Transcript Strength of Materials I EGCE201 กำลังวัสดุ 1

Strength of Materials I
EGCE201 กำลังวัสดุ 1
Instructor:
ดร.วรรณสิ ริ พันธ์อุไร (อ.ปู)
ห้องทำงำน: 6391 ภำควิชำวิศวกรรมโยธำ
E-mail: [email protected]
โทรศัพท์: 66(0) 2889-2138 ต่อ 6391
Beam Deflections
As a beam is loaded,
• different regions are subjected to V and M
• the beam will deflect
x
x
Recall the curvature equation
The slope (q) & deflection (y) at any spanwise location
can be derived
A cantilever with a point load
• The moment is M(x) = -Px
• The curvature equation for
this case is
• The curvature is  at A and
finite at B
• Obviously, the curvature can
be related to displacement
The slope & deflection equation
• The following relations
are established
dy
tan q 
; ds  dq
dx
• For small angles,
dy
q  ; s x
dx
2
1 dq d y M ( x )

 2 
 ds dx
EI
Double Integration method
• Multiply both sides
d 2 y M ( x)


2
of
 this
EI
dx equation
1
by EI and integrate
x
dy
Slope eqn
EI
 EIq   M ( x )dx  C1
0
dx
• Where C1 is a constant of integration, Integrating again
EIy  
x
0
 xM ( x )dx  C dx  C
1
2
 0

x
x
0
0
  dx  M ( x )dx  C1 x  C2
Deflection eqn
C1 and C2 must be determined from boundary conditions
Boundary conditions
• Cantilever beam
• Overhanging beam
• Simply supported beam
Statically indeterminate beam
• The reactions at the supports cannot be completely
determined from the 3 eqns.
F
x
0
F
y
0
M 0
• The degree to which a beam is statically indeterminate
is identified by a no. of reactions that can’t be
determined.
• To determine all reactions, the beam deformations must
be considered.
Statically indeterminate to 1st degree
• From the FBD shown, one can write
• Obviously not enough to solve for Ay, MA,
and B
Begin by drawing a FBD diagram for an arbitrary segment of beam
Ay  wL  B
1 2
M A  wL  BL
2
(1)
For beam segment AC, summing moment about C we write
M  MA 
1 2
wx  Ay x  0
2
The curvature equation is determined from
d2y
1
EI 2  M   wx 2  Ay x  M A
2
dx
Integrating twice in x, we have
dy
1
1
 EIq   wx 3  Ay x 2  M A x  c1
dx
6
2
1
1
1
EIy   wx 4  Ay x 3  M A x 2  c1 x  c2
24
6
2
EI
Using the B.C.s x=0, q0, and y=0 the constants C1 = C2 = 0
EIy  
1
1
1
wx 4  Ay x 3  M A x 2
24
6
2
Using the 3rd B.C. (x=L, y=0) the above equation becomes
0
or
1
1
1
wL4  Ay L3  M A L2
24
6
2
0  3 M A  Ay L 
1 2
wL
4
(2)
The 2 equations generated from the original FBD of the beam and
the equation generated by considering a BC at x=L from the
curvature equation:
Ay  wL  B
MA 
0  3 M A  Ay L 
1 2
wL
4
1 2
wL  BL
2
2 equations and 2 unknowns, they are now solved to determine the
reactions at all supports
Ax  0
Ay 
5
wL
8
MA 
1 2
wL
8
B
3
wL
8
Ans
Statically indeterminate to 2nd degree
• From the FBD shown, one can write
• Obviously not enough to solve for Ay, MA,
By and MB
Ay  wL  B y
MA 
1 2
wL  B y L  M B
2
For beam segment AC, summing moment about C we write
M  MA 
1 2
wL  Ay x  0
2
The curvature equation is determined from
d2y
1
EI 2  M   wx 2  Ay x  M A
2
dx
Integrating twice in x, we have
dy
1
1
 EIq   wx 3  Ay x 2  M A x  c1
dx
6
2
1
1
1
EIy   wx 4  Ay x 3  M A x 2  c1 x  c2
24
6
2
EI
Using the B.C.s x=0, q0, and y=0 the constants C1 = C2 = 0
1
1
EIq   wx 3  Ay x 2  M A x
6
2
EIy  
1
1
1
wx 4  Ay x 3  M A x 2
24
6
2
Using the 3rd B.C. (x=L, q=0, y=0) the above equation becomes
1
1
0   wL3  Ay L2  M A L
6
2
1
1
1
0   wL4  Ay L3  M A L2
24
6
2
or
1
1
M A   wL2  Ay L
6
2
1
3 M A   wL2  Ay L
4
1 2 1
M A   wL  Ay L
6
2
Ay  wL  B y
1 2
3 M A   wL  Ay L
4
1 2
M A  wL  B y L  M B
2
2 equations and 2 unknowns, they are now solved to determine the
reactions at all supports
Ax  0
Ay  B y 
1
wL
2
M A  MB 
1
wL2
12
Ans
Superposition
• The slope & deflection at any point of the beam may
be determined by superposing the slopes and
deflections caused by the distributed and concentrated
loads.
Superposition (cont.)
Superposition (cont.)
Superposition (cont.)
• The method of superposition can be used for many
loading conditions, provided the equations for the slope
and deflection associated with a specific type of load
are known. For example,
Tables
• Click on each beam in which u are interested to
see the curvature equation, max deflection, and
slope for each loading condition (You should
start working on it before seeing!)
Singularity functions
• Instead of using the curvature equation, one can
also use what so called the singularity functions.
For the beam shown above, the bending moment in
terms of singularity functions by
• Integrate twice to get
Next Week
1. Quiz
2. Continuation of Beam Deflections
- theorems of area moment method
- moment diagrams by parts