Strength of Materials I
EGCE
201
ก
ำลังวัสดุ
1
Instructor:
ดร
.
วรรณสิริ พันธ์อุไร
(
อ
.
ปู
)
ห้องท ำงำน
:
6391
ภำควิชำวิศวกรรมโยธำ
E

mail:
egwpr@mahidol.ac.th
โทรศัพท์
:
66
(
0
)
2889

2138
ต
่
อ
6391
Design of beam for bending
Design of beam for bending
Steps in designing a beam for bending
Assume that E,G,
s
all
, and
t
all
for the material selected are
known
1.
Construct the diagrams
corresponding to the specified
loading conditions for the beam
and define V
max
and
M
max
.
2.
Assume that the design of the
beam is controlled by the
normal stress at +,

c in the
section and determine S
min
.
3.
From available tables, select
beams with S>S
min
.
Considerations:

small weight per unit length

small displacement (show in the latter chapter)
1.
2
.
Design Example
1. Modeling distributed loads as equivalent concentrated force
2. Use the knowledge gained from lecture 4 (either by forming
equations or by inspection) to form moment diagram.
1. the shear force is linear, the moment varies
parabolically.
2. the area under the shear diagram is
–
so is
the correspondent moment
A =
0.5
(
9
)(

2250
) =

10
,
125
M =

10
,
125
Design Example (continued)
M
max
= 18,000 lb

ft =216,000 lb

in
V
max
= 3750 lb
3. From 2, we find
4.
Complete Beam Analysis Example
•
For the beam loaded and supported as shown,
determine the max tensile and compressive
stresses and where they occur.
Given
1
. Determine the location of the centroid.
•
The area of the cross section (A) = 18.75 in
2
2. Determine the area moment of inertia w.r.t. the centroidal axis (as
shown in figure on the lower left) and use transfer formula to
calculate I of the section.
3
. Begin the analysis by defining reactions at A and B using FBD,
one writes the equations of equilibrium.
4
. Next, the shear diagram can be
constructed (see previous example
or lecture
4
for more details).
5. Now the moment diagram can be
constructed.
The moment is the area under the shear
force diagram and the three areas are
A1 = 0.5(3.5ft)(

17.5kip)
=

30.63 kip

ft
A2 = 0.5(3.5ft)(52.5

35 kip)+(3.5ft)(35 kip)
= 153.13 kip

ft
A3 = 0.5(7ft)(

35kip)
=

122.5 kip

ft
The moment diagram is as shown.
A1+A2=
A1=
A1+A2+A3=0
Recall
Positive moment, top beam is in compression
and the bottom is under tension
6. Next, we compute stresses
At x=3.5 ft, moment is negative so
the top is in tension while
the bottom is under compression.
7.At x=7 ft, moment is positive
so the top is under compression
while the bottom is in tension.
Next
Shear Stress in Beams
A cutting plane is passed
through a beam at an arbitrary
spanwise location, the internal
reactions are required for
Equilibrium are a bending
moment and a shear force.
The moment and shear force
as shown are considered positive.
The shear and normal stresses
acting on an element of area are
represented as forces by multiplying
them by the area (dA)
3
out of
6
equations of equilibrium involve the normal force
s
x
dA
Pure Bending
3 out of 6 equations of equilibrium involve the shearing force
t
xy
dA,
t
xz
dA
1.
2.
From 1.
Vertical shearing stresses exist in a transverse section of the beam
if a shear force exists at that section
Shear stress on a horizontal plane
H
Horizontal shear derivation
•
Observing that x is
constant over the
cross section, the
expression for H is
written as
Q is the first area moment w.r.t. to the N.A. of
that part of the section located above the line y=y1
H
PQ
I
x
Shear Flow, q
H
x
PQ
I
P
V
But
Along a horizontal plane a distance y1 above the NA,
the
horizontal shear per unit length
of beam
The ratio of
H/x
is termed the
shear
flow
and is denoted by “q”
where
q
VQ
I
Example
•
Determine the shear flow (
q)
of the
following cross section
Solution
•
The cross section is broken into
3
sections and
the second area moment of inertia
•
Given V =
8000
lb, one compute
•
The shear flow can now be expressed as
q
Q
0
558
.
•
Compute Q
Q is the first area moment w.r.t the N.A.
A is the area of the cross section above the plane for which q is
being determined.
is the location of that area w.r.t the N.A (+,

).
Q
A
y
y
Transverse shear stresses in beam
•
The horizontal shear flow at C
•
A shear force exists on a horizontal
plane passing C
q
VQ
I
For a narrow rectangular beam
b < h/4
For an I

beam
web
flange
Example
•
A shear force acts on a cross
section. The cross section
shown is made by nailing
planks together.
1. Use shear flow to define the
required nail spacing if each
nail supports 700 lb shear
before failure.
2. Compute shear stress at
various locations in the cross
section.
1. Compute the moment of inertia
2. Compute the nail spacing
3. Compute stress distribution
Displacements in beams
Next week
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