A beam is a horizontal structural member used to support loads in buildings by spanning between supports. Common beam shapes include I-beams, angle beams, and channel beams, which are usually made of steel or wood. The key parts of a beam are the flanges and web. Beams can support concentrated or distributed loads. Beam theory involves calculating the beam's moment of inertia and deflection under loads using properties like the modulus of elasticity. More complex beam shapes like box beams and H-beams calculate moment of inertia by dividing the cross-section into positive and negative areas.

1. 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. 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. 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.
Flanges
Web
Flanges
Web

4. 4
Introduction to Beams
• Beams are supported in structures via
different configurations
Source: Statics (Fifth Edition), Meriam and Kraige, Wiley

5. 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. 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. 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. 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. 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.
12
bh3
x 
I
b
h
X-axis (passing
through centroid)

10. 10
Beam Theory
• Example: Calculate the moment of inertia about the X-
axis for a yardstick that is 1” high and ¼” thick.
12
bh3
x 
I
b = 0.25”
h = 1.00”
  
12
in
1.00
in
0.25
3
x 
I
4
x in
0.02083

I
X-Axis
Y-Axis

11. 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
12
bh3
y 
I
  
12
in
0.25
in
1.00
3
y 
I
4
y in
0.00130

I

12. 12
Beam Theory
• Suppose a concentrated load, P, is applied
to the center of the simply supported
beam.
P
L

13. 13
Beam Theory
• The beam will bend downward as a result
of the load P.
P

14. 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. 15
Beam Theory
• The deflection (Δ) of a simply supported, center loaded
beam can be calculated from the following formula:
I
48E
PL
Δ
3

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. 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. 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”
12
bh3

I
I
48E
PL
Δ
3


18. 18
Beam Theory
• Step 1: Calculate the moment of inertia, I.
12
bh3

I
  
12
in
2
in
3
3

I
4
in
2

I

19. 19
Beam Theory
• Step 2: Calculate the deflection, Δ.
I
48E
PL
Δ
3

  
 
4
2
6
3
in
2
in
lb
10
x
30
48
in
36
lb
500
Δ







  
 
4
2
6
3
in
2
in
lb
10
x
30
48
in
46656
lb
500
Δ







in
0.0081
Δ 

20. 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.

21. 21
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-axis
X-axis

22. 22
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

23. 23
Beam Theory
bi = 3.5 in.
ho = 6 in.
hi = 5.5 in.
bo = 4 in.
X-axis
12
h
b
3
o
o
pos 
I
12
h
b
3
i
i
neg 
I
  
12
in
6
in
4
3
pos 
I
  
12
in
5.5
in
3.5
3
neg 
I

24. 24
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.
neg
pos
box -I
I
I 
     
12
in
5.5
in
3.5
12
in
6
in
4
3
3
box 

I
4
box in
23.5

I
12
h
b
12
h
b
3
i
i
3
o
o
box 

I
     
12
in
166.4
in
3.5
12
in
216
in
4 3
3
box 

I

25. 25
Beam Theory
• The moment of inertia of an I-beam can be calculated in
a similar manner.

26. 26
Beam Theory
• Identify the positive and negative areas…
Positive Area Negative Area

27. 27
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
12
h
b
2
12
h
b
3
i
i
3
o
o


beam
I
I

28. 28
Beam Theory
• The moment of inertia of an H-beam can be calculated in
a similar manner…

29. 29
Beam Theory
• The moment of inertia of an H-beam can be calculated in
a similar manner…

30. 30
Beam Theory
• …however, the H-beam is divided into three positive
areas.
b2
b1
h2 h1
b1
h1
12
h
b
12
h
b
12
h
b
3
1
1
3
2
2
3
1
1
beam
-
H 


I
12
h
b
12
h
b
2
3
2
2
3
1
1
beam
-
H 

I

31. 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. 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
12
h
b
2
12
h
b
3
i
i
3
o
o


beam
I
I
ho = 5 in. hi = 4 in.

33. 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
4in
1.75in
2
12
5in
4in
3
3


beam
I
I
ho = 5 in. hi = 4 in.
4
in
23.0

beam
I
I

34. 34
Beam Theory
• Next, calculate the deflection (Esteel = 30 x 106 psi).
I
48E
PL
Δ
3

L = 8 ft
P = 5000 lbf

35. 35
Beam Theory
• Calculate the deflection, Δ.
I
48E
PL
Δ
3

  
 
4
2
f
6
3
in
23
in
lb
10
x
30
48
in
96
lb
5000
Δ







  
 
4
2
6
3
in
23
in
lb
10
x
30
48
in
884736
lb
5000
Δ







in
0.134
Δ 

36. 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. 37
Beam Theory
• Volume = (Area) x (Length)
   
8ft
4in
0.5in
3
V 
AL
V 
  
in
96
2.0in
3
V 2

3
in
576
V 

38. 38
Beam Theory
• Convert to cubic feet…
3
3
12in
1ft
in
576
V 














 3
3
3
1728in
1ft
576in
V
3
ft
0.333
V 

39. 39
Beam Theory
• Calculate mass of the beam
• Mass = Density x Volume
ρV
m 
 
3
3
m
0.333ft
ft
lb
490
m 






m
lb
163.3
m

40. 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. 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. 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. 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.