1. SFD &BMD
(POINT LOAD & UDL)
By: Mechanical Mania

2. Shearing Force (S.F.)
It is the internal resistance developed at any section to maintain free body equilibrium of either left or right part of the section.
The algebraic sum of the vertical forces on either side of the section of a loaded beam, is called shearing force (S.F)
Sign Convention
Shear force having an upward direction to the right hand side of section or downward direction to the left hand side of section will be taken positive and vice-versa.
It may be horizontal or vertical. Shear force at any section is algebraic sum of all transverse forces either from left or right of that section.

3. Bending Moment (B.M.)
Bending moment at any section is the internal reaction due to all the transverse force either from left side or from left side or from left side or from right side of that section.
The algebraic sum of the moments of the forces on either side of the section of a loaded beam, is called bending moment (EM.).
It is equal to algebraic sum of moments at that section either from left or from right side of that section. Bending moment is different from twisting moment.
Sign convention of Bending moment
A bending moment causing concavity upward will be taken as negative and called sagging bending moment. similarly a B.M. causing convexity upward will be taken as positive and called hogging B.M.

4. Relationship Between Bending Moment (M), shear force (S) and Loading Rate (W)
Rate of change of shear force is equal to load
dS/dX=W
Here, w = Load per unit length
Negative slope represents downward loading.
Rate of change of bending moment along the length of beam is equal to shear force.
dM/dX= S_X
At hinge, bending moment will be zero. Bending moment is maximum or minimum when shear force is zero or changes sign at a section. If degree of loading is curve = n then Degree of shear force curve = n + 1 And degree of bending moment curve = n+2 Point of contra-flexture/inflexion is that point where bending moment changes its sign.

5. SIMPLY SUPPORTED BEAM WITH POINT LOAD EXAMPLE
Draw shear force and bending moment diagram of simply supported beam carrying point load. As shown in figure below.

6. First find reactions of simply supported beam.
Both of the reactions will be equal. Since, beam is symmetrical. i.e.,
R1 = R2 = W/2 = 1000 kg.
Now find value of shear force at point A, B and C.
When simply supported beam is carrying point loads. Then find shear force value in sections. Shear force value will remain same up to point load. Value of shear force at point load changes and remain same until any other point load come into action.
Shear force between ( A – B ) = S.F (A-B) = 1000 kg
Shear force between (B – C) = S.F (B -C) = 1000 – 2000
S.F (B – C) = – 1000 kg.

7. Shear Force Diagram

8. Bending Moment
In case of simply supported beam, bending moment will be zero at supports. And it will be maximum where shear force is zero.
Bending moment at Point A and C = M(A) = M(C) = 0
Bending moment at point B = M(B) = R1 x Distance of R1 from point B.
Bending moment at point B = M (B) = 1000 x 2 = 2000 kg.m

10. SIMPLY SUPPORT BEAM WITH UDL & POINT LOAD EXAMPLE
Draw shear force and bending moment diagram of simply supported beam carrying uniform distributed load and point loads. As shown in figure.

11. First find reactions R1 and R2 of simply supported beam.
Reactions will be equal. Since, beam is symmetrical.
R1 = R2 = W/2 = ( x4)/2 = 1000kg
Hence, R1 = R2 = 1000 kg.
Shear Force
Shear force between  section A – B = S.F (A – B) = 1000 kg.
Shear force at right side of point B = S.F (B) = 1000 – 600
S. F (B) right = 400 kg.
Now shear force at left side of point C. Because of uniform distributed load, value of shear continuously varies from point B to C.
Shear force at point C (Left) = S.F (L) = 400 – (200×4)
Shear force at point C (Left) = S.F (L) = -400 kg
Shear force between section C – D = S.F (C-D) = -400 – 600
Shear force between section C – D = S.F (C-D) = kg.

13. From Shear force, one can see;
Shear force is maximum at point A and remain same until point load.
At point B shear force value decreases, because of point load.
From B to C shear force continuously decreases, because of udl.
At point C shear force gradually falls, because of point load.
From point C to D, shear force remain same, because no other point load is acting in this range.

14. Bending Moment B.M WILL BE ZERO AT SUPPORTS. I.E., M(A) = M(D) = 0
B.M at points B and C = M(B) = M(C) = 1000 x2 = 2000 kg.m
Now, how to find maximum bending moment?
Bending moment will be maximum at point, where shear force is zero.  Hence, bending moment will be maximum at mid point.
M (max) = 1000×4 – 600×2 -200×2(2/2)
M (max) = 2400 kg.m

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