6. Internal shear force causes a clockwise rotation of the beam
segment; and the internal moment causes compression in the top
fibers of the segment.
7. SIGN CONVENTIONS
• A force that tends to bend the beam
downward is said to produce a
positive bending moment. A force
that tends to shear the left portion of
Positive
Bending
Negative
Bending
the beam upward with respect to the
right portion is said to produce a
positive shearing force.
Positive
Shear
Negative
Shear
8. PROCEDURE
1.
Draw the free-body-diagram of the beam with sufficient room
under it for the shear and moment diagrams
(if needed, solve for support reactions first).
1.
Draw the shear diagram under the free-body-diagram.
•
The change in shear ∆V equals the negative area under the distributed
loading.
•
Label all the loads on the shear diagram
∆V = − ∫ w( x )dx
9. PROCEDURE
3. Draw the moment diagram below the shear diagram.
• The shear load is the slope of the moment and point moments result in
jumps in the moment diagram.
• The area under the shear diagram equals the change in moment over the
segment considered (up to any jumps due to point moments).
• Label the value of the moment at all important points on the moment
diagram.
∆M = ∫ V ( x )dx
10. Relations between Distributed
Load, Shear and Moment
Distributed Load
Slope of the shear
diagram
Slope of the
shear diagram
dV
= − w(x)
dx
dM
=V
dx
Negative of distributed
load intensity
Shear moment diagram
Change in shear
∆M BC = ∫ Vdx
Area under shear
diagram
Change in moment
∆VBC = − ∫ w( x)dx
Area under shear
diagram
11. Procedure for Analysis
Support Reactions
•
•
Find all reactive forces and couple moments acting on the beam
Resolve them into components
Shear and Moment Reactions
•
•
•
•
Specify separate coordinates x
Section the beam perpendicular to its axis
V obtained by summing the forces perpendicular to the beam
M obtained by summing moments about the sectioned end
Shear and Moment Reactions
•
•
Plot (V versus x) and (M versus x)
Convenient to plot the shear and the bending moment diagrams below the FBD
of the beam
15. Rules of Thumb/Review
• Moment is dependent upon the shear diagram
the area under the shear diagram = change in the moment
(i.e. A shear 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
18. Concentrated Load
•
Find reactions
•
Cut through beam to the left of the
load P (a distance x from the left
end), FBD
– Equilibrium yields V and M for
the left side of the beam
•
Cut through the beam to the right of
P, FBD
– Equilibrium yields V and M for
the right side of the beam.
19. Concentrated Load Moment
Diagram
Bending Moment Diagram
•
The bending moment in the left
side increases linearly from zero
at the support to P(ab/L) at the
concentrated load x=a
•
In the right side, the bending
moment is again a linear function
of x, varying from P(ab/L) at x=a
to zero at the support x=L.
•
The maximum bending moment is
therefore P(ab/L), which occurs at
the concentrated load.
In this example a=b=L/2
20. a
b
A
R Ay =
y
P
Example 1 :
C
P ⋅b
( a + b)
R By =
x
P
a
∑M
z
Mxz
A
P ⋅b
( a + b)
Qxy
=0 ;
+ Mxz + P ( x − a )
∴ Mxz
Where
Pb
( x) = 0
−
( a + b)
Pb
( x ) − P ( x − a)
=
( a + b)
( x − a)
x
B
can only be +VE or ZERO.
P⋅a
( a + b)
21. y
P
a
b
A
Pb
( a + b)
x
B
Pa
( a + b)
x
(i) When x ≤ a :
(ii) When x > a :
BMD:
C
Mxz =
Mxz
Mxz
Pab
( a + b)
Pb
( x ) − P ( x − a)
( a + b)
0
Pb
( x ) − P ( x − a)
=
( a + b)
Eq. 1
2
Eq. 2
+ve
0 A
C
1
B
23. Q xy = P; Mxz = −P( L − x )
P
L
y
P.L
Mxz
A
B
x
Qxy
x
P
P
Qxy
Mxz
+ve
Shear Force
Diagram (SFD)
0
0
-P.L
-ve
Bending
Moment
Diagram (BMD)
24. V & M Diagrams
8 kips
12 ft
P = 20 kips
12 kips
20 ft
8 kips
V
x
(kips)
What is the area of the
blue rectangle?
96 ft-kips
M
(ft-kips)
-12 kips
96 ft-kips
b
What is the area of
the green rectangle?
-96 ft-kips
a
c
x
25. BMD for simply supported beam
with UDL:
Parabolic, max moment at mid span of value WL2/8 ,
where w is the distributed load and L the length of the beam.
26.
27. UNIFORM LOAD
•
The beam and its loading is
symmetric, the reactions are equal to
wL/2
•
The slope of the shear diagram at each
point equals the negative distributed
load intensity at each point
dV
= −w( x )
dx
28. UNIFORM LOAD
•
Therefore, the shear force and bending
moment at a distance x from the left end
are:
dV
dM
=− ( x ),
w
=V
dx
dx
•
These equations are valid through the
length of the beam and can be plotted as
shear and bending moment diagrams.
The maximum value at the midpoint
where
dM
= V = 0.
dx
Mmax= wL2/8
29. Distributed Load w
per unit length
Example : Distributed Load
y
x
Mxz=
wL2
A
2
B
L
x
RAy=wL
wL2
Mxz
wx
2
Qxy
wL
Mxz
Qxy
∑F = 0 ;
∑M = 0 ;
y
z
wL − wx − Q xy = 0
⇒ Q xy = w ( L − x )
wL2
x
− wL ( x ) + wx = 0
Mxz +
2
2
30. ⇒ Q xy = w ( L − x )
⇒ Mxz
wx 2 wL2
= wLx −
−
2
2
Mxz
BMD:
L
0
-ve
x
-wL2
2
wL2
=−
2
@ x = 0;
Mxz
@ x = L;
Mxz = 0
L
@x = ;
2
Mxz
wL2
=−
8
31.
32.
33. Draw Some Conclusions
• The magnitude of the shear at a point equals the slope of the
moment diagram at that point.
• The area under the shear diagram between two points equals the
change in moments between those two points.
• At points where the shear is zero, the moment is a local maximum
or minimum.
