1

Introduction to Beams

• A beam is a horizontal structural member used to support loads

• Beams are used to support the roof and floors in buildings

2

Introduction to Beams

• Common shapes are

I Angle Channel

• Common materials are steel and wood

Source: Load & Resistance Factor Design (First Edition), AISC

3

Introduction to Beams• The parallel portions on an I-beam or H-beam are

referred to as the flanges. The portion that connects the flanges is referred to as the web.

FlangesWeb

Flanges

Web

4

Introduction to Beams

• Beams are supported in structures via different configurations

Source: Statics (Fifth Edition), Meriam and Kraige, Wiley

5

Introduction to Beams

• Beams are designed to support various types of loads and forces

Concentrated Load Distributed Load

Source: Statics (Fifth Edition), Meriam and Kraige, Wiley

6

Beam Theory• Consider a simply supported beam of

length, L. The cross section is rectangular, with width, b, and depth, h.

L

b

h

7

Beam Theory• An area has a centroid, which is similar to a center of

gravity of a solid body.• The centroid of a symmetric cross section can be easily

found by inspection. X and Y axes intersect at the centroid of a symmetric cross section, as shown on the rectangular cross section.

h/2

h/2

b/2 b/2

X - Axis

Y - Axis

8

Beam Theory• An important variable in beam design is the moment of

inertia of the cross section, denoted by I.• Inertia is a measure of a body’s ability to resist rotation.• Moment of inertia is a measure of the stiffness of the

beam with respect to the cross section and the ability of the beam to resist bending.

• As I increases, bending and deflection will decrease.• Units are (LENGTH)4, e.g. in4, ft4, cm4

9

Beam Theory• I can be derived for any common area using calculus.

However, moment of inertia equations for common cross sections (e.g., rectangular, circular, triangular) are readily available in math and engineering textbooks.

• For a rectangular cross section,

• b is the dimension parallel to the bending axis. h is the dimension perpendicular to the bending axis.

12bh3

x =I

b

h

X-axis (passingthrough centroid)

10

Beam Theory• Example: Calculate the moment of inertia about the X-

axis for a yardstick that is 1” high and ¼” thick.

12bh3

x =I

b = 0.25”

h = 1.00” ( ) ( )12

in1.00in 0.25 3

x =I

4x in0.02083=I

X-Axis

Y-Axis

11

Beam Theory• Example: Calculate the moment of inertia about the Y-

axis for a yardstick that is 1” high and ¼” thick.

h = 0.25”

b = 1.00”

X-Axis

Y-Axis

12bh3

y =I

( ) ( )12

in0.25in 1.00 3

y =I

4y in0.00130=I

12

Beam Theory• Suppose a concentrated load, P, is

applied to the center of the simply supported beam.

P

L

13

Beam Theory• The beam will bend downward as a result

of the load P.

P

14

Beam Theory• The deflection (Δ) is the vertical

displacement of the of the beam as a result of the load P.

Deflection, Δ

L

15

Beam Theory• The deflection (Δ) of a simply supported, center loaded

beam can be calculated from the following formula:

I48EPL

Δ3

=

where,

P = concentrated load (lbs.)

L = span length of beam (in.)

E = modulus of elasticity (lbs./in.2)

I = moment of inertia of axis perpendicular to load P (in.4)

L

P

16

Beam Theory• Modulus of elasticity, E, is a property that indicates the

stiffness and rigidity of the beam material. For example, steel has a much larger modulus of elasticity than wood. Values of E for many materials are readily available in tables in textbooks. Some common values are

Material Modulus of Elasticity (psi)

Steel 30 x 106

Aluminum 10 x 106

Wood ~ 2 x 106

17

Beam Theory• Example: Calculate the deflection in the steel beam

supporting a 500 lb load shown below.

P = 500 lb

L = 36”

b = 3”

h = 2”

12bh3

=II48E

PLΔ

3

=

18

Beam Theory• Step 1: Calculate the moment of inertia, I.

12bh3

=I

( ) ( )12

in2in 3 3

=I

4in2=I

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Beam Theory• Step 2: Calculate the deflection, Δ.

I48EPL

Δ3

=

( ) ( )( )4

26

3

in 2inlb

10 x 3048

in36lb 500Δ

=

( ) ( )( )4

26

3

in 2inlb

10 x 3048

in46656lb 500Δ

=

in0.0081Δ =

20

Beam Theory• These calculations are very simple for a solid, symmetric

cross section.• Now consider slightly more complex symmetric cross

sections, e.g. hollow box beams. Calculating the moment of inertia takes a little more effort.

• Consider a hollow box beam as shown below:

4 in.

6 in.

0.25 in.

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Beam Theory• The same equation for moment of inertia, I = bh3/12, can

be used.• Treat the outer dimensions as a positive area and the

inner dimensions as a negative area, as the centroids of both are about the same X-axis.

Positive Area Negative Area

X-axisX-axis

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Beam Theory• Calculate the moment of inertia about the X-axis for the

positive area and the negative area using I = bh3/12. The outer dimensions will be denoted with subscript “o” and the inner dimensions will be denoted with subscript “i”.

bi = 3.5 in.

ho = 6 in.

hi = 5.5 in.

bo = 4 in.

X-axis

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Beam Theory

bi = 3.5 in.

ho = 6 in.

hi = 5.5 in.

bo = 4 in.

X-axis

12hb 3

oopos =I

12hb 3

iineg =I

( ) ( )12

in6in 4 3

pos =I( ) ( )

12in5.5in 3.5 3

neg =I

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Beam Theory• Simply subtract Ineg from Ipos to calculate the moment of

inertia of the box beam, Ibox

4 in.

6 in.

0.25 in.

negposbox - III =

( ) ( ) ( ) ( )12

in5.5in 3.512

in6in 4 33

box −=I

4box in23.5=I

12hb

12hb 3

ii3

oobox −=I

( ) ( ) ( ) ( )12

in166.4in 3.512

in216in 4 33

box −=I

25

Beam Theory• The moment of inertia of an I-beam can be calculated in

a similar manner.

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Beam Theory• Identify the positive and negative areas…

Positive Area Negative Area

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Beam Theory• …and calculate the moment of inertia similar to the box

beam (note the negative area dimensions and that it is multiplied by 2).

bo

hi

bi bi

ho

12hb2

12hb 3

ii3

oo −=−beamII

28

Beam Theory• The moment of inertia of an H-beam can be calculated in

a similar manner…

29

Beam Theory• The moment of inertia of an H-beam can be calculated in

a similar manner…

30

Beam Theory• …however, the H-beam is divided into three positive

areas.

b2

b1

h2 h1

b1

h1

12hb

12hb

12hb 3

113

223

11beam-H ++=I

12hb

12hb2 3

223

11beam-H +=I

31

Beam Theory• Example: Calculate the deflection in the I-beam shown

below. The I-beam is composed of three ½” x 4” steel plates welded together.

L = 8 ft

P = 5000 lbf

½” x 4” steel plate (typ.)

32

Beam Theory• First, calculate the moment of inertia for an I-beam as

previously shown, i.e. divide the cross section of the beam into positive and negative areas.

bo = 4 in.bi = bi

12hb2

12hb 3

ii3

oo −=−beamII

ho = 5 in. hi = 4 in.

33

Beam Theory• First, calculate the moment of inertia for an I-beam as

previously shown, i.e. divide the cross section of the beam into positive and negative areas.

bo = 4 in.bi =1.75in bi

( ) ( ) ( ) ( )12

4in1.75in212

5in4in 33

−=−beamII

ho = 5 in. hi = 4 in.

4in 23.0=−beamII

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Beam Theory• Next, calculate the deflection (Esteel = 30 x 106 psi).

I48EPL

Δ3

=

L = 8 ft

P = 5000 lbf

35

Beam Theory• Calculate the deflection, Δ.

I48EPL

Δ3

=

( ) ( )( )4

2f6

3

in 23inlb

10 x 3048

in96lb 5000Δ

=

( ) ( )( )4

26

3

in 23inlb

10 x 3048

in884736lb 5000Δ

=

in0.134Δ =

36

Beam Theory• Example: Calculate the volume and mass of the beam if

the density of steel is 490 lbm/ft3.

L = 8 ft ½” x 4” steel plate (typ.)

37

Beam Theory• Volume = (Area) x (Length)

( ) ( ) ( )8ft4in0.5in3V =ALV =

( )( )in 962.0in3V 2=3in 576V =

38

Beam Theory• Convert to cubic feet…

33

12in1ft

in 576V

=

= 3

33

1728in1ft

576inV

3ft 0.333V =

39

Beam Theory• Calculate mass of the beam• Mass = Density x Volume

ρVm=

( )33m 0.333ft

ftlb

490m

=

mlb 163.3m=

40

Materials

• Basswood can be purchased from hobby or craft stores. Hobby Lobby carries many common sizes of basswood. DO NOT purchase balsa wood.

• 1201 teams must submit a receipt for the basswood.

• The piece of basswood in the Discovery Box WILL NOT be used for Project 2.

• Clamps and glue are provided in the Discovery Box. Use only the glue provided.

41

Assembly

• I-beams and H-beams: Begin by marking the flanges along the center where the web will be glued.

• Box beams: No marking is necessary.

• I-beams and H-beams: Apply a small amount of glue along the length of the web and also to the flange.

• Box beams: Apply a small amount of glue two one side and the bottom to form an “L” shaped section.

42

Assembly• I-beams and H-beams: Press

the two pieces together and hold for a couple of minutes.

• Box beams: Press the two pieces together into an “L” shape and hold for a couple of minutes.

• I-beams and H-beams: Clamp the pieces and allow the glue to cure as instructed on the bottle.

• Box beams: Clamp the pieces and allow the glue to cure as instructed on the bottle.

• YOU MUST ALLOW EACH GLUE JOINT TO CURE COMPLETELY BEFORE CONTINUING ON TO ADDITIONAL GLUE JOINTS!

43

Assembly• Stiffeners may be applied to the

web of an I or H beam or between sides of a box beam.

• Stiffeners are small pieces of wood that aid in gluing and clamping the beam together.

• Stiffeners may be necessary if the pieces of wood that the team has chosen are thin (less than ~3/16”). Thin pieces of wood may collapse when clamped together.

• Stiffeners keep the flanges stable during clamping and glue curing.