SHEAR FORCE | BENDING MOMENT | SIGN CONVENTION | TYPES OF BEAM | SIMPLY SUPPORTED | TYPES OF LOAD | SHEAR FORCE BENDING MOMENT DIAGRAM |
SHEAR FORCE & BENDING MOMENT
Shear force: - Sharing force at any point along a loaded beam is algebraic sum of all the vertical forces acting at any one side of the point.
A graphical representation of the variation of the shear force along the length of the beam is known as Shear Force diagram.
Sign convention: -
When the section move from left side to right side -
Upward forces are taken as (+ve)
Downward forces are taken as (-ve)
When the section move from right side to left side -
Downward forces are taken as (+ve)
Upward forces are taken as (-ve)
Fig.1 | Sign convention of Shear Force |
Share force causing anti-clockwise moment (-ve)
[Note – Sign convention may change, there is no effect on result]
Bending Moment: - Bending moment at any section or point of a beam is the algebraic sum of the moments about the section or point of all the vertical forces acting at any one side of the section or point.
Graphical representation of the variation of the BM along the length of the beam is known as bending moment diagram.
Sign convention: -
When the section move from left side to right side: -
Clockwise moments are taken as (+ve)
Anti-clockwise moments are taken as (-ve)
When the section move from right side to left side: -
Anti-clockwise moments are taken as (+ve)
Clockwise moments are taken as (-ve)
Fig.2 | Sign convention of Bending Moment |
Rate of change of shear force (or slope of the share force curve) is equal to intensity of loading.
i.e. dF/ dX = w
Rate of change of bending moment is equal to share force.
i.e. dM/ dX = F
Types of beam:
Cantilever: - One end fixed and another in free.
Fig.3 | Cantilever Beam |
Simply supported beam: - The ends of a beam are made to freely rest on a support.
Fig.4 | Simply Supported Beam |
Overhanging beam: - A beam having its end portion extended beyond the support.
Fig.5 | Overhanging Beam |
A beam may be overhanging on one side or both side.
Fixed beam: -The Beam is fixed at both ends.
Continuous beam: -A beam supported on more than two supports is known as continuous beam.
Fig.6 | Continuous Beam |
Concentrated or point load: - A load which is acting at a point of a beam is called Point Load
Fig.7 | Point Load |
Uniformly Distributed Load: - It is a distributed load in which load is spread in such a manner that intensity of load per unit length is constant throughout the length, up to which loading is spread.
Fig.8 | Uniformly Distributed Load |
Uniformly Varying Load: - It is a distributed load in which load is spread over a beam in such a manner that it varies uniformly on each unit length.
Fig.9 | Uniformly Varying Load |
- Bending moment is maximum, where share force is zero or changes the sign.
- Point of contraflexure or inflection- where the bending moment changes sign.
- In a simply supported beam the BM is zero where SF is maximum and vice versa.
- The variation of shear force between two sections is equal to the area under the load distribution diagram.
- The variation of bending moment between two sections is equal to the area under Shear Force diagram.
Shape of share force diagram and bending moment diagram in different loading condition: -
Load | Shear Force diagram | Bending moment diagram |
Point loads | Straight line | Inclined line |
UDL | Inclined line | Parabolic curve |
UVL | Parabolic curve | Cubic curve |
Shear Force and Bending Moment for beam:
Cantilever beam with a point load at its free end:
Cantilever beam with a point load at its free end |
SFx = W
SFmax = W
BMx = W.x
BMmax = W.L (At the fixed end)
Cantilever beam with uniformly distributed load over its entire length:
Fig.11 | Cantilever beam with UDL |
SFx = w.x
At x = 0; SFmin = 0
At x = L; SFmax = w.L
BMx = (w.x2)/2
At x = 0; BMmin = 0
At x = L; BMmax = (w.L2)/2
Cantilever beam with uniformly varying load over its entire length (Zero at its free end and maximum at its fixed end):
Fig. 12 | Cantilever with UVL |
SFx = ½.x.(w/L).x [1/2.base. height]
At x = 0; SFmin = 0
At x = L; SFmax = (w.L)/2
BMx = (w.x2/2L). (x/3) [SFx. Distance from CG]
At x = 0; BMmin = 0
At x = L; BMmax = (w.L2)/6
Simply supported beam with a point load at its mid-point: -
Fig.13 | Simply supported with point load |
Simply supported beam with uniformly distributed load over its entire length: -
Fig.14 | Simply supported with UDL |
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