4-1 types of beams - eng.uowasit.edu.iq · 4-1 types of beams Beams are members supporting...

Chapter Four---------------------------------------------------------------------------------------------------------------Steel Beams =============================================================================================

------------------------------------------------------------------------------------------------------------------------------------------ Assist. Prof. Dr.Thaar S. Al-Gasham , Wasit University, Eng. College 64

4-1 types of beams

Beams are members supporting transverse loads. They are probably

thought of as being used in horizontal position and subjected to gravity or

vertical loads, but there are frequent exceptions- roof rafter, for example.

Below the well-known types of beams;

Joist: - they are closely spaced beams supporting the floors and roofs

of buildings.

Lintels: - they are beams over the openings in masonry walls, such as

windows and doors.

Spandrel: - they support the exterior walls of building and perhaps a

part of floor or hallway loads.

Stringer: - they are beams in the floor bridge running parallel to the

roadway.

Floor beams: - they are perpendicular to the roadway of the bridge and

are used to transfer the loading from the stringers to the supporting

girder or trusses.

Girders:- they are the largest beams in the bridge floor, by them the

loadings are transferred from the floor beams to the columns or

supporting soil.



4-2 sections used as beams

The most economical shapes to be used as beams are W-shapes.

Sometimes channels are used as beams when light loadings are applied or

small flanges are required due to available of narrow clearness. The W-

shapes have more steel concentrated in their flange than do S-shapes and

thus have larger moment of inertia and resisting moment for the same

weights. They are relatively wide and have appreciable lateral stiffness.

4-3 bending stress and the plastic moment

Initially, when the moment is applied to a beam, the stress will vary linearly

from the neutral axis to the extreme fiber of the beam as shown in fig (b), if

the applied moment is increased, there will to be a linear variation in the

stress until the yielding stress is reached in the outermost beam fiber as

shown in part c, this moment is known as the yielding moment.



If the moment is increased beyond the yielding moment, the outermost fiber

will have same yield stress, and the duty of providing the necessary

additional resisting moment will fall on the fiber nearer to the neutral axis.

This procedure will continue, with more and more fibers stressed to the

yielding stress as shown in part d and e. until finally plastic distribution is

approached, as shown in part f. note that the strain variation form the neutral

axis to the outer fiber remain linearly for all of these cases.

The applied moment is called plastic moment, when all beam fibers reached

to the yielding. The ratio of plastic to yielding moment is referred as shape

factor, which is 1.5 for rectangular sections and ranging from 1.1 to 1.2 for

standard rolled- beam section.

4-4 PLASTIC HINGES

This section is devoted to a description of the development of a plastic hinge

as in the simple beam shown in Fig. The load shown is applied to the beam

and increased in magnitude until the yield moment is reached and the

outermost fiber is stressed to the yield stress. The magnitude of the load is

further increased, with the result that the outer fibers begin to yield. The

yielding spreads out to the other fibers, away from the section of maximum



moment, as indicated in the figure. The distance in which this yielding occurs

away from the section in question is dependent on the loading conditions

and the member cross section. For a concentrated load applied at the center

line of a simple beam with a rectangular cross section, yielding in the extreme

fibers at the time the plastic hinge is formed will extend for one-third of the

span. For a W shape in similar circumstances, yielding will extend for

approximately one-eighth of the span. During this same period, the interior

fibers at the section of maximum moment yield gradually, until nearly all of

them have yielded and a plastic hinge is formed, as shown in Fig.

Although the effect of a plastic hinge may extend for some distance along

the beam, for analysis purposes it is assumed to be concentrated at one

section. For the calculation of deflections and for the design of bracing, the

length over which yielding extends is quite important.

For plastic hinges to form, the sections must be compact. The compact

sections are defined as being those which have sufficiently stocky profiles

such that they are capable of developing fully plastic stress distributions

before they buckle locally.

The student must realize that for plastic hinges to develop the members must

not only be compact but also must be braced in such a fashion that lateral

buckling is prevented.

Finally the effects of shear, torsion, and axial loads must be considered. They

may be sufficiently large as to cause the members to fail before plastic

hinges can form. In the study of plastic behavior, strain hardening is not

considered.



When steel frames are loaded to failure, the points where rotation is

concentrated (plastic hinges) become quite visible to the observer before

collapse occurs.

4-5 elastic design

Until recent years, almost all steel beams were designed on the basis of the

elastic theory. The maximum load that a structure could support was

assumed to equal the load that first caused a stress somewhere in the

structure to equal the yield stress of the material. The members were

designed so that computed bending stresses for service loads did not exceed

the yield stress divided by a safety factor. Engineering structures have been

designed for many decades by this method, with satisfactory results. The

design profession, however, has long been aware that ductile members do

not fail until a great deal of yielding occurs after the yield stress is first

reached. This means that such members have greater margins of safety

against collapse than the elastic theory would seem to indicate.



4.6 The Plastic Modulus

The yield moment My equals the yield stress times the elastic modulus. The

elastic modulus equals (I/c) or (bd2/6) for a rectangular section, and the yield

moment equals (Fy b d2 /6). This same value can be obtained by considering

the resisting internal couple shown in Fig.

The elastic section modulus can again be seen to equal (bd2/6) for a

rectangular beam.

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

The result is the so-called plastic moment, Mp It is also the nominal moment

of the section, Mn. This plastic, or nominal, moment equals T or C times the

lever arm between them. For the rectangular beam of Fig.



The plastic moment is said to equal the yield stress times the plastic section

modulus. From the foregoing expression for a rectangular section, the plastic

section modulus Z can be seen to equal bd2/4. The shape factor, which

equals Mp/My or Z/S, is 1.50 for a rectangular section. A study of the plastic

section modulus determined here shows that it equals the statically moment

of the tension and compression areas about the plastic neutral axis. 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. As all fibers are

considered to have the same stress (Fy) 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.

----------------------------------------------------------------------------------------------------

Example (1):- determine My, Mp, and shape factor for the steel beam of

cross-sectional area shown in figure below, then determine the nominal

distributed load that can be placed on 12-ft simple span beam. Fy=50 ksi



Solution

1- Determine the centroid of section y' from top

�̅� =12 (

12) (

14) + 6(

12

)(2)(312

)

12 (12) + 6(

12

)(2)= 1.875 𝑖𝑛

2- Determine the inertia moment (I)

𝐼𝑥 =12(

12

)3

12+ 12 (

1

2) (1.875 −

1

4)

2

+

12

(6)3

12+ 6 (

1

2) (2) (1.875 − 3

1

2 )

2

= 49.81 𝑖𝑛4

3- Determine the elastic modulus section



𝑠 =𝐼

𝑐

c for compression= 1.875, and for tension=6.5-1.875=4.625'' (control)

𝑆𝑥 =𝐼𝑥

𝑐=

49.81

4.625= 10.77 𝑖𝑛3

4- Determine My

𝑀𝑦 = 𝐹𝑦𝑆 = 50(10.77) = 538.5 𝑖𝑛 − 𝑘 = 44.875 𝑓𝑡 − 𝑘

5- Determine the yielding load

𝑀𝑦 =𝑤𝑛𝑙2

8

𝑤𝑛 =44.875 ∗ 8

122= 2.49 𝑘/𝑓𝑡

6- Determine the plastic centroid, in which the tension and compression

areas are equal. In this example, the plastic centroid located at fiber

separating the web and flange.

7- Determine the plastic section modulus

𝑍𝑥 = (12) (1

2) (

1

4) + 2(6) (

1

2) (3) = 19.5 𝑖𝑛3

8- Determine the plastic moment

𝑀𝑝 = 𝐹𝑦𝑍𝑥 = 50(19.5) = 975 𝑖𝑛 − 𝑘 = 81.25 𝑓𝑡 − 𝑘

9- Determine the plastic load

𝑀𝑝 =𝑤𝑛𝑙2

8

𝑤𝑛 =81.25 ∗ 8

122= 4.51 𝑘/𝑓𝑡

10- Determine the shape factor

𝑠ℎ𝑎𝑝𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 =𝑍

𝑠=

𝑀𝑝

𝑀𝑦=

19.5

10.77= 1.81



4-7 The Virtual-Work Method

Although a plastic hinge may have formed in a statically indeterminate

structure, the load can still be increased without causing failure if the

geometry of the structure permits. The plastic hinge will act like a real hinge

insofar as increased loading is concerned. As the load is increased, there is

a redistribution of moment, because the plastic hinge can resist no more

moment. As more plastic hinges are formed in the structure, there will

eventually be a sufficient number of them to cause collapse. Actually, some

additional load can be carried after this time, before collapse occurs, as the

stresses go into the strain hardening range, but the deflections that would

occur are too large to be permissible.













4-8 analysis and design of steel beams

The steel beams are classified into three categories; compact, non-compact

and slender sections.

1- Compact sections; they are sections having sufficiently stocky profile

so that it is capable of developing fully plastic stress distribution before

their web or flange buckle locally.

2- Non-compact sections: they are sections for which the yield stress can

be reached in some, but in all, of its compression elements.







4-8-1 Flexural strength of compact sections

These section can be failed in three zones depending on the

unbraced length as shown in figure below

The term Lb is used throughout this chapter to describe the length

between points which are either braced against lateral displacement of the

compression flange or braced against twist of the cross section.



4.8.1.1 Zone 1;-

If 𝐿𝑏 ≤ 𝐿𝑃 𝑜𝑟

If elastic analysis approached is used

If plastic analysis approached is used



Example :-

Solution

1- Specify the type of beam

2- Specify the failure zone

The beam is continuous lateral support Lb=0, thus it will fail in zone 1

3- Calculate Mn

From table (1-1) W16x31 have Zx=54.0



𝑀𝑛 = 𝑀𝑝 = 𝐹𝑦𝑍𝑥 =50(54.0)

12= 225 𝑓𝑡 − 𝑘

LRFD 𝑀𝑢 = 0.9 𝑀𝑛 = 0.9𝑥225 = 202.5 𝑓𝑡 − 𝑘

ASD 𝑀𝑎 =𝑀𝑛

1.67=

225

1.67= 134.7 𝑓𝑡 − 𝑘

Note;- you can use table (3-2) instead of table (1-1) for W-shapes with

fy=50 kip.

4- Computing the applied moment

𝑤𝑑 = 450 + 31 = 481𝐼𝑏

𝑓𝑡

For LRFD

𝑤𝑢 = 1.6 𝑤𝑙 + 1.2𝑤𝑑 = 1.6(550) + 1.2(481) = 1457.2𝐼𝑏

𝑓𝑡= 1.457𝑘/𝑓𝑡



𝑀𝑎𝑝𝑝𝑙𝑖𝑒𝑑 =𝑤𝑙2

8=

1.457(30)2

8= 163.9 𝑓𝑡 − 𝑘𝑖𝑝𝑠 < 202.5

∴ 𝑡ℎ𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑠𝑎𝑡𝑖𝑠𝑓𝑎𝑐𝑡𝑜𝑟𝑦 𝑎𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 𝐿𝑅𝐹𝐷 𝑚𝑒𝑡ℎ𝑜𝑑

For ASD

𝑤𝑢 = 𝑤𝑙 + 𝑤𝑑 = (550) + (481) = 1031𝐼𝑏

𝑓𝑡= 1.031𝑘/𝑓𝑡

𝑀𝑎𝑝𝑝𝑙𝑖𝑒𝑑 =𝑤𝑙2

8=

1.031(30)2

8= 115.987 𝑓𝑡 − 𝑘𝑖𝑝𝑠 < 134.7

∴ 𝑡ℎ𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑠𝑎𝑡𝑖𝑠𝑓𝑎𝑐𝑡𝑜𝑟𝑦 𝑎𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 𝐴𝑆𝐷 𝑚𝑒𝑡ℎ𝑜𝑑

----------------------------------------------------------------------------------------------------

4.8.1.2 zone 2

In this zone inelastic buckling will occur

Note;- table (3-2) lists value of Lr for W-shapes with fy=50 kis

If table (3-1) is used the equation below can be used rather than above

equation.



Cb is called bending coefficient and can be calculated as following





Example:-



4.8.1.3 zone 3

In this zone elastic buckling will occur;

----------------------------------------------------------------------------------------------------

Example:-

Determine the flexural strength of a W 14x68 of A242 steel subjected to

A. Continuous lateral support.

B. Unbraced length of 20 ft with Cb=1.0

C. Unbraced length of 30 ft with Cb=1.0



Solution

From table(1-1)

𝑍𝑥 = 115, 𝑟𝑦 = 2.46, 𝑟𝑡𝑠 = 2.80, 𝐽 = 3.01, 𝑆𝑥 = 103, ℎ𝑜 = 13.3

A. Continuous lateral support.

Lb=0 , the beam is in zone 1

𝑀𝑛 = 𝑀𝑝 = 𝐹𝑦 𝑍𝑥

𝑀𝑛 = 𝑀𝑝 =50𝑥115

12= 5750 𝑘𝑖𝑝 − 𝑖𝑛 = 479.2 𝑘𝑖𝑝 − 𝑓𝑡

LRFD 𝑀𝑢 = 0.9 𝑀𝑛 = 0.9𝑥479.2 = 431.28 𝑘𝑖𝑝 − 𝑓𝑡


1.67= 479.2/1.67 = 286.9 𝑘𝑖𝑝 − 𝑓𝑡

B. Unbraced length of 20 ft with Cb=1.0

Lb must be compared with both Lp and Lr to indicate the failure zone

𝐿𝑏 = 1.76 𝑟𝑦√𝐸

𝐹𝑦= 1.76𝑥2.46√

29000

50= 104.3 𝑖𝑛 = 8.692 𝑓𝑡



𝐿𝑟 = 1.95 𝑟𝑡𝑠

𝐸

0.7𝐹𝑦 √

𝐽𝑐

𝑆𝑥ℎ𝑜

√1 + √1 + 6.76 (0.7𝐹𝑦𝑆𝑥ℎ𝑜

𝐸𝐽𝑐)

2

C=1.0 for doubly symmetric I-section

𝐿𝑟

= 1.95 (2.80)29000

0.7(50) √

(3.01)(1.0)

(103)(13.3)√1 + √1 + 6.76 (

0.7(50)(103)(13.3)

(29000)(3.01)(1.0))

2

= 351.24𝑖𝑛 = 29.27 𝑓𝑡

𝐿𝑏 < 20 < 𝐿𝑟 ∴ 𝑡ℎ𝑒 𝑏𝑒𝑎𝑚 𝑤𝑖𝑙𝑙 𝑓𝑎𝑖𝑙 𝑖𝑛 𝑧𝑜𝑛𝑒 2

𝑀𝑛 = 𝐶𝑏 [𝑀𝑝 − (𝑀𝑝 − 0.7𝐹𝑦𝑆𝑥) (𝐿𝑏 − 𝐿𝑝

𝐿𝑟 − 𝐿𝑝)] ≤ 𝑀𝑝

𝑀𝑛 = 1.0𝑥 [5750 − (5750 − 0.7𝑥50𝑥103) (20 − 8.692

29.27 − 8.692)] = 4571.3 𝑘 − 𝑖𝑛

< 5750 𝑜. 𝑘

𝑀𝑛 =4571.3

12= 380.94 𝑘 − 𝑖𝑛

LRFD 𝑀𝑢 = 0.9 𝑀𝑛 = 0.9𝑥380.94 = 342.8 𝑘𝑖𝑝 − 𝑓𝑡


1.67= 380.94 /1.67 = 228.1 𝑘𝑖𝑝 − 𝑓𝑡

C. Unbraced length of 30 ft with Cb=1.0

Since 𝐿𝑟 < 30 the beam will fail in zone 3

𝐹𝑐𝑟 =𝐶𝑏𝜋2𝐸

(𝐿𝑏 𝑟𝑡𝑠⁄ )2√1 + 0.078

𝐽𝑐

𝑆𝑥ℎ𝑜(

𝐿𝑏

𝑟𝑡𝑠)

2



𝐹𝑐𝑟 =1.0𝑥𝜋2𝑥29000

(30𝑥12 2.80⁄ )2√1 + 0.078

3.01𝑥1.0

(103)(13.3)(

30𝑥12

2.8)

2

= 33.9 𝑘𝑠𝑖

𝑀𝑛 = 𝐹𝑐𝑟𝑆𝑥 = 33.9𝑥103 = 3491.7 𝑘 − 𝑖𝑛 = 290.98 𝑘 − 𝑓𝑡 < 479.2 𝑜. 𝑘

LRFD 𝑀𝑢 = 0.9 𝑀𝑛 = 0.9𝑥290.98 = 261.9 𝑘𝑖𝑝 − 𝑓𝑡


1.67= 290.98 /1.67 = 174.2 𝑘𝑖𝑝 − 𝑓𝑡

Important note: - you can use Table (3-2) to directly find Mp, Lb, and Lr .

Since section above have yielding stress of 65 ksi

----------------------------------------------------------------------------------------------------

4-9 Flexural stress of non-compact sections

Most W, M, S, and C shapes are compact. A few are non-compact

because of flange width-thickness ratio, but none are slender.

For I-shape, the webs are always compact. Thus, the checking will be done

on flange only



Example:-

Referring to Table (3-2), the section W14x90 have footnote ''f'', which means

it is non compact section due to flange.





For ASD solution

𝑀𝑎 =5320

1.67= 3185.6 𝑘 − 𝑖𝑛 = 265.5 𝑘𝑖𝑝 − 𝑓𝑡

< 354.4 𝑡ℎ𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑑𝑜𝑠𝑒 𝑛𝑜𝑡 ℎ𝑎𝑣𝑒 𝑎𝑑𝑒𝑞𝑢𝑎𝑡𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑎𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔

𝑡𝑜 𝐴𝑆𝐷 𝑚𝑒𝑡ℎ𝑜𝑑

4-10 summary of procedure adopting to evaluate of Flexural

stress of steel sections





4-11 shear strength of steel beams





=========================================================

Example:-



4-12 block shear



=========================================================

Example:-



=========================================================

4-13 deflection



=========================================================



4-14 design of steel beams

=========================================================

Example:-





Another way of checking as follows

The applied moment is 821.4 ft-k is smaller than LRFD flexural strength of

W24x84 that is 840 ft-k as referred in Table (3-2)







Important note:-



4-15 floor and roof framing system



Example:-

LRFD Solution

1. Evaluating the wu as follows;

𝑤𝑢 = 1.6 𝐿. 𝐿 + 1.2 𝐷. 𝐿 = 1.6(8)(100) + 1.2(8)(150)5

12= 1880

𝐼𝑏

𝑓𝑡

= 1.88𝑘𝑖𝑝

𝑓𝑡



2. Evaluating the applied moment

𝑀𝑢 =𝑤𝑢𝑙2

8=

1.88(20)2

8= 94𝑘𝑖𝑝 − 𝑓𝑡

3. Referring to table (3-2) to select the lightest section with flexural

strength equal to or greater than Mu

Select W10x22

4- check the beam self-weight

𝑤𝑢 = 1880 + 1.2(22) = 1906.4𝐼𝑏

𝑓𝑡= 1.906

𝐼𝑏

𝑓𝑡



𝑀𝑢 =𝑤𝑢𝑙2

8=

1.906 (20)2

8= 95.3 𝑘𝑖𝑝 − 𝑓𝑡 < 97.5 𝑜. 𝑘

5- Checking for shear

𝑉𝑢 =𝑤𝑢𝑙

2=

1.906 (20)

2= 19.06 𝑘𝑖𝑝 < 73.2 𝑜. 𝑘

---- use section W10x22 according to LRFD method.

ASD Solution

1. Evaluating the wa as follows;

𝑤𝑎 = 𝐿. 𝐿 + 𝐷. 𝐿 = (8)(100) + (8)(150)5

12= 1300

𝐼𝑏

𝑓𝑡= 1.3

𝑘𝑖𝑝

𝑓𝑡

2. Evaluating the applied moment

𝑀𝑎 =𝑤𝑎𝑙2

8=

1.3(20)2

8= 65𝑘𝑖𝑝 − 𝑓𝑡

3. Referring to table (3-2) to select the lightest section with flexural

strength equal to or greater than Mu

Both above sections do have ASD moment greater than the applied

one, but W12x22 is the lightest one

Select W12x22

4- check the beam self-weight

𝑤𝑢 = 1300 + (22) = 1322𝐼𝑏

𝑓𝑡= 1.322

𝐼𝑏

𝑓𝑡

𝑀𝑎 =𝑤𝑢𝑙2

8=

1.322 (20)2

8= 66.1 𝑘𝑖𝑝 − 𝑓𝑡 < 73.1 𝑜. 𝑘

1- Checking for shear



𝑉𝑎 =𝑤𝑎𝑙

2=

1.322 (20)

2= 13.22 𝑘𝑖𝑝 < 64.0 𝑜. 𝑘

---- use section W12x22 according to ASD method.



4-15 design of steel beams using manual charts

In part 3 of manual there are figures named as Table (3-1). As shown in

figure below.







Example:-











Example:-















========================================================

Example:-









Date post:	27-Jun-2020
Category:	Documents
Upload:	others
View:	24 times
Download:	0 times

4-1 types of beams - eng.uowasit.edu.iq · 4-1 types of beams Beams are members supporting...

Documents