# Design of beam for bending

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15 Νοε 2013 (πριν από 4 χρόνια και 8 μήνες)

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Strength of Materials I

EGCE
201

ำลังวัสดุ
1

Instructor:

ดร
.
วรรณสิริ พันธ์อุไร

(

.
ปู
)

ห้องท ำงำน
:

6391
ภำควิชำวิศวกรรมโยธำ

E
-
mail:
egwpr@mahidol.ac.th

โทรศัพท์
:
66
(
0
)
2889
-
2138

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
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
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

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