# EME 231 Enginering Statics PreTest/PostTest

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```					EGR 236 Properties and Mechanics of Materials                    Spring 2012
Lecture 18: Flexure in Composite Beams

Today:
-- Homework questions:
-- New Topics:
-- How to handle beams made from multiple materials.
Work Problems from Chap 6: 114, 116, 119

Following today's class you should be able to:
-- understand that the strain of a composite beam is proportional to the distance
from the neutral axis, regardless of material.
-- be able to change the structure of a composite beam into an equivalent beam
of homogeneous material for calculation of stress.
-- be able to compute maximum stress values in each material of a composite
beam.

Composite Beams:
A composite beam is a beam constructed of two or more different materials.
Common examples include beams that have a center section of wood
sandwiched between two outer bands of steel or concrete beams which have
steel reinforcing rods. Steel rods or bands are often used to help provide
support to materials which are weaker in tension but very strong in
compression. By placing the steel where the beam is subjected to the tensile
bending loads, can reduce the overall size and weight of the beam.
In a beam made of two materials, the deformed plane of the material is
expected to remain plane if the materials behave elastically.
However since one of the materials has a larger modulus of elasticity than the
other, it will handle a larger percentage of the load for similar deformation.

Less Stiff Material

Stiffer Material

Strain           Stress
Distribution       Distribution

Notice how the stiffer material, generally sees the higher stress distribution for
one such typical beam.

One method to calculate the stress values of a composite beam is to transform
the beam into an equivalent beam consisting entirely of one of the two
materials. To do this you need to either replace the stiffer material with a
wider piece of the softer material, or you need to replace the softer material
with a thinner piece of the stiff material.

Consider the wood-metal composite beam:

                   

Transformed                  Original        Transformed
to Soft                     Composite        to Stiff
Material                     Beam            Material

In this example, you would either replace the steel with a wider piece of wood
or replace the wood with a narrower piece of steel. The ratio of width can be
found by a ratio of the modulus of elasticity of the materials.
E1
n
E2
Example:
A 6” x 8” wood beam is reinforced by a 0.5” x 6” steel plate on the bottom,
and a 1” x 2” steel bar on the top. If the allowable stress in the wood and steel
are 2000 and 18,000 psi respectively, and the elastic modulus of the steel is 20
times larger than that of the wood, find the maximum bending moment that can
be applied.

Solution:
Esteel
n            20
Ewood

Let's transform the wood into an
equivalent width of steel:

wwood 6 in
wsteel               03 in
.
n     20

For a beam of uniform material, the
bending stress is given as

Mc
                                                                      2x1
I

To find c and I start by locating the                                       0.3 x 8
centroid of the section.
ycent
yi Ai
ycent                                                                         6 x 0.5
Ai

( 025 )( 6  05 )  ( 45 )( 03  8 )  ( 9 )( 2 1 )
.           .        .     .
                                                         3.993 in
( 6  05 )  ( 03  8 )  ( 2 1 )
.        .
Next find the moment of inertia about the centroid.

I total  I1  I2  I3                                            A3 =2 x 1

d3
A2=0.3 x 8
d2
where
Ii  Ii _cent  Adi 2
i                       d1
ycent=3.993

so                                                                    A1=6 x 0.5
1
I3     ( 2 )( 1 )3  ( 2 1 )( 90  3993 )2  503in4
.    .         .
12
1
I2     ( 03 )( 8 )3  ( 03  8 )( 45  3993 )2  13.4in4
.               .         .   .
12
1
I1     ( 6 )( 05 )3  ( 6  05 )( 025  3993 )2  421in4
.              .    .     .          .
12

so
I total  503  13.4  421  105 .8 in 4
.            .

The distance to the outer fibers of the steel are cst_top = 9.5-3.993 = 5.507 in
and cst_bot = 3.993 in

The maximum distance to the wood fibers are cst_top = 8.5-3.993 = 4.507 in
and cst_bot = 3.993-0.5 = 3.493 in

For the Steel (at the top and for maximum stress of 18,000 psi):
Mc
 
I
 I ( 18000lb / in2 )( 105.8in 4 )
M                                    346 ,000 lb  in
c         ( 5507in )
.

 28,833 lb  ft
To check out the maximum moment that stays below the max stress of the
wood, recall that the strain of the materials must match at the same distance
from the neutral axis.

 wood   steel

 wood        steel                     Ewood           1
                    wood            steel   steel
Ewood        Esteel                      Esteel          n

1         1 Mc
 wood   steel 
n         n I

n wood I 20( 2000lb / in2 )( 1058in4 )
.
M                                          939,000 lb  in
c                4507in
.

 78,250 lb  ft

Since the wood can support a larger moment than the steel, the beam is limited
by the stress in the steel.

Maximum moment to be applied to this beam is the 28,800 lb-ft.
EGR 236 Mechanics of Materials              HW Set 19                      Spring 2011
Problem 6:114

The composite beam consists of a Douglas Fir wood core and three plates of A-36 steel. If
the allowable bending stress for wood is 20 MPa and for the steel is 130 MPa, determine the
maximum moment that can be applied to the beam.

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