internal forces 7 engineering mechanics: statics in si units, 12e copyright © 2010 pearson...
TRANSCRIPT
Copyright © 2010 Pearson Education South Asia Pte Ltd
Internal Forces77
Engineering Mechanics: Statics in SI Units, 12e
Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Objectives
• Method of sections for determining the internal loadings in a member
• Develop procedure by formulating equations that describe the internal shear and moment throughout a member
• Analyze the forces and study the geometry of cables supporting a load
Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Outline
1. Internal Forces Developed in Structural Members
2. Shear and Moment Equations and Diagrams
3. Relations between Distributed Load, Shear and Moment
4. Cables
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.1 Internal Forces Developed in Structural Members
• The design of any structural or mechanical member requires the material to be used to be able to resist the loading acting on the member
• These internal loadings can be determined by the method of sections
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.1 Internal Forces Developed in Structural Members
• Force component N, acting normal to the beam at the cut session
• V, acting tangent to the session are normal or axial force and the shear force
• Couple moment M is referred as the bending moment
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.1 Internal Forces Developed in Structural Members
• For 3D, a general internal force and couple moment resultant will act at the section
• Ny is the normal force, and Vx and Vz are the shear components
• My is the torisonal or twisting moment, and Mx and Mz are the bending moment components
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.1 Internal Forces Developed in Structural Members
Procedure for AnalysisSupport Reactions• Before cut, determine the member’s support reactions• Equilibrium equations used to solve internal loadings
during sectioning
Free-Body Diagrams• Keep all distributed loadings, couple moments and
forces acting on the member in their exact locations• After session draw FBD of the segment having the
least loads
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.1 Internal Forces Developed in Structural Members
Procedure for AnalysisFree-Body Diagrams (Continue)• Indicate the z, y, z components of the force, couple
moments and resultant couple moments on FBD• Only N, V and M act at the section• Determine the sense by inspection
Equations of Equilibrium• Moments should be summed at the section • If negative result, the sense is opposite
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 7.3
Determine the internal force, shear force and the bending moment acting at point B of the two-member frame.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Support ReactionsFBD of each memberMember AC
∑ MA = 0;-400kN(4m) + (3/5)FDC(8m)= 0
FDC = 333.3kN
+→∑ Fx = 0;-Ax + (4/5)(333.3kN) = 0
Ax = 266.7kN
+↑∑ Fy = 0;Ay – 400kN + 3/5(333.3kN) = 0
Ay = 200kN
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Support Reactions
Member AB
+→∑ Fx = 0; NB – 266.7kN = 0
NB = 266.7kN
+↑∑ Fy = 0; 200kN – 200kN - VB = 0
VB = 0
∑ MB = 0; MB – 200kN(4m) – 200kN(2m) = 0
MB = 400kN.m
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.2 Shear and Moment Equations and Diagrams
• Beams – structural members designed to support loadings perpendicular to their axes
• A simply supported beam is pinned at one end and roller supported at the other
• A cantilevered beam is fixed at one end and free at the other
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.2 Shear and Moment Equations and Diagrams
Procedure for AnalysisSupport 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
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.2 Shear and Moment Equations and Diagrams
Procedure for AnalysisShear and Moment Reactions (Continue)• Plot (V versus x) and (M versus x)• Convenient to plot the shear and the bending moment
diagrams below the FBD of the beam
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 7.7
Draw the shear and bending moments diagrams for the shaft. The support at A is a thrust bearing and the support at C is a journal bearing.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Support Reactions
FBD of the shaft
mxkNMM
kNVFy
.5.2;0
5.2;0
mkNxM
xkNmxkNMM
kNV
VkNkNFy
.)5.210(
0)(5.2)2(5 ;0
5.2
055.2 ;0
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Shear diagram
Internal shear force is always positive within the shaft AB.
Just to the right of B, the shear force changes sign and remains at constant value for segment BC.
Moment diagram
Starts at zero, increases linearly to B and therefore decreases to zero.
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.3 Relations between Distributed Load, Shear and Moment
Distributed Load• Consider beam AD subjected to an arbitrary load
w = w(x) and a series of concentrated forces and moments
• Distributed load assumed positive when loading acts downwards
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.3 Relations between Distributed Load, Shear and Moment
Distributed Load• A FBD diagram for a small segment of the beam
having a length ∆x is chosen at point x along the beam which is not subjected to a concentrated force or couple moment
• Any results obtained will not apply at points of concentrated loadings
• The internal shear force and bending moments assumed positive sense
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.3 Relations between Distributed Load, Shear and Moment
Distributed Load• Distributed loading has been replaced by a resultant
force ∆F = w(x) ∆x that acts at a fractional distance k (∆x) from the right end, where 0 < k <1
2)()(
0)()(;0
)(
0)()(;0
xkxwxVM
MMxkxxwMxVM
xxwV
VVxxwVFy
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.3 Relations between Distributed Load, Shear and Moment
Distributed Load
Vdx
dM
)(xwdx
dVSlope of the
shear diagramNegative of distributed
load intensity
Slope of shear diagram
Shear moment diagram
VdxM BC
dxxwVBC )(
Change in shearArea under
shear diagram
Change in moment Area under shear diagram
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.3 Relations between Distributed Load, Shear and Moment
Force and Couple Moment• FBD of a small segment of the beam• Change in shear is negative• FBD of a small segment of the beam located at the
couple moment• Change in moment is positive
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 7.9
Draw the shear and moment diagrams for the overhang beam.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
The support reactions are shown.
Shear Diagram Shear of –2 kN at end A of the beam is at x = 0.
Positive jump of 10 kN at x = 4 m due to the force.
Moment Diagram mkN 8420
04
MMM
xx
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.4 Cables
• Cables and chains used to support and transmit loads from one member to another
• In force analysis, weight of cables is neglected • Assume cable is perfectly flexible and inextensible• Due to its flexibility cables has no resistance to bending• Length remains constant before and after loading
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.4 Cables
Cable Subjected to Concentrated Loads
• For a cable of negligible weight, it will subject to constant tensile force
• Known: h, L1, L2, L3 and loads P1 and P2
• Form 2 equations of equilibrium • Use Pythagorean Theorem to relate the three
segmental lengths
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 7.11
Determine the tension in each segment of the cable.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 7.11
FBD for the entire cable.
kNE
EkNkNkNkN
F
kNA
mknmkNmkNmA
M
EA
F
y
y
y
y
y
E
xx
x
10
0315412
;0
12
0)2(3)10(15)15(4)18(
;0
0
;0
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 7.11
Consider leftmost section which cuts cable BC since sag yC = 12m.
kNT
TkNkN
F
kNT
F
kNEA
mkNmkNmA
M
BCBC
BCBC
y
BCBC
x
xx
x
C
2.10,6.51
0sin412
;0
033.6cos
;0
33.6
0)5(4)8(12)12(
;0
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.4 Cables
Cable Subjected to a Distributed Load• Consider weightless cable subjected to a loading
function w = w(x) measured in the x direction
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.4 Cables
Cable Subjected to a Distributed Load• For FBD of the cable having length ∆• Since the tensile force changes continuously, it is
denoted on the FBD by ∆T• Distributed load is represented by second integration,
dxdxxwF
yH
)(1
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 7.12
The cable of a suspension bridge supports half of the uniform road surface between the two columns at A and B. If this distributed loading wo, determine the maximum force developed in the cable and the cable’s required length. The span length L and, sag h are known.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Note w(x) = wo
Perform two integrations
Boundary Conditions at x = 0
Therefore,
Curve becomes
This is the equation of a parabola
Boundary Condition at x = L/2
dxdxwF
y oH
1
21
2
2
1CxC
xw
Fy o
H
0/,0,0 dxdyxy
021 CC
2
2x
F
wy
H
o
hy
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
For constant,
Tension, T = FH/cosθ
Slope at point B
Therefore
Using triangular relationship
22
2 4 and
8x
L
hy
h
LwF oH
2
4 222
max
LwFT oH
)cos( maxmax
HFT
H
o
LxH
o
Lx F
Lw
F
w
dx
dy
2tantan 1
max
2/
max2/
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
For a differential segment of cable length ds,
Determine total length by integrating,
Integrating yields,
dxxL
hds
L
2/
0
2
2
812
L
h
h
L
L
hL 4sinh
4
41
21
2
dxdx
dydydxds
222 1
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.4 Cables
Cable Subjected to its Own Weight• When weight of the cable is considered, the loading
function becomes a function of the arc length s rather than length x
• FBD of a segment of the cable is shown
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.4 Cables
Cable Subjected to its Own Weight• Apply equilibrium equations to the force system
• Replace dy/dx by ds/dx for direct integration
dsswFdx
dy
dsswT
FT
H
H
)(1
)(sin
cos
12
22
dx
ds
dx
dy
dydxdsT
Copyright © 2010 Pearson Education South Asia Pte Ltd
7.4 Cables
Cable Subjected to its Own Weight• Therefore
• Separating variables and integrating
2/12
2 )(1
1 dsswF
dsx
H
2/1
2
2)(
11
dsswFdx
ds
H
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 7.13
Determine the deflection curve, the length, and the maximum tension in the uniform cable. The cable weights wo = 5N/m.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
For symmetry, origin located at the center of the cable.
Deflection curve expressed as y = f(x)
Integrating term in the denominator
2/121
2/11 CswF
dsx
oH
2/122/11 dswF
dsx
oH
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Substitute
So that
Perform second integration
or
1/1 CswFu oH
dsFwdu Ho )/(
21sinh Cu
w
Fx
o
H
211 1
sinh CCswFw
Fx o
Ho
H
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Evaluate constants
or
dy/dx = 0 at s = 0, then C1 = 0
To obtain deflection curve, solve for s
dswFdx
dyo
H
1
1
1Csw
Fdx
dyo
H
x
F
w
w
Fs
H
o
o
H sinh
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Hence
Boundary Condition y = 0 at x = 0
For deflection curve,
This equations defines a catenary curve.
x
F
w
dx
dy
H
osinh
3cosh CxF
w
w
Fy
H
o
o
H
o
H
w
FC 3
1cosh x
F
w
w
Fy
H
o
o
H
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Boundary Condition y = h at x = L/2
Since wo = 5N/m, h = 6m and L = 20m,
By trial and error,
1cosh x
F
w
w
Fh
H
o
o
H
1
50cosh
/56
H
H
F
N
mN
Fm
NFH 9.45
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
For deflection curve,
x = 10m, for half length of the cable
Hence
Maximum tension occurs when θ is maximum at
s = 12.1m
m
mmN
mN
mN
mxy
2.24
1.12109.45
/5sinh
/5
9.45
2
1109.0cosh19.9
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
NNF
T
N
mmN
dx
dy
H
ms
9.758.52cos
9.45
cos
8.52
32.19.45
1.12/5tan
maxmax
max
max1.12
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
1. In a multi-force member, the member is generally subjected to an internal _________.
A) Normal force B) Shear force
C) Bending moment D) All of the above.
2. In mechanics, the force component V acting tangent to, or along the face of, the section is called the _________ . A) Axial force B) Shear force
C) Normal force D) Bending moment
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
3. A column is loaded with a vertical 100 N force. At which sections are the internal loads the same?
A) P, Q, and R B) P and Q
C) Q and R D) None of the above.
4. A column is loaded with a horizontal 100 N force. At which section are the internal loads largest?
A) P B) Q
C) R D) S
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
5. Determine the magnitude of the internal loads
(normal, shear, and bending moment) at point C.
A) (100 N, 80 N, 80 N m)
B) (100 N, 80 N, 40 N m)
C) (80 N, 100 N, 40 N m)
D) (80 N, 100 N, 0 N m )
2. A column is loaded with a horizontal 100 N force. At which section are the internal loads the lowest?
A) P B) Q
C) R D) S