Lecture 33: Prismatic Beam Design

cageysyndicateUrban and Civil

Nov 15, 2013 (3 years and 4 months ago)

75 views

E
GR

23
6

Properties and
Mechanics of Materials

Spring 20
1
3

Lecture
3
3

Prismatic Beam Design


Today:


--

Homework questions:


--

New Topics:


--

Beam Design Problems




--

Homework: Read Section
11:1
-
3






Work Problems
from
Chap 11: 12

and
27


Following today's class you should be able to:


--

work through a step by step design process to design either a steel or


prismatic wooden beam for a given loading.



Section Modulus
:

Recall that the equation
for

bending stress is given by


bending
Mc
I




another way of expressing this equation is to introduce a

variable called the
s
ection modulus that is entire depend
s
upon cross section
al

geomet
r
y such that


bending
M
S




This implies that the sec
tion

modulus is defined by the cross section

s moment
of in
ert
ia,

I
, and the distance of the maximum fiber form the neutral axis,
c
, as


I
S
c


This will be conv
enient
for our use when designing pri
s
matic beams because it
rolls the ge
ometric
considerations all
shape parameters into a
sing
le

variable.


Design
Pr
ocess for Steel or Prismatic Wood
Beams:

1) From reference tables, det
ermine the values of allowable normal and shear
for the selected material

which are based on
yield strength for ductile materials

or

based on ultimate strength for brittle materials.


2) Solve for external load reac
tions of the beam system.


3) Draw the

Load, Shear, and Moment diagrams. From these
diagrams,
determine the location of the maximum shear and bending loads.


4) Assume beam acts as a long beam and that bending stress is the primary
stress

(meaning: ignore the transverse shear stress for now
.)

Calculate the
required section modulus for the given material and maximum internal bending
moment.



max
min
all
M
S




5) Use the calculation of

S
min
and other beam shape considerations to select an
acceptable sized
beam. Other consideration might include


min
imum allowed
weight to length ratio
,

maximum allowable
beam
depth,

maximum
beam
width,
and/or
maximum allowa
ble beam
deflection.


6) Check the shear stress for the selected beam to verify that the maximu
m

t
ransverse shear
stress is not exceeded

using

the shear stress formula


max NA
max
V Q
It



For
a
prismatic
beam with a
rectangular cross sec
tion
this
can be simplified to


max NA max
max
V Q V
It A

 
3
2


(for prismatic beam)

For S
-
and W
-

shaped I beams
, the entire shear load can be
approximated by
using the average shear force over the web area and is given by the formula


max
max
web
V
A




(for I beam)


I
f

the maxim
u
m shear load is not exceeded, th
e chosen beam should be fine. If
the maximum shear load is exceeded

by these calculations,
then you
should

g
o
back and
pick a different beam in step 5.

Then repeat step 6.




7)

Fo
r S
-
b
eam
s

and W
-
Beams, the principle stresses at the junction between
the web and the flange should also be checked. Sometime
s

the combined
normal
-
shear load at this junction may be larger than the individual maximum
normal or shear stresses.

E
xample 1:


from
Problem
11:4

a) The simply supported beam is made of timber. Determine its dimensions if it is to be
rectangular and have a height to weight ratio of
h/b

= 1.25.

b) Repeat this design but pick a steel I
-
beam for this same span and load.

______________
______________________________________________________

Solution:










































3000N/m


E
xample 1

Solution
:


from
Problem
11
:
4

a)
The simply supported beam is made of timber
.
Determine its dimensions if it is to be
rectangular
and have a height to
weight
ratio of
h/b

= 1.25.

b) Repeat this design but pick a steel I
-
beam for this same span and load.

____________________________________________________________________

Solution:

Step 1:

Define strength of the material



Let's use strength values for Douglas fir


σ
U
_tensile
=2.1 MPa



σ
U
_compresion
= 26 MPa.

Assume this material yields as a brittle material.

Maximum Normal Stress Failure
and apply a

safety factor of 2.


σ
allow

=
2.1 / 2 = 1.05

MPa


2)
External
load reactions
in equilibrium
.

0
A
M





4500 (2 ) (6 ) 0
B
N m F m
  


1500
B
F N



0
x
F




0
x
F




0
Ax
F


450 0
Ay B
F N F
  


1.5
B
F kN


4500 1500 3
Ay
F kN
  


3) Draw the Load, Shear, and Moment diagrams.



Max shear force is
300 N



To find the m
ax
b
ending stress



find where

V

= 0.








0
y
F




then

(1000 )( )
3000 0
2
x x
V
  




2
3000 500
V x
 






for
V

= 0 then
6 2.45
x m
 



0
M



2
3000( ) (500 )(/3) 0
M x x x
  


then fo
r


x

= 2.45 m


3 3
3000 166.7 3000(2.45) 166.7(2.45) 4899
M x x x N m
     


therefore at point C, the maximum ben
ding stress is 4899 N
-
m




4500 N

F
Ax

F
Ay

F
B

M

V

3000 N/m

3kN

1.5 kN

Load

3 kN

x

V

M

1000
x

3000N/m

4)
Calculate the required section modulus (based only on bending)


max
min
all
M
S




min
N m MPa mm
S mm
.MPa N/mm m

 
3
2
4899 1000
4657140
105 1 1


5) Use the calculation of
S
min
and other beam shape considerations to sele
ct an acceptable
sized beam.
In this case it's to be rectangular and have a weight to height ratio of
h/b

= 1.25.


therefore:


1.25
h b


and


( )
0.2604
min
bh
I b.b
S b
c h/
   
3
2
3
1
12
2
125
6

or


4657140
0.2604 0.2604
min
S
mm
b mm
  
3
3
3
262

and


1.25 1.25 (262mm) =327 mm
h b
 


6)
Checking the max. shear stress for
V
max

= 300 N at the left support by use of the
prismatic beam equation:.


1.5
3 1.5(3000N)
2 (262mm)(327mm)
max max
max
V V
MPa
.MPa
A bh N/mm

   
2
00525



This seem much smaller than the allowed normal stresses an
d should be fine.


So a wooden beam of Douglas fir or similar strength wood with a cross section of
262 mm
by 327

mm
or larger
should serve well for this design.

T
his translates to 10.3" x 12.9"
.


---------------------------------------------------------
---------------------------------------------------

Part b) Repeat the design for a steel I
-
beam:


Step 1) for structural steel (A
-
36)
σ
y

=

2
50 MPa


Assume this material yields as a ductile material.

Maximum Shear Stress Failure Theory applies

using a safety factor of 2.
σ
allow


<

σ
y
/

2


=
use a Normal Stress Failure and apply a

safety factor of 2.
σ
allow



< 250

/ 2 =
1
25 MPa



Step 2 and 3) do not change: so


V
max

= 300
0

N and
M
max

= 4899 N
-
m,

Step 4) The minimal section modulus for the I beam is



min
N m MPa mm
S mm
MPa N/mm m

 
3
2
4899 1000
39120
125 1 1

converting to inches so that we can use your I
-
beam table
.


3912 2.397
25.4
min
in
S mm in
mm
 
3
3 3

Use the section value to identify acceptable
S
-
beam cross sections.






























6) Check the shear stress
in the web.

where web area :



( 2 ) (0.193in)(4.000-2(0.293))
web web f
A t d - t.in mm
   
2 2
0659 425


then



max
max
web
V
N
.MPa
A mm

  
2
3000
705
425


which appears to be well within acceptable values.



So use the

S4x 7.7

I beam

EGR
23
6

Mechanics of Materials

H
W

Set
3
3

Spring 201
3

Problem
11
:
12


Select the lightest weight steel wide
-
flange overhand
beam from Appendix B that will safely
support the loading. Assume the support at A is a pin and the support at B is a roller. The
allowable bending stress is 24 ksi and the allowable shear stress is 14 ksi.

_______________________________________________
___________________________


Solution:






























EGR
23
6

Mechanics of Materials

H
W

Set
3
3

Spring 201
3

Problem
11
:
27


The

box beam has an allowable bending stress of 10 MPa and allowable shear stress of 775
kPa. Determine the maximum intensity of
w

of the distributed loading that it can safely
s
upport. Also determine the maximum safe nail span for each third of the length of the
beam. Each nail can resist a shear force of 200 N.

--------------------------------------------------------------------------------------------------------------

Sol
ution: