beam lecture

43
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

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Page 1: Beam Lecture

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

Page 2: Beam Lecture

2

Introduction to Beams

• Common shapes are

I Angle Channel

• Common materials are steel and wood

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

Page 3: Beam Lecture

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

Page 4: Beam Lecture

4

Introduction to Beams

• Beams are supported in structures via different configurations

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

Page 5: Beam Lecture

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

Page 6: Beam Lecture

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

Page 7: Beam Lecture

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

Page 8: Beam Lecture

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

Page 9: Beam Lecture

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)

Page 10: Beam Lecture

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 in0.02083I

X-Axis

Y-Axis

Page 11: Beam Lecture

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 in0.00130I

Page 12: Beam Lecture

12

Beam Theory• Suppose a concentrated load, P, is

applied to the center of the simply supported beam.

P

L

Page 13: Beam Lecture

13

Beam Theory• The beam will bend downward as a result

of the load P.

P

Page 14: Beam Lecture

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

Page 15: Beam Lecture

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

Page 16: Beam Lecture

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

Page 17: Beam Lecture

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

Page 18: Beam Lecture

18

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

12bh3

I

12

in2in 3 3

I

4in2I

Page 19: Beam Lecture

19

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

Page 20: Beam Lecture

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.

Page 21: Beam Lecture

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-axisX-axis

Page 22: Beam Lecture

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

Page 23: Beam Lecture

23

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

Page 24: Beam Lecture

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.

negposbox - III

12

in5.5in 3.512

in6in 4 33

box I

4box in23.5I

12hb

12hb 3

ii3

oobox I

12

in166.4in 3.512

in216in 4 33

box I

Page 25: Beam Lecture

25

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

a similar manner.

Page 26: Beam Lecture

26

Beam Theory• Identify the positive and negative areas…

Positive Area Negative Area

Page 27: Beam Lecture

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

12hb2

12hb 3

ii3

oo beamII

Page 28: Beam Lecture

28

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

a similar manner…

Page 29: Beam Lecture

29

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

a similar manner…

Page 30: Beam Lecture

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

Page 31: Beam Lecture

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.)

Page 32: Beam Lecture

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.

Page 33: Beam Lecture

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

Page 34: Beam Lecture

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

I48EPL

Δ3

L = 8 ft

P = 5000 lbf

Page 35: Beam Lecture

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

Page 36: Beam Lecture

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.)

Page 37: Beam Lecture

37

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

8ft4in0.5in3V ALV

in 962.0in3V 23in 576V

Page 38: Beam Lecture

38

Beam Theory• Convert to cubic feet…

33

12in1ft

in 576V

3

33

1728in1ft

576inV

3ft 0.333V

Page 39: Beam Lecture

39

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

ρVm

33m 0.333ft

ftlb

490m

mlb 163.3m

Page 40: Beam Lecture

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.

Page 41: Beam Lecture

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.

Page 42: Beam Lecture

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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!

Page 43: Beam Lecture

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.