full size .. |
Draw the Shear and bending moment diagrams for the beam AB .
Solution :
Find RA and RB
∑Fy = 0 i.e. RA + RB = (1.8 x 2.6) + 4 kN = 8.68 kN
∑MB = 0 i.e. - 4 RA + 2.4 x 4 + ( 1.8 x 2.6) x ( 4 - 1.3 ) = 0
4 RA = 9.6 + 12.63 = 22.23;
RA = 5.56 kN
RB = 8.68 - 5.56 = 3.12 kN
∑Fy = 0 i.e. RA + RB = (1.8 x 2.6) + 4 kN = 8.68 kN
∑MB = 0 i.e. - 4 RA + 2.4 x 4 + ( 1.8 x 2.6) x ( 4 - 1.3 ) = 0
4 RA = 9.6 + 12.63 = 22.23;
RA = 5.56 kN
RB = 8.68 - 5.56 = 3.12 kN
Common Relationships :
To check your answers are correct, you can use this shear force and bending moment diagram plotter http://learntoengineer.com/beam
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