Transcript ENGR-36_Lec-18_Fa12_Beams-1

Engineering 36
Chp 10:
Beams-1
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-18_Beams-1.pptx
Introduction
 In Previous Chapters We Examined
• Determining External Forces Acting
On A Structure
– Loads & Reactions
• Determining Forces Which Hold Together
The Various Members Of A Structure
– Trusses & Machines (at PIN Joints)
 Next, We Learn How to Determine The
Internal Forces Which Hold Together
The Various Parts Of A Given Member
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Bruce Mayer, PE
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Introduction (2)
 The Four Types of INTERNAL Forces
Present in Structural Members
• Tension or Compression
• Shear
• Bending
• Torsion/Twisting
 The Subsquent Analyses do Not
Consider Torsion Loads
• For More Torsion Info See ENGR45
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Introduction (3)

Examine in Detail Two Important
Types Of Engineering Structures:
1. BEAMS - usually long, straight, prismatic
(constant cross-section) members
designed to support loads applied at
various points along the member
2. CABLES - flexible members capable of
withstanding only tension, designed to
support concentrated or distributed loads
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Bruce Mayer, PE
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Internal Forces in Members
 (a) Straight twoforce member AB is
in equilibrium under
application of
F and −F.
 (b) Internal Forces
equivalent to F and
−F are required for
equilibrium of
free-bodies AC and
CB.
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Also
Called N
VIRTUAL
Section
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Internal Forces in Members (2)
 Multiforce member
ABCD is in equilibrium under the
application of cable
and member (pin)
contact forces.
FBD
 INTERNAL forces
equivalent to a FORCECOUPLE (F/V-M) system are
necessary for equilibrium of
free-bodies JD and ABCJ
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VIRTUAL
Section J
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Internal Forces in Members (3)
 An internal FORCE-COUPLE (F/V-M or N/V-M)
system is required for equilibrium of TWOFORCE members which are NOT STRAIGHT
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Bruce Mayer, PE
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Example: Beam in Frame

Given 3-Member
Structure at Left

Determine the
INTERNAL forces in
a) member ACF at
point J
b) member BCD at
point K

Note
  arctan4.8m 2  2.7m
  41.7 
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-18_Beams-1.pptx
Example: Beam in Frame
• Cut member ACF at J.
– The internal forces at J are
represented by equivalent
force-couple (F/V-M or N/VM) system which is
determined by considering
equilibrium of either cut part.
• Cut member BCD at K.
 Solution Plan:
• Compute Rcns and
Forces at
connections for
each member
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– Determine force-couple
(F/V-M or N/V-M) system
equivalent to internal forces
at K by applying equilibrium
conditions to either cut part.
Bruce Mayer, PE
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Example: Beam in Frame
 First Determine External Rcns
& Connection-Forces
 Consider the ENTIRE Frame
as a rigid Free Body
M
E
 0  2400N 3.6 m  F 4.8 m  0
 F  1800N
F
y
 0  2400N  1800N  E y  0
 E y  600N
F
x
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 0  Ex  0
Bruce Mayer, PE
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Example: Beam in Frame
 Consider Link Member BCD
as a Free Body
M
B
0
 2400N 3.6 m   C y 2.4 m   0
 C y  3600N
M
C
0
 2400N 1.2 m   By 2.4 m   0
 By  1200N
F
x
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0
 Bx C x  0
Don’t Know (yet)
Bruce Mayer, PE
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Example: Beam in Frame
 Consider ABE as a Free Body
M A  0:
 Fx  0 :
Bx 2.7 m  0
Bx  0
Bx  Ax  0
Ax  0
 Fy  0 :
 Ay  B y  600 N  0
Ay  1800 N
 Recall from Member BCD
1800 N
F
0
x
• But from Above Bx = 0
0
3600 N
C x  0
 ALL Forces on ACF are
now KNOWN
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 0  Bx C x  0
Bruce Mayer, PE
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Example: Beam in Frame
 Cut member ACF at J
• The internal forces at J are
represented by an equivalent
force-couple system
 Consider Free Body AJ
MJ  0:
 1800 N 1.2 m  M  0
 Fx  0 :
F  1800 N  cos 41.7  0
M  2160 N  m
F  N  1344N
 Fy  0 :
V  1800 N sin 41.7  0
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V  1197 N
Bruce Mayer, PE
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Example: Beam in Frame
 Cut member BCD at K
• Determine a force-couple system
equivalent to internal forces at K
 Consider Free Body BK
MK  0:
1200 N1.5 m  M  0
M  1800 N  m
 Fx  0 :
F N 0
 Fy  0 :
 1200 N  V  0
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V  1200 N
Bruce Mayer, PE
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3D Internal Forces
 Full 3D Loading:
• Axial Force, Ny
• Torsional Moment,
My
• TOTAL Shear
Magnitude
Vtot  Vx2  Vz2
 For Structural
Analysis (ENGR45)
need:
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• TOTAL Bending
Moment Magnitude
M tot  M x2  M z2
Bruce Mayer, PE
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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 & Material Best Suited To
Resist SHEARING-Forces and BENDING-Moments
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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
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Shear & Bending-Moment
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 Goal = determine bending
moment and shearing force at
any point, say C, in a beam
subjected to concentrated
and distributed loads
 Determine reactions at
supports by treating whole
beam as a 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
[email protected] • ENGR-36_Lec-18_Beams-1.pptx
2D V & M Sign Conventions
 Consider a TypicalCase (Gravity)
Loaded SimplySupported Beam
with the
X-Axis 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 takes
BOWL Shape
Bruce Mayer, PE
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2D V & M Sign Conventions (2)
 Positive Shear
• Right Member
Pushes DOWN on
Left Member
 Positive Bending
• Beam Concave UP
 POSITIVE Internal
Forces, V & M
• Note that at a Virtual
Section the V’s & M’s
MUST Balance
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Bruce Mayer, PE
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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
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LOAD Diagram
“Kinks” at LoadApplication Points
SHEAR Diagram
MOMENT Diagram
Bruce Mayer, PE
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Build V&M Diagram
a) Determine reactions at supports
b) Cut beam at C (dist x) and
consider left member AC
V   P 2 M  P 2x
c) Cut beam at E and consider
right member EB
V '   P 2 M '  P 2L  x
d) Plot V vs x
e) Plot M vs x
 Note: For a beam subjected to
CONCENTRATED LOADS,
shear is CONSTANT between
loading points and moment
Bruce Mayer, PE
Engineering-36: Engineering Mechanics - Statics varies LINEARLY
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Example: Torqued Beam
 Solution Plan
 For the Given Load
& Geometry, Draw
the shear and
bending moment
diagrams for the
beam AB
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• Taking entire beam
as free-body,
calculate reactions at
A and B.
• Determine
equivalent internal
force-couple systems
at sections cut within
segments AC, CD,
and DB
• Plot Results
Bruce Mayer, PE
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Example: Torqued Beam
 Taking entire beam as a
free-body, calculate
reactions at A and B.
M A  0:
By 32 in.  480 lb 6 in.  400 lb 22 in.  0
B y  365 lb
MB  0:
480 lb 26 in.  400 lb 10 in.  A32 in.  0
A  515 lb
 Fx  0 :
Bx  0
• Note that the 400 lb load at E may be REPLACED
by a 400 lb force and 1600 in-lb couple at D
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Bruce Mayer, PE
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Example: Torqued Beam
 Evaluate equivalent internal
force-couple systems at
sections cut within segments
AC, CD, and DB
• For AC use Cut-1
 Fy  0 :
515  40x  V  0
V  515 40x
 M1  0 :  515x  40x12 x  M  0
M  515x  20x 2
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-18_Beams-1.pptx
Example: Torqued Beam
 Evaluate equivalent internal
force-couple systems at
sections cut within segments
AC, CD, and DB
• For CD use Cut-2
 Fy  0 :
515  480  V  0
V  35 lb
 M 2  0 :  515x  480x  6  M  0
M  35 lb x  2880 lb  in
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-18_Beams-1.pptx
Example: Torqued Beam
 Evaluate equivalent internal
force-couple systems at
sections cut within segments
AC, CD, and DB
• For DB Use Cut-3
 Fy  0 :
M
3
 0:
515  480  400  V  0
V  365 lb
 515x  480 x  6  1600  400 x  18  M  0
M  11,680lb  in  365lbx
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-18_Beams-1.pptx
Example: Torqued Beam
 Plot Results
• From A to C:
V  515 40x
M  515x  20x 2
• From C to D
V  35 lb
M  35 x  2880
• From D to B
V  365 lb
M  11,680  365 x
– Note that over A-C The Bending-Moment Equation is
SECOND Order, and Concave DOWN
Bruce Mayer, PE
Engineering-36: Engineering Mechanics - Statics
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Axial Forces
 In Civil Engineering Most
Beams are loaded Transversely
Relative to the Beam Axis
• Most Beams do NOT have AXIAL Loads
 In ME however, many beam-like
structures (e.g., Shafts) have significant
axial loads that accompany the
transverse Shear-Force and
Bending-Moment
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Bruce Mayer, PE
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Axial Force Diagram
 When Significant
Axial Loads are
present in a Beam,
An AXIAL-Force
Diagram is added to
the typical V&M
Diagrams
 The “N” Diagram is
typically placed
ABOVE the
V-diagram
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Bruce Mayer, PE
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Axial-Force
Diagram
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Bruce Mayer, PE
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WhiteBoard Work
Let’s Work
These Nice
Problems
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Bruce Mayer, PE
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Engineering 36
Appendix
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
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Bruce Mayer, PE
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Bent Bar Prob
 Determine at Pt-C the
magnitudes of
• Axial Force
• Shear Force
• Bending Moment
• Torsional Moment
– The Moment on the
AC Axis
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Bruce Mayer, PE
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Beam Prob
 For the Beam with
Loading a shown:
• Draw the SHEAR
and BENDINGMOMENT
Diagrams
• Determine the Largest-Magnitude BendingMoment and its Location
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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