types of beam
TRANSCRIPT
BendingBending
Shear and Moment Diagram, Graphical method to construct shear
and moment diagram, Bending deformation of a straight member, The flexure formula
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2
Shear and moment diagramShear and moment diagram
3
Axial load diagram
Torque diagram
Both of these diagrams show the internal forces acting on the members. Similarly, the shear and moment diagrams show the internal shear and moment acting on the members
Type of BeamsType of Beams
Statically Determinate
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Simply Supported Beam
Overhanging Beam
Cantilever Beam
Type of BeamsType of Beams
Statically Indeterminate
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Continuous Beam
Propped Cantilever Beam
Fixed Beam
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Example 1Example 1
Equilibrium equation for 0 ≤ x ≤ 3m:
A B
* internal V and M should be assumed +ve
kNV
VF
Fy
9
0
0
−==−−
=∑
)(9
0
0
kNmxM
MVx
M
−==+
=∑
M
VF
x
Shear DiagramShear Diagram
Lecture 1 8
Sign convention: V= -9kN
M
VF
x
Shear DiagramShear Diagram
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Sign convention: M= -9x kNmX=0: M= 0X=3: M=-27kNm
M=-9x
M = -9x kN.m
V = -9 kNF
x
10
V=9kN
M=X
At cross section A-A
X
At section A-A
Example 2Example 2
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kN5
01050F
kN5
0)2()1(10;0
y
=
=−+=Σ
=
=−=Σ
y
y
y
yA
A
A
C
CM
1) Find all the external forces
12
)(5
0
0
10
downkNV
VF
F
mx
y
==−
=<≤
∑
)(5
0
0
10
ccwkNmVxM
VxM
M
mx
A
===−
=<≤
∑
)(5
010
0
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upkNV
VF
F
mx
y
==+−
=≤≤
∑
)()510(
10
0)1(10
0
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ccwkNmxM
VxM
MVx
M
mx
A
−=−=
=+−
=≤≤
∑
Force equilibrium
Force equilibrium
Moment equilibrium
Moment equilibrium
)(5
)(5
10
ccwxM
downkNV
mx
==
<≤
)(510
)(5
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ccwxM
upkNV
mx
−==
<≤
M=5xM=10-5x
Boundary cond for V and M
Solve itSolve it
Draw the shear and moment diagrams for simply supported beam.
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Distributed LoadDistributed Load
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For calculation purposes, distributed load can be represented as a single load acting on the center point of the distributed area.
Total force = area of distributed load (W : height and L: length)Point of action: center point of the area
ExampleExample
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ExampleExample
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Solve itSolve it
Draw the shear and moment diagrams the beam:
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Solving all the external loads
kN
WlF
48)6(8 ===
Distributed load will be
Solving the FBD
012
364
)3(48
0)3(4
0
==
==
=−
=∑
xy
y
x
A
AkNA
kNB
FB
M
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Boundary Condition 40 <≤ x
xV
Vx
FY
812
0812
0
−==−−
=∑
2
2
2
412
0412
04)812(
04
0
xxM
xxM
xxxM
xVxM
M A
−==+−
=−−−=−−
=∑
Equilibrium eq
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Boundary Condition 64 <≤ x
xV
Vx
FY
848
083612
0
−==−−+
=∑
144484
0414448
04)4(36)848(
0)2/(8
0
2
2
2
−+−=
=++−=−+−−
=−−
=∑
xxM
xxM
xxxM
xxVxM
M A
Equilibrium eq
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40 <≤ x
x=0 V= 12 kNx=4 V=-20 kN
xV 812−=
2412 xxM −=
x=0 M= 0 kNx=4 V=-16 kN
64 <≤ xxV 848−=
x=4 V= 16kNx=6 V= 0 kN
144484 2 −+−= xxM
x=4 V= -16kNx=6 V= 0 kN
Graph based on equationsGraph based on equations
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y = c Straight horizontal line
y = mx + cy=3x + 3 y=-3x + 3
y=3x2 + 3y=-3x2 + 3
y = ax2 + bx +c
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Graphical methodGraphical method
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( )( ) xxwV
VVxxwV
Fy
∆−=∆=∆+−∆−
=Σ↑+
0)(
:0
• Relationship between load and shear:
( ) ( )[ ] ( )( ) ( ) 2
0
:0
xkxwxVM
MMxkxxwMxV
M o
∆−∆=∆
=∆++∆∆+−∆−=Σ
• Relationship between shear and bending moment:
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Dividing by ∆x and taking the limit as ∆x0, the above two equations become:
Regions of distributed load:
Slope of shear diagram at each point
Slope of moment diagram at each point
= − distributed load intensity at each point
= shear at each point
)(xwdx
dV −= Vdx
dM =
ExampleExample
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∫=∆ dxxwV )( ∫=∆ dxxVM )(
The previous equations become:
change in shear =
Area under distributed load
change in moment =
Area under shear diagram
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+ve area under shear diagram
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Bending deformation of a straight Bending deformation of a straight membermember
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Observation: - bottom line : longer - top line: shorter- Middle line: remain the same but rotate (neutral line)
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Strains
sss ∆
∆−∆=>−∆
'lim
0ε
Before deformation
xs ∆=∆
After deformation, ∆x has a radius of curvature ρ, with center of curvature at point O’ θρ∆=∆=∆ xs
Similarly θρ ∆−=∆ )(' ys
ρε
θρθρθρε
y
ys
−=
∆∆−∆−=
>−∆
)(lim
0
Therefore
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ρε c−=maxMaximum strain will be
max
max
)(
)/
/(
εε
ρρ
εε
c
y
c
y
−=
−=
max)( σσc
y−=
-ve: compressive state+ve: tension
The Flexure FormulaThe Flexure Formula
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The location of neutral axis is when the resultant force of the tension and compression is equal to zero.
∑ == 0FFR
NotingdAdF σ=
∫
∫
∫∫
−=
−=
==
A
A
A
ydAc
dAc
y
dAdF
max
max)(
0
σ
σ
σ
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Since , therefore
Therefore, the neutral axis should be the centroidal axis
0max ≠c
σ 0=∫A
ydA
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∑= ZZR MM )(
∫
∫
∫ ∫
=
=
==
A
A
A A
dAyc
dAc
yy
dAyydFM
2max
max)(
σ
σ
σ I
Mc=maxσ
Maximum normal stress
Normal stress at y distance
I
My=σ
Line NA: neutral axis Red Line: max normal stress c = 60 mmYellow Line: max compressive stress c = 60mm
I
Mc=maxσ
I
Mc−=maxσ
Line NA: neutral axis Red Line: Compressive stress y1 = 30 mmYellow Line: Normal stress y2 = 50mm
I
My11 −=σ
I
My22 =σ
Refer to Example 6.11 pp 289
I: moment of inertial of the cross I: moment of inertial of the cross sectional areasectional area
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3bhI xx =−
644
44 DrI xx
ππ ==−
Find the stresses at A and B
I: moment of inertial of the cross I: moment of inertial of the cross sectional areasectional area
Locate the centroid (coincide with neutral axis)
mm
AA
AyAy
A
Ayy n
ii
n
iii
5.237
)300)(50()300)(50(
)300)(50(325)300)(50(15021
2211
1
1
=++=
++=
=∑
∑
=
=
I: moment of inertial of the cross I: moment of inertial of the cross sectional areasectional area
Profile I
4633
)10(5.11212
)300(50
12mm
bhI I ===
A A
46
2623
)10(344.227
)5.87)(300)(50()10(5.11212
)(
mm
Adbh
I AAI
=
+=+=−
I about Centroidal axis
I about Axis A-A using parallel axis theorem
Profile II
46
23
23
)10(969.117
)5.87)(50)(300(12
)50)(300(
12)(
mm
Adbh
I AAII
=
+=+=−
Total I46
466
)10(313.345
)10(969.117)10(344.227
)()(
mm
mm
III AAIIAAIAA
=+=
+= −−−
* Example 6-12 to 6-14 (pp 290-292)
Solve itSolve it
If the moment acting on the cross section of the beam is M = 6 kNm, determine the maximum bending stress on in the beam. Sketch a three dimensional of the stress distribution acting over the cross sectionIf M = 6 kNm, determine the resultant force the bending stress produces on the top board A of the beam
46
32
3
)10(8.786
12
)300(40])170)(40)(300(
12
)40(300[2
mm
I I
=
++=
Total Moment of Inertia
Max Bending Stress at the top and bottom
MPaII
McM top 45.1
)190()10(6000 3
−=−=−=
MPaM bottom 45.1=
Bottom of the flange
MPaII
McM topf 14.1
)150()10(6000 3
_ −=−=−=
MPaM bottomf 14.1_ =
1.45MPa
1.14MPa
6kNm
Resultant F = volume of the trapezoid
1.45MPa
1.14MPa
40 mm
300 mm
kN
NFR
54.15
15540)300)(40(2
)14.145.1(
=
=+=
Solve itSolve it
The shaft is supported by a smooth thrust load at A and smooth journal bearing at D. If the shaft has the cross section shown, determine the absolute maximum bending stress on the shaft
Draw the shear and moment diagram
kNF
kNF
F
M
A
D
D
A
3
3
)25.2(3)75.0(3)3(
0
==
+=
=∑
External Forces
Absolute Bending Stress
Mmax = 2.25kNm
MPa
I
Mc
8.52
)2540(4
)40()10(2250
44
3
max
=
−== πσ