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
Symmetric Bending of Beams
• A beam is any long structural member on which loads
act perpendicular to the longitudinal axis.
Learning objectives
• Understand the theory, its limitation and its applications
for strength based design and analysis of symmetric
bending of beams.
• Develop the discipline to visualize the normal and shear
stresses in symmetric bending of beams.
Pure Bending
Independent of
material model
Deformation of symmetric member
Under action of M and M’, the member will bend but will
remain symmetric with respect to the plane containing the
couples.
except
compressed
(compression)
elongated
(tension)
There exists a surface // to the upper and lower faces of the member
(see as a line on the cross-section) where no elongation and the bending
normal stress is zero. This surface is called the neutral axis.
The length of arc DE for both deformed and undeformed
L = rq
Consider an arc some distance y above the neutral surface
The length of arc JK can be expressed as
L’ = (r- y)q
The deformation d of JK
d = L’-L = (r- y)q - rq = - yq
x  The normal strain is max when y is the largest  max 
y
r
c
r
Derive flexure formula
Because
 x  E x
y
 x  -  max
c
From earlier,
y
 x  -  max
c
Derive flexure formula (continued)
m
 y
 Fx  0:  dF    x dA   -  c   mdA  -( c ) ydA
This equation can be satisfied only if
 ydA  0
The first area moment of the cross section about its NA = 0
Derive flexure formula (continued)
Taking the moment about the z axis = 0
 y
dF   x dA  -   m dA
 c
 m  2
 y
M z   - y x dA)   - y -   m dA     y dA
 - ydF  M z

c
 c 
I z   y 2 dA is the 2nd area moment of the cross section w.r.t. z axis
 max
Mc
I
y
 x  -  max
c
Mz y
x  Iz
Discuss flexure stress
Compression
Top Surface (+y)
Mz y
x  Iz
Tension
Bottom Surface (-y)
x  -
M z  - y
Iz
Not only bending about z axis produces a normal
stress in x direction but also bending about y axis.
If the bending moment is about the y axis, a similar
relationship exists.
x  -
Myz
Iy
 max  -
Myd
Iy
Example
Compute the area moment of inertia
A normal stress in the x direction due to Mz
A normal stress in the x direction due to My
Transverse Deformations due to bending
Member of several materials
• Assume a bar is made of two different materials
bonded together. The bar will deform as previously
shown.
• The normal strain in x direction will vary linearly with
distance from the N.A.
Method of transformed section

 E1 y 
dF1  -

 r 
 E y
dF2  - 2  ndA)
 r 
E 
n  - 2 
 E1 
The resistance to bending would be
the same if each section were made
of the same material ,where the 2nd
material was multiplied by n
Example – transformed section to all aluminum
Season’s Greeting!
Try it for yourself at home
Transform section to all steel
Beams bending analysis
•Beams carry loads perpendicular to their longitudinal axis.
•Internal shear forces and
bending moments develop
along the span of a beam.
In designing a beam, it is critical to determine the internal
shear force (V) and bending moment distribution (M). This
is accomplished by constructing shear and bending moment
diagrams.
Steps in constructing a V and M diagram
In general, the load distribution across the width of the
beam is assumed to be applied uniformly. Therefore,
a beam can be analyzed in 2 dimensions rather than 3.
1. Determine the reactions at each support.
Review of support conditions
Shear forces and bending moments in beams
Neither half is in equilibrium
The force imbalance that exists must be counteracted
so that static equilibrium is maintained. This is done
Through internal forces and moments.
Calculating internal V and M as a function of x by isolating
a segment of beam a distance x from the left end whose
width is dx.
=
eliminate last term = 0
Diagram by inspection
V = constant
M = f(x)
V = f(x)
M = f(x2)
V = no effect
M = spike