INTRODUCTION TO BEAMS

• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering

CHAPTER

8b

ENCE 355 - Introduction to Structural Design

Department of Civil and Environmental Engineering University of Maryland, College Park

INTRODUCTION TO BEAMS

Part II – Structural Steel Design and Analysis

FALL 2002 Dr . Ibrahim. Assakkaf^By

CHAPTER 8b. INTRODUCTION TO BEAMS Slide No. 1

ENCE 355 ©Assakkaf

Elastic Design

Q

For many years the elastic theory has been the basis for steel structural design and analyses. This theory is based on the yield stress of a steel structural element.

Q

However, nowadays, it has been

replaced with a more rational and

realistic theory, the ultimate stress

design that is based on the plastic

capacity of a steel structure.

Elastic Design

Q

In the elastic theory, the maximum load that a structure could support is

assumed to equal the load that caused a stress somewhere in the structure equal the yield stress F

of the material.

Q

The members were designed so that computed bending stresses for service loads did not exceed the yield stress divided a factor of safety (e.g., 1.5 to 2)

CHAPTER 8b. INTRODUCTION TO BEAMS Slide No. 3

ENCE 355 ©Assakkaf

Elastic Design

Q

Elastic Versus Ultimate-based Design of Steel Structures

• According to ASD, one factor of safety (FS) is used that accounts for the entire uncertainty in loads and strength.

• According to LRFD (probability-based), different partial safety factors for the different load and strength types are used.

ASD

FS

∑

=

≥ ^m

i

n Li

R

LRFD

∑

=

≥ ^m

i i i

n L

R γ

φ

Elastic Design

Q

Engineering structures have been

designed for many years by the allowable stress design (ASD), or elastic design with satisfactory results.

Q

However, engineers have long been aware that ductile members (e.g., steel) do not fail until a great deal of yielding occurs after yield stress is first reached.

Q

This mean that such members have greater margin of safety against collapse than the elastic theory would seem to suggest.

CHAPTER 8b. INTRODUCTION TO BEAMS Slide No. 5

ENCE 355 ©Assakkaf

The Elastic Modulus

Q

The yield moment M

equals the yield stress F

times the elastic modulus S:

where

S F

M

=

⁽¹⁾

section cross

of fiber outer to

N.A.

from distance

inertia of

moment

=

=

=

c I

c S I

The Elastic Modulus

Q

The elastic modulus for a rectangular section b × d as shown in Fig. 1 can be computed by using:

– The flexural formula, or – The internal couple method

d

b

Figure 1

CHAPTER 8b. INTRODUCTION TO BEAMS Slide No. 7

ENCE 355 ©Assakkaf

The Elastic Modulus

Q

Using the Flexural Formula

– Rectangular Cross Section:

6

6 2

/ 12 / 2

12 ,

/

2

2 3

3

bd S F

F M

bd d

bd c S I c d

I bd

S F M

S M c I M I

c F M

y y

y y y

y y

y y

=

=

∴

=

=

=

⇒

=

=

=

=

=

=

The Elastic Modulus

Q

Using the Internal Couple Method:

– Rectangular Section:

N.A. ²

d

2 d

Fy

Fy

4 2

2

1 Fdb

d b F

C _y = ^y



 

 ×

=

4 2

2

1 Fdb

d b F

T _y = ^y



 

 ×

=

d

b

6 3

2 arm 4

moment Force

bd2

d F db

M_y F^y = ^y



 



×



 



=

×

=

3d 2

Figure 2

CHAPTER 8b. INTRODUCTION TO BEAMS Slide No. 9

ENCE 355 ©Assakkaf

The Plastic Modulus

Q

The resisting moment at full plasticity can be determined in a similar manner.

Q

The result is the so-called plastic moment M

.

Q

It is also the nominal moment of the section, M

n

p

M

M = ⁽²⁾

The Plastic Modulus

Q

The plastic ( or nominal) moment equals T or C times the lever arm between

them as shown.

N.A. ²

d

2 d

Fy

Fy

2 2

db b F F d

C _y = ^y



 

 ×

=

4 2

db b F F d

T _y = ^y



 

 ×

=

d

b

3d 2

Figure 3

4 2

2 2

arm 2 lever Force

bd2

d F db d F

d C T

M_p ^y × = _y

 



=

=

=

×

=

CHAPTER 8b. INTRODUCTION TO BEAMS Slide No. 11

ENCE 355 ©Assakkaf

The Plastic Modulus

Q

The plastic moment is equal to the yield stress F

times the plastic modulus Z.

Q

From the foregoing expression for a rectangular section, the plastic modulus Z can be seen to equal bd

/4.

4

4

2

2

Z bd

F bd Z F

M_{p y y}

=



 



= 

=

The Plastic Modulus

Q

The shape factor, which is equal

Q

Is also equal to

S Z S F

Z F M

M

y y y

p = =

5 . 1 6 Factor 4

Shape ₂

2

=

=

= bd

bd S Z

So, for rectangular section, the shape factor equal 1.5.

CHAPTER 8b. INTRODUCTION TO BEAMS Slide No. 13

ENCE 355 ©Assakkaf

The Plastic Modulus

Q

Shape Factor

– Definition

“The shape factor of a member cross section can be defined as the ratio of the plastic moment M_p to yield moment M_y”.

– The shape factor equals 1.50 for

rectangular cross sections and varies from about 1.10 to 1.20 for standard rolled- beam sections

The Plastic Modulus

Q

Shape Factor

The shape factor Z can be computed from the following expressions:

S Z M M

y P

=

=

Factor Shape

from Or

Factor Shape

₍₃₎

(4)

CHAPTER 8b. INTRODUCTION TO BEAMS Slide No. 15

ENCE 355 ©Assakkaf

The Plastic Modulus

Q

Neutral Axis for Plastic Condition

– The neutral axis for plastic condition is different than its counterpart for elastic condition.

– Unless the section is symmetrical, the neutral axis for the plastic condition will not be in the same location as for the elastic condition.

– The total internal compression must equal the total internal tension.

The Plastic Modulus

Q

Neutral Axis for Plastic Condition

– As all fibers are considered to have the same stress F_y in the plastic condition, the areas above and below the plastic neutral axis must be equal.

– This situation does not hold for

unsymmetrical sections in the elastic condition.

CHAPTER 8b. INTRODUCTION TO BEAMS Slide No. 17

ENCE 355 ©Assakkaf

The Plastic Modulus

Q

Plastic Modulus

– Definitions

“The plastic modulus Z is defined as the ratio of the plastic moment M_pto the yield stress F _Y.”

“It can also be defined as the first moment of area about the neutral axis when the areas above and below the neutral axis are equal.”

The Plastic Modulus

Q

Example 1

Determine the yield moment M_y, the plastic or nominal moment M_p (M_n), and the plastic modulus Z for the simply supported beam having the cross section shown in Fig. 4b.

Also calculate the shape factor and nominal load P_n acting transversely

through the midspan of the beam. Assume that F_Y = 50 ksi.

CHAPTER 8b. INTRODUCTION TO BEAMS Slide No. 19

ENCE 355 ©Assakkaf

The Plastic Modulus

Q

Example 1 (cont’d)

P_n

12 ft 12 ft

15 in.

8 in.

17 in.

15 in.

1 in.

1 in.

1 in.

Figure 4

(a) (b)

The Plastic Modulus

Q

Example 1 (cont’d)

Elastic Calculations:

( ) ( ) ( )

( )( ) ( )( ) ( )( ) _{9.974in fromlower base}

38

5 . 0 1 8 5 . 8 1 15 5 . 16 1 15

in 38 1 8 1 15 1

15 ²

+ =

= +

= + +

= yC

A

15 in.

8 in.

17 in.

15 in.

1 in.

1 in.

1 in.

N.A.

9.974 in.

( ) ( ) ( ) ( )

4

3 3

3 3

in 64 . 672 , 1

3 026 . 6 14 3

026 . 7 15 3

974 . 8 7 3 974 . 9 8

=

− +

−

x= I

( ) _69875ft-kip

12 7 . 167 50

in 7 . 974 167 . 9

64 . 672 ,

1 3 M F S .

c

S= I = = _y= _Y = =

CHAPTER 8b. INTRODUCTION TO BEAMS Slide No. 21

ENCE 355 ©Assakkaf

The Plastic Modulus

15 in.

8 in.

17 in.

15 in.

1 in.

1 in.

1 in.

N.A.

yN. A₁

A₂

Q

Example 1 (cont’d)

Plastic Calculations:

• The areas above and below the neutral axis must be equal for plastic analysis

( ) ( )( ) ( ) ( )

in 11

22 8 15 15 2

8 15

15

1 1

8 1 15

1 15

2 1

=

=

− +

=

+

=

− +

+

=

− +

=

N N

N N

N N

y y

y y

y y

A A

The Plastic Modulus

Q

Example 1 (cont’d)

Plastic Calculations (cont’d):

15 in.

8 in.

17 in.

15 in.

1 in.

1 in.

1 in.

N.A.

11 in.

A₁

A₂

( )(1 11.5) 11( )( )1 5.5 15( )( ) ( )( )1 4.5 41 2 228in³

8 + + + =

= Z

( )

12 228

50 =

=

=

=M F Z M_{p n y}

36 . 75 1 . 698 Factor 950

Shape = = =

y n

M M

36 . 7 1 . 167 Factor 228

Shape

from calculated be

also can factor shape the Note,

=

=

= S Z

CHAPTER 8b. INTRODUCTION TO BEAMS Slide No. 23

ENCE 355 ©Assakkaf

The Plastic Modulus

Q

Example 1 (cont’d)

– In order to find the nominal load P_n, we need to find an expression that gives the maximum moment on the beam. This

maximum moment occurs at midspan of the simply supported beam, and is given by

2

4

/

L M P

M

=

=

P_n

12 ft 12 ft

L

The Plastic Modulus

Q

Example 1 (cont’d)

( )

( )

24 950 4

Therefore, 4 950 24

_/2 4

=

=

=

=

=

n n

n L

P

P P

L M P

M

P_n

12 ft 12 ft

L

15 in.

8 in.

17 in.

15 in.

1 in.

1 in.

1 in.

N.A.

11 in.

A₁

A₂

CHAPTER 8b. INTRODUCTION TO BEAMS Slide No. 25

ENCE 355 ©Assakkaf

Theory of Plastic Analysis

Q

The basic theory of plastic analysis is considered a major change in the distribution of stresses after the

stresses at certain points in a structure reach the yield stress F

.

Q

The plastic theory implies that those

parts of the structure that have been

stressed to the yield stress F

cannot

resist additional stresses.

Theory of Plastic Analysis

Q

They instead will yield the amount required to permit the extra load or

stresses to be transferred to other parts of the structure where the stresses are below the yield stress F

, and thus in the elastic range and able to resist

increased stress.

Q

Plasticity can be said to serve the

purpose of equalizing stresses in cases of overload.

CHAPTER 8b. INTRODUCTION TO BEAMS Slide No. 27

ENCE 355 ©Assakkaf

Theory of Plastic Analysis

Q

Idealized Stress-Strain Diagram for Steel

– The stress-strain diagram is assumed to have the idealized shape shown in Fig. 5.

– The yield stress and the proportional limit are assumed to occur at the same point for this steel.

– Also, the stress-strain diagram is assumed to be a perfectly straight line in the plastic range.

– Beyond the plastic range there is a range of strain hardening.

Theory of Plastic Analysis

Figure 5. Stress-Strain Diagram for Steel

Unit Strain, ε Unit Str

ess, f

Fy

(

)

Slope ε E= = f Plasticity

Strain hardening

CHAPTER 8b. INTRODUCTION TO BEAMS Slide No. 29

ENCE 355 ©Assakkaf

Theory of Plastic Analysis

Q

Idealized Stress-Strain Diagram for Steel

– The strain hardening range could theoretically permit steel members to withstand additional stress.

– However, from a practical standpoint, the stains occurring are so large that they cannot be considered.

– Furthermore, inelastic buckling will limit the ability of a section to develop a moment greater than M_p, even if strain hardening is significant.