Engineering 36
Chp 7:
Beams-2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Beam – What is it?
 Beam  Structural member
designed to support loads applied
at various points along its length
 Beams can be subjected to
CONCENTRATED loads or
DISTRIBUTED loads or a
COMBINATION of both.
 Beam Design is 2-Step Process
1. Determine Axial/Shearing Forces and Bending Moments
Produced By Applied Loads
2. Select Structural Cross-section and Material Best Suited To
Resist the Applied Forces and Bending-Moments
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Beam Loading and Supports
 Beams are classified according to the Support
Method(s); e.g., Simply-Supported, Cantilever
 Reactions at beam supports are Determinate
if they involve exactly THREE unknowns.
• Otherwise, they are Statically INdeterminate
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Shear & Bending-Moment
Engineering-36: Engineering Mechanics - Statics
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 Goal = determine bending
moment and shearing force at
any point in a beam subjected
to concentrated and
distributed loads
 Determine reactions at
supports by treating whole
beam as free-body.
 Cut beam at C and draw freebody diagrams for AC and CB
exposing V-M System
 From equilibrium
considerations, determine
M & V or M’ & V’.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
V & M Sign Conventions
 Consider a
Conventionally
(Gravity) Loaded
Simply-Supported
Beam with the XAxis Origin
Conventionally
Located at the LEFT
C
P
x
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 Next Consider a
Virtual Section
Located at C
 DEFINE this Case
as POSITIVE
• Shear, V
– The Virtual Member
LEFT of the Cut is
pushed DOWN by the
Right Virtual Member
• Moment, M
– The Beam BOWS
UPwardBruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
V & M Sign Conventions (2)
 Positive Shear
• Right Member
Pushes DOWN on
Left Member
 Positive Bending
• Beam Bows UPward
 POSITIVE Internal
Forces, V & M
• Note that at a Virtual
Section the V’s & M’s
MUST Balance
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
V & M Diagrams
 With the Signs of
V&M Defined we
Can now Determine
the MAGNITUDE
and SENSE for V&M
at ANY arbitrary
Virtual-Cut Location
 PLOTTING V&M vs.
x Yields the Stacked
Load-Shear-Moment
(LVM) Diagram
Engineering-36: Engineering Mechanics - Statics
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LOAD Diagram
“Kinks” at LoadApplication Points
SHEAR Diagram
MOMENT Diagram
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Relations Between Load and V
 On Element of Length Δx
from C to C’; ΣFy = 0
V  V  V   wx  0
dV
V
 lim
 w
dx x 0 x
 Separating the Variables and
Integrating from Arbitrary
Points C & D
VD
 dV  V
D
VC
xD
 VC    w x  dx 
xC
 area under LOAD curve 
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Relations Between Ld and M
 Now on C-C’ take ΣMC’ = 0
M  M   M  Vx  wx x  0

2

dM
M
 lim
 lim V  12 wx  V
dx x 0 x x 0
0
 Separating the Variables and
Integrating from Arbitrary
Points C & D
xD
M D  M C   V x  dx  area under SHEAR curve 
xC
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Summary: Load, V, M Relations
 The 1st Derivative of
V is the Negative of
the Load
dV
dx
  w x0 
x  x0
 The 1st Derivative of
M is the Shear
dM
dx
 V  x0 
x  x0
 The Shear is the
Negative of the Area
under the Ld-Curve
 The Moment is the
Area under the
V-Curve
V    wx dx  C
M   V x dx  C
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Recall: Derivative = SLOPE
 The SLOPE of the
V-Curve is the
Negative-VALUE of
the Load Curve
dV
dx
 mV  x0    w x0 
x  x0
• Note that w is a
POSITIVE scalar;
i.e.; it is a Magnitude
 The SLOPE of the
M-Curve is the
Positive-VALUE of
the Shear Curve
dM
dx
 mM  x0   V  x0 
x  x0
y
x
dy dx  m
dy dx  y x
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Calculus Summary
 The SLOPE of the V-curve is the
negative MAGNITUDE of the w-Curve
 The SLOPE of M-Curve is the VALUE
of the V-Curve
 The CHANGE in V between Pts a&b is
the DEFINATE INTEGRAL between Pts
a&b of the w-Curve
 The CHANGE in M between Pts a&b is
the DEFINATE INTEGRAL between Pts
a&b of the V-Curve

Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Calculus Summary
 When w is down
• The negative VALUE
of the w-Curve is the
SLOPE of the V-curve
• The Negative AREA Under
w-Curve is the CHANGE in
the V-Value
• The VALUE of the V-Curve
is the SLOPE of M-Curve
• The AREA under the VCurve is the CHANGE in
the M-Value
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Example  V&M by Calculus
 Solution Plan
 For the Given Load
& Geometry, Draw
the shear and
bending moment
diagrams for the
beam AE
Engineering-36: Engineering Mechanics - Statics
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• Taking entire beam
as free-body,
calculate reactions at
Support A and D.
• Between
concentrated load
application points,
dV/dx = −w = 0, and
so the SLOPE is
ZERO, and Thus
Shear is Constant
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Example  V&M by Calculus
• Between concentrated
load application points,
dM/dx = mM = V = const.
• The CHANGE IN
MOMENT between load
application points is
equal to AREA UNDER
 Solution Plan (cont.)
SHEAR CURVE between
• With UNIFORM
Load-App points
loading between D
• With a LINEAR shear
and E, the shear
variation between D and
variation is LINEAR
E, the bending moment
diagram is a PARABOLA
– mV = −1.5 kip/ft
(i.e., 2nd degree in x).
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Example  V&M by Calculus
 Taking entire beam as a
free-body, determine
reactions at supports
MA  0:
D24 ft   20 kips 6 ft   12 kips 14 ft 
 12 kips 28 ft   0
 Fy 0 :
D  26 kips
Ay  20 kips  12 kips  26 kips  12 kips  0
F
x
0  Ax  0
Engineering-36: Engineering Mechanics - Statics
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Ay  18 kips
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Example  V&M by Calculus
 The VERTICAL Reactions
Ay  18 kips
D  26 kips
 Between concentrated load
application points, dV/dx = 0,
and thus shear is Constant
 With uniform loading
between D and E, the shear
variation is LINEAR.
• SLOPE is constant at −w
(−1.5 kip/ft in this case)
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Example  V&M by Calculus
6 ft
8 ft
10 ft
8 ft
 Between concentrated load
application points, dM/dx = V
= Const. And the change in
moment between load
application points is equal to
AREA under the SHEAR
CURVE between points.
M B  M A  108 M B  108 kip  ft
M C  M B  16
M C  92 kip  ft
M D  M C  140 M D  48 kip  ft
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Example  V&M by Calculus
6 ft
8 ft
10 ft
8 ft
 With a Linear Shear variation
between D and E, the
bending moment diagram is
PARABOLIC.
M D  48 kip  ft
M E  M D  48 M E  0
 Note that the FREE End of a
Cantilever Beam Cannot
Support ANY Shear or
Bending-Moment
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
WhiteBoard Work
Let’s Work
This Problem
w/ Calculus
& MATLAB
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Engineering 36
Appendix
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-19_Beams-2.pptx