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UNIT-II
Shear Force Diagrams
and
Bending Moment Diagrams
Lecture Number -6
Prof. M. J. Naidu
Mechanical Engineering Department
Smt. Kashibai Navale College of Engineering, Pune-41
Strength of Materials
Reverse Cases
• Points to remember to draw…
•
1.
2.
3.
4.
SFD from BMD…
SFD is zero where BMD is horizontal straight line.
If BMD is inclined straight line then, V= Slope of BMD.
If BMD is a parabola, then consider M= ax2 +bx+c.
Find values of a,b,c from values of M or from dM/dx for diff.
points
5. SFD is inclined Straight line in this region.
6. Vertical step in BMD implies presence of external couple.
Strength of Materials
Reverse Cases…
• Load diagram from SFD…
1. Straight horizontal line – No load region.
2. Inclined straight line in SFD
rate of loading =
Slope of SFD
UDL on load diagram
3. Parabola in SFD
consider V=ax2+bx+c
Get a, b, c from V or from dV/dx at different points
UVL in load diagram region.
4. Vertical step in SFD
Point load on load diagram
Strength of Materials
Reverse Cases…
Illustrative Problem 3:
In the following problem, draw bending moment and load
diagrams corresponding to the given shear diagram. Specify
values at all change of load positions and at all points of zero
shear.
Strength of Materials
Reverse Cases…
• To draw the Load Diagram
• Upward concentrated load at A is 10 kN.
• The shear in AB is a 2nd-degree curve, thus the load in AB is
uniformly varying.
• In this case, it is zero at A to 2(10 + 2)/3 = 8 kN at B. No load in
segment BC.
• A downward point force is acting at C in a magnitude of 8 - 2 =
6 kN.
• The shear in DE is uniformly increasing, thus the load in DE is
uniformly distributed and upward.
• This load is spread over DE at a magnitude of 8/2 = 4 kN/m.
Strength of Materials
Reverse Cases…
• To draw the Bending Moment Diagram
• To find the location of zero shear, F:
x2/10 = 32/(10 + 2)
x = 2.74 m
• MA = 0
• MF = MA + Area in shear diagram
MF = 0 + 2/3 (2.74)(10) = 18.26 kN·m
• MB = MF + Area in shear diagram
MB = 18.26 - [1/3 (10 + 2)(3) - 1/3 (2.74)(10) - 10(3 - 2.74)]
MB = 18 kN·m
Strength of Materials
Reverse Cases…
• MC = MB + Area in shear diagram
MC = 18 - 2(1) = 16 kN·m
• MD = MC + Area in shear diagram
MD = 16 - 8(1) = 8 kN·m
• ME = MD + Area in shear diagram
ME = 8 - ½ (2)(8) = 0
• The moment diagram in AB is a second degree curve, at BC
and CD are linear and downward. For segment DE, the
moment diagram is parabola open upward with vertex at E.
Strength of Materials
Reverse Cases…
Strength of Materials
Reverse Cases…
Workout numerical 1:
In the following problem, draw bending moment and load
diagrams corresponding to the given shear diagram. Specify
values at all change of load positions and at all points of
zero shear. [ consider the values in SI units directly without
converting ]
Strength of Materials