Bending Moment and Shear Force Diagrams
dv = - w(x)
dx
Slope of the shear = - distributed load
Force diagram at
Intensity at each point
Each point
dM = V
dx
Slope of Moment
Diagram at each point
= Shear at each point
Different Slopes Description
Slope = Zero
Constant positive
slope
Constant negative
slope
Negative slope
increasing
Negative slope
decreasing
Positive slope
decreasing
Positive slope
increasing
dv = - w(x)
dM = V
dx
dx
•The slope of the shear force
diagram, at each point, is
equal to the negative of the
intensity of distributed
loading.
•Distributed loading is
positive (w is positive when
it act downward) and
increases from zero to WB.
Thus the slope of the shear
force diagram will be
negative and increases from
0 to –WB.
dv = - w(x)
dM = V
dx
dx
•The slope of the Moment
Diagram at each point is
equal to the shear force (V).
•The slope of the moment
diagram start with a value
of VA then decreases to
zero (tangent become
horizontal) and then
become negative (i.e
shear force is negative) to
a value of –VB.
Regions of concentrated Force and Moment
•When a concentrated force acts downward so the shear force
diagram will jump downward at that particular point.
•When a concentrated moment Mo is applied clockwise, the moment
diagram will jump upward. When Mo acts counterclockwise, the
moment diagram will jump downward.
dM= V dX
∆MAB = MB - MA= VLAB = 4.8 kN (6 m) = 28.8 KN.m
Since MA = 0 thus: MB=28.8 KN.m
MAB is a straight line of slope 4.8
∆MBC = MC - MB= VLBC = -3.2 (2) = -6.4 KN.m
MC = 28.8 – 6.4 = 22.4 KN.m
∆MCD = MD - MC= VLBC= -11.2 (2) = -22.4 KN.m
MD = 22.4-22.4 = 0
•There is no distributed loading (w=0)
•Slope of the shear force at any point (dv /dx)= w=0
•Thus the shear force has to be a horizontal strain
line (slope = 0)
•V=p (shear force is positive since it cause a
clockwise rotation, check sign notations)
(dM/dx)=V=P
Thus the moment diagram has a constant slope of value P,
thus it has to be a straight line with a positive slope
At x=0, M= -PL
x=L , M=0