FLOW CHART FOR DESIGN OF BEAMS - WEC...

FLOW CHART FOR DESIGN OF BEAMS

Write Known Data

Estimate self-weight of the member.

a. The self-weight may be taken as 10 percent

of the applied dead UDL or dead point load

distributed over all the length.

b. If only live load is applied, self-weight may

be taken equal to 5 percent of its magnitude.

c. In case only factored loads are given, self-

wt. may be taken equal to 3 % of the given loads.

Calculate Factored Loads

Draw B.M. and S.F. Diagrams

Calculate Cb For Each Unbraced Segment

Find Mu,max, Vu,max, Lb for each segment and guess

which segment is the most critical.

Design this segment first and then check for

others.

Assume the section to be compact without LTB in

the start and calculate Zx accordingly.

Assumed Zx,req = φb = 0.9 ;

Fy = 250 MPa for A36 steel

yb

u

F

M

φ610×

Zsel ≥ Zreq

Selection of Section Minimum weight

d ≥ dmin

dmin = For ∆max = L/3605500

lyF

= L /22 for A36 steel and simply supported beams

For ∆max required to be lesser than L /360, like L /500

or L /800, find (Ix)req from the deflection formula, with

only the live load acting, and select section such that Ix≥ (Ix)req.

Method 1 Use Of Selection Tables

1. Enter the column headed Zx and find a value

equal to or just greater than the plastic

section modulus required.

2. The beam corresponding to this value in the

shape column and all beams above it have

sufficient flexural strength based on these

parameters.

3. The first beam appearing in boldface type

(top of a group) adjacent to or above the

required Zx is the lightest suitable section.

4. If the beam must have to satisfy a certain

depth or minimum moment of inertia

criterion, proceed up the column headed

“Shape” until a beam fulfilling the

requirements is reached.

5. If Cb > 1.0, use Lm in place of Lp for the

approximate selection.

6. If Lb is larger than Lm of the selected section,

use the unbraced design charts.

7. Apply moment capacity, shear capacity,

deflection and all other checks.

8. The column headed φbMp may also be used

in place of the Zx column in the above

method.

Method 2: Use Of Unbraced Design Charts

This method is applicable in cases where the

above method is not fully applicable and Lb ≥ Lp.

The design charts are basically developed for

uniform moment case with Cb = 1.0.

Following notation is used to separate full plastic,

inelastic LTB, and elastic LTB ranges:

Solid Circle represents Lp

Hollow Circle represents Lr

1. According to Mu in kN-m units and Lb in

meters, enter into the charts.

2. Any section represented by a curve to the

right and above ( ) the point selected

in No. 1 will have a greater allowed unbraced

length and a greater moment capacity than

the required values of the two parameters.

3. A dashed line section is not an economical

solution.

If dashed section is encountered while

moving in top-right direction, proceed

further upwards and to the right till the first

solid line section is obtained.

Select the corresponding section as the trial

section, and it will be the lightest available

section for the requirements.

4. If Cb > 1.0, use Mu,req = Mu / Cb but check

that the selected section has φbMp > Mu.

Check the three conditions of compact section for

internal stability, namely,

1. web continuously connected with flange,

2. flange stability criterion, and

3. web stability criterion.

If any one out of the above three is not satisfied,

revise the section.

Either calculate Lp, Lr, and Lm or find their values

from beam selection tables.

Lp = (mm)

= 0.05 ry (m) for A36 steel

yfy FEr76.1

Lr =

27.0

76.6117.0

95.1

++cJ

hS

E

F

hS

cJ

F

Er oxy

oxy

ts

Mr = 0.7Fy Sx/106 kN-m

= =

for doubly symmetric I-sections

2

tsrx

wy

S

CI

x

oy

S

hI

2

c = 1.0 for a doubly symmetric I-shape

ho = d – tf

BF =pr

rp

LL

MM

−−

Lm = Lp + ≤ Lr

−b

bp

C

C

BF

M 1

Calculate design flexural strength:

1- If Lb ≤ Lm Mn = Mp = Zx Fy / 106 (kN – m)

2- If Lm < Lb ≤ Lr Mn = Cb [Mp – BF(Lb – Lp)]

≤ Mp (kN – m)

3- If Lb > Lr Mn = CbFcrSx ≤ Mp

where Fcr =

2

2

2

078.01

+

ts

b

ox

ts

b

b

r

L

hS

cJ

r

L

EC π

Bending strength check:

Mu ≤ φbMn OK

If not satisfied revise the trial selection.

Shear check:

For ≤ 2.24 (= 63.4 for A36 steel)

Cv = 1.0wt

hywFE /

φv Vn = Fyw Aw Cv (kN)1000

6.09.0 ×If not satisfied, revise the section.

Deflection check:

Find ∆act due to service live loads.

∆act ≤ L/360 or other specified limit OK

Check self-weight:

Calculated self weight ≤ 1.2 × assumed self

weight OK

Otherwise, revise the loads and repeat the

calculations.

Write final selection using standard designation.

1- For buildings, L/d ratio is usually limited to a

maximum of 5500 / Fy.

L /d ≤ 5500 / Fy

∴ dmin = Fy L /5500 (L /22 for A36 steel)

2- For bridge components and other beams

subjected to impact or vibratory loads,

L /d ≤ 20

3- For roof purlins,

L /d ≤ 6900 / Fy (27.5 for A36 steel,

sometimes relaxed to a value equal to 30)

DEFLECTIONS

The actual expected deflections may be calculated

using the mechanics principles.

However, results given in Manuals and Handbooks

may also be used directly.

Some of the typical deflection formulas are

reproduced here.

1- For uniformly loaded and simply supported

beams,

∆max =

EI

w

384

5 4ll

2- For uniformly loaded continuous beams,

∆midspan = [Mc – 0.1(Ma + Mb)]

Where Mc = magnitude of central moment,

Ma, Mb = magnitude of end moments.

EI48

5 2l

3- For simply supported beams subjected to

point load (refer to Figure 4.16),

∆midspan = where a ≤ L/ 2( )22

43

12a

EI

aP −l

a b

P

4- For overhanging part of beam subjected to

UDL,a

No Loadw per meter

l

∆max ≅ ( )aEI

aw34

24

3 +l

5. For the above case, with UDL also present

within supports,

∆max ≅ ( )332 3424

aaEI

aw +− ll

6- For overhanging part of beam subjected to

point load,

PNo load

al

∆max = ( )EI

aaP

3

2 +l

Example 4.1: BEAMS WITH CONTINUOUS

LATERAL SUPPORT

Design a 7m long simply supported I-section beam

subjected to service live load of 5 kN/m and

imposed dead load of 6 kN/m, as shown in Figure

4.18. The compression flange is continuously

supported.

Use (a) A36 steel and (b) steel with fy = 345 MPa

and permissible live load deflection of span / 450.

Solution:

In beams with continuous lateral support,

unbraced length is not applicable or it may be

assumed equal to zero in calculations.

Assumed self weight = 10% of

superimposed DL

= 0.6 kN/m

wu = 1.2D + 1.6L

= 1.2 × 6.6 + 1.6 × 5

= 15.92 kN/m

97.51 kN-m

B.M.D.

15.92 kN/m

55.72 kN55.72 kN7 m

55.72 kN

55.72 kN

S.F.D.

If the beam is continuously braced, Cb value is not

applicable but may be considered equal to 1.0 in case it is

required in the formulas.

Mu = Mmax = 97.51 kN-m

Vu = Vmax = 55.72 kN

Compression flange continuously braced.

(a) A36 Steel

Assuming the section to be internally compact and

knowing that there is no LTB,

(Zx)req = = = 433.4 × 103 mm3

yb

u

F

M

φ610×

2509.0

1051.97 6

××

dmin = = 318 mm22

7000

Zsel ≥ Zreq

Selection of section Min. weight section

d ≥ dmin

Consulting beam selection tables (Reference-1),

following result is obtained:

W310 × 32.7 provides sufficient strength but

depth is little lesser.

Select W360 × 32.9; Zx = 544 × 103 mm3

Check internal compactness of section as under:

1. web is continuously connected OK

2. = 7.5 < λp = 10.8 OK

3. = 53.3< λp =107 OK

f

f

t2

b

wt

h

∴ The section is internally compact.

Continuously braced conditions imply: Lb < Lp

and Cb = not included in formulas

φbMn = φbMp

= 0.9 × 544 × 103 × 250 / 106 ≅ 121 kN-m

Mu = 97.51 kN-m < φbMp OK

h/tw =53.3 < 63.4

⇒ shear yield formula is applicable

φvVn = FywAwCv

= × 250 × 349 × 5.8 × 1.0

= 273.27 kN

1000

6.09.0 ×

1000

6.09.0 ×

Vu = 55.72 (kN) < φvVn OK

∆act due to service live load =

= = 9.44 mm

EI

w 4

384

5 ll×( )

4

4

108280200000

70005

384

5

××××

∆all = L/360 = 7000 / 360 = 19.44 mm

∆act < ∆all OK

Self-weight = = 0.32 kN/m1000

81.99.32 ×

This is lesser than assumed self-weight of 0.6 kN/m in

the start. OK

Final Selection: W360 × 32.9

(b) Steel With Fy = 345 MPa



(Zx)req = = = 314.0 × 103 mm3

yb

u

F

M

φ610×

3459.0

1051.97 6

××

dmin = = 438 mm16

7000

This dmin is much larger, so the direct deflection criteria

will be used.

EI

wL

4

384

5

450

ll ==

= 5024 × 104 mm4

E

wI L

req

3

384

4505 l×=200000

70005

384

4505 3××

Zsel ≥ Zreq


I ≥ Imin

Consulting beam selection tables (Reference-1),

following result is obtained:

W310 × 28.3 provides sufficient strength and

moment of inertia.

Ix = 5410 × 104 mm4 ; Zx = 405 × 103 mm3

Check internal compactness of section as under:

1. web is continuously connected OK

2. bf / 2tf = 5.7 < λp = 9.1 OK

3. h / tw= 46.2 < λp =90.5 OK∴ The section is internally compact.

Continuously braced conditions imply: Lb < Lp

and Cb = N.A.

φbMn = φbMp

= 0.9 × 405 × 103 × 345 / 106

≅ 125.8 kN-m

Mu = 97.51 kN-m < φbMp OK

h/tw =53.3 < = 53.9

⇒ shear yield formula is applicable

yFE /24.2

φvVn = FywAwCv

= × 345 × 309 × 6.0 × 1.0

= 345.40 kN

1000

6.09.0 ×1000

6.09.0 ×

Vu = 55.72 (kN) < φvVn OK

∆act < ∆all Already satisfied OK

Self-weight = = 0.28 kN/m1000

81.93.28 ×

This is lesser than assumed self-weight of 0.6 kN/m in

the start. OK

Final Selection: W310 × 28.3

Example 4.2: BEAMS WITH COMPRESSION

FLANGE LATERALY SUPPORTED (BRACED)

AT REACTION POINTS ONLY

Design the beam of Figure 4.19 with the lateral

bracing only provided at the reaction points.

wD = 8 kN/m

wL = 10 kN/m

3.5m3.5m

PD = 60 kN

PL = 50 kN

Solution:

Self load = 0.1

= 1.7 kN/m

+7

608

wu = 1.2 × 9.7 + 1.6 × 10

= 27.64 kN/m

Pu = 1.2 × 60 + 1.6 × 50

= 152 kN

The factored loading and the shear force and bending

moment diagrams are shown in Figure 4.20.

172.74 kN

172.74 kN S.F.D.76

76

B.M.D.

169.30 kN-m

266.00 kN-m

wu = 27.64 kN/m

172.74 kN172.74 kN 3.5m3.5m

Pu = 152 kN

Calculation of Cb:

MB = Mmax = 435.3 kN-m

MA = MC = 172.4 × 1.75 – 27.64 × 1.752/2

= 259.97 kN-m

Cb = Rm = 1.24

≤ 3.0 (where Rm = 1.0)97.259630.4355.6

30.4355.12

×+××

Mmax = 435.3 kN-m : Vmax = 172.74 kN : Lb = 7 m

Assuming the section to be internally compact with no LTB,

(Zx)req =

= = 1935 × 103 mm3

dmin. = L/22 = 7000/22 = 318 mm

yb

u

F

M

φ610×

2509.0

103.435 6

××

Zsel ≥ Zreq


d ≥ dmin

Using beam selection tables of Reference-1,

Trial section: W610 × 82

Mp = 549.00 kN-m, Mr = 327.25 kN-m,

BF = 64.99, Lr = 5.10 m

Lp = 1.67 m : Lm = 1.67 + = 3.33 m

−24.1

124.1

99.64

00.549

Lb >> Lm ⇒ revise the section using beam

selection charts of Reference-1

For Lb = 7m and Mu,eq = 435.3 / 1.24 = 351.05 kN-

m, from design charts of Reference-1,

Select W410 × 100

φbMp = 479.25 kN-m > Mu OK

d > dmin OK

Check internal compactness of section:

1- web is continuously connected OK

2- = 7.7< λp = 10.8 OK

3- = 35.9 < λp = 107 OK

∴ The section is internally compact.

f

f

t

b

2

wt

h

Lp = 3.11 m

Lr = 10.00 m

BF = 28.53 kN

Mp = 532.50 kN-m

Lm = Lp + ≤ Lr

Lm = 3.11 + = 6.72 m

−b

bp

C

C

BF

M 1

−24.1

124.1

53.28

50.532

Lm < Lb ≤ Lr:φbMn = Cb × φb [Mp – BF(Lb − Lp)] ≤ φbMp

= 1.24 × 0.9 [532.50 – 28.53(7.00 − 3.11)]= 470.44 kN-m

Mu = 435.3 kN-m ≤ φbMn OK

h/tw = 35.9 < 63.4 ⇒φvVn = Fyw Aw Cv

= × 250 × 415 × 10.0 × 1.0

= 560.25 kN

1000

6.09.0 ×1000

6.09.0 ×

Vu = 172.74 kN < φvVn OK

∆act = ( ) ( ) ( )( )22

44

4

3500700075.0103970020000012

350050000

1039700200000

700010

384

5 −×××××+××

××

= 3.94 + 4.50 = 8.44 mm

L/360 = 19.44 mm

∆act < L/360 OK

Self weight = = 0.981 kN/m

< 1.7 kN/m assumed in the start OK

1000

81.9100×

Final Selection: W410 × 100

Example 4.3: BEAMS WITH COMPRESSION

FLANGE LATERALLY SUPPORTED AT

INTERVALS

Design the beam of Figure 4.21, ignoring self-weight

and deflection check. Compression flange is laterally

supported at points A, B and C.

Solution:

The factored loading and the shear force and bending

moment diagrams are shown in Figure 4.21.

7m 8m 6m

A C

BD

PD =35 kN

PL =30 kN

PD =60 kN

PL =50 kNwD =3 kN/m

wL =2 kN/m

Pu =90 kNPu =152 kN

wu =6.8 kN/m

208.04 kN81.56 kN

9090

118.04118.04

33.9681.56

S.F.D.

(−)

(+)

540.00 kN-m

404.32 kN-m

B.M.D.

Mu = Mmax = 540 kN-m

Mu = 404.32 kN-m for portion AC

Mu = 540.00 kN-m for portions CB and BD

Vu = Vmax = 118.04 kN

Lb for portion AC = 7 m

Lb for portion CB = 8 m

Lb for portion BD = 6 m

Calculation of Cb:

i) Portion AC

M = 81.56 x – 6.8 x2/2

Mmax = 404.32 kN-m

MA = 132.32 kN-m (x = 1.75 m)

MB = 243.81 kN-m (x = 3.50 m)

MC = 334.48 kN-m (x = 5.25 m)

Cb = Rm

≤ 3.0 (where Rm = 1.0)

= 1.49

48.334381.243432.132332.4045.2

32.4045.12

×+×+×+××

ii) Portion CB

M = 540 – 118.04 x (x starts from the end B)

Mmax = 540.00 kN-m

MA = 303.92 kN-m (x = 2 m)

MB = 67.84 kN-m (x = 4 m)

MC = 168.24 kN-m (x = 6 m)

Cb = Rm

≤ 3.0 (where Rm = 1.0)

= 2.22

24.168384.67492.303300.5405.2

00.5405.12

×+×+×+××

iii) Portion BD

Cb = 1.0

Portion (Mu)max

(kN-m)

(Vu)max

(kN)

Lb

(m)

Cb

AC 404.32 81.56 7 1.49

CB 540.00 118.04 8 2.22

BD 540.00 90.00 6 1.00

Greater value of Mu makes a segment more

critical.

Similarly, a longer unbraced length reduces the

member capacity.

However, smaller value of Cb is more critical for a

particular segment.

The effect of these three factors together

determines whether a selected segment is actually

more critical than the other segments.

The decision about which part of the beam is

more critical for design depends on experience

but may not always be fully accurate.

If a wrong choice is made for this critical section,

the procedure will correct itself.

Only the calculations will be lengthy in such cases

with the end result being always the same.

For this example, assume that portion BD is more

critical and first design it. Later on, check for the

other two portions.

dmin ≈ = = 682 mm22

l

22

15000

Depth for portion AB may be relaxed a little due to the

presence of cantilever portion.

However, the portion BD may have its own larger depth

requirements.

Portion BD

Assuming the section to be internally compact

with no LTB,

Assumed (Zx)req = =

= 2400 × 103 mm3yb

u

F

M

φ610×

2509.0

10540 6

××

From beam selection tables, the trial section is:

W610 × 92

Lp = 1.75 m, Lr = 5.33 m, Lb = 6 m,

Lm = Lp (as Cb = 1.0)

Lb >> Lm ⇒ use beam selection charts

For Mu = 540 kN-m and Lb = 6m

From unbraced beam curves ⇒ W610 × 125

is the first choice but select W690×125 to

satisfy dmin.

Trial Section: W690×125

Actual d = 678 (still a little lesser but within

permissible limits)

Lp = 2.62 m ; Lr = 7.67 m ; BF = 77.0 kN ; Lm = Lp ;

Mp = 999.50 kN-m ; Mr = 610.75 kN-m

Compactness Check:

1- web is continuously connected OK

2- bf/2tf = 7.8 ≤ λp = 10.8 OK

3. h/tw = 52.7 ≤ λp = 107 OK∴ Section is internally compact.

Lm < Lb ≤ Lr

φbMn = Cb × φb [Mp – BF(Lb – Lp)] ≤ φbMp

= 1.0 × 0.9 [999.50 – 77.0(6 – 2.62)]

= 665.22 kN-m ≤ 899.55 kN-m

Mu = 540 kN-m < φbMn OK

h/tw = 52.7 ≤ 63.4 ⇒φvVn = × 250 × 678 × 11.7 × 1.0

= 1070.9 kN

1000

6.09.0 ×

Vu = 118.04 kN ≤ φvVn OK

Check for Portion AC

Lb = 7m ; Cb = 1.49

Lp < Lb ≤ Lr

φbMn = Cb × φb [Mp – BF(Lb – Lp)] ≤ φbMp

= 1.49 × 0.9 [999.50 – 77.0(7 – 2.62)]

= 887.92 kN-m ≤ 899.55 kN-m

∴ φbMn = 887.92 kN-m

Mu = 404.32 kN-m < φbMn OK

For W690×125:

Sx = 3490×103 mm3

Mp = 999.50 kN-m ; Mr = 610.75 kN-m

Check For Portion CB

φbMp = 899.55 kN-m

J = 117 ×104 mm4 ; ry = 52.6 mm

Lb = 8m ; Cb = 2.22

Iy = 4410 ×104 mm4 ; ry = 52.6 mm

d = 678 mm ; tf = 16.3 mm

ho = d − tf = 678 − 16.3 = 661.7 mm

tsrx

oy

S

hI

2

3

4

1034902

7.661104410

××××

=

= = 64.66 mm

Lb > Lr

= 123.86

φbMn = φb × Sx / 106 ≤ φbMp

59.64

8000=ts

b

r

L

2

2

2

078.01

+

ts

b

ox

ts

b

b

r

L

hS

cJ

r

L

EC π

632

32

2

10/10349086.1237.661103490

11170000078.01

86.123

20000022.29.0 ×××××

×+××× π=

φbMn = 1137.10 kN-m ≤ 899.55 kN-m

Hence φbMn = φbMp = 899.55 kN-m and portion BD

is most critical as expected earlier.

Mu = 540 kN-m < φbMn OK

Deflection Check:

Requires detailed calculations for ∆actual that is not

asked for in this example. Very approximate

calculations are performed as under:

∆max for cantilever ≈where a is the cantilever length and l is the span

=

( )EI

aPa

3

2 +l

( )4

2

10000,119000,2003

000,6000,15000,6000,30

×××+×

∆max = 31.76 mm (will be reduced by loads on span ‘l’)

Also ∆max considering one end of the overhang as

fixed

≈=

EI

Pa

3

3

4

3

10000,119000,2003

000,6000,30

××××

= 9.08 mm

l/360 = 6000/360 = 16.67mm

∴ Deflection may be critical, detailed

calculations are recommended.

Self weight = = 1.23 kN/m

(asked to be ignored in the problem statement)

1000

81.9125×

Final Selection: W690×125

Example 4.4: Select suitable trial section for a 8m long

simply supported W-section beam subjected to service

live load of 10 kN/m and imposed dead load of 4 kN/m.

The compression flange is continuously supported in the

lateral direction and the deflection is limited to

span/1500.

Solution:

Assumed self weight = 10% of superimposed

dead load

= 0.4 kN/m

say 0.8 kN/m as live load is greater and deflection

requirements are strict.

wu = 1.2D + 1.6L

= 1.2 × 4.8 + 1.6 × 10 = 21.76 kN/m

Mu = Mmax = = 174.08 kN-m8

876.21 2×



(Zx)req = =

= 773.7 × 103 mm3

dmin = = 364 mm

yb

u

F

M

φ610×

2509.0

1008.174 6

××

22

8000

∆act due to service live load = EI

w 4

384

5 ll×

= 8,000 / 1500( )

min

4

000,200

000,810

384

5

I×××

Imin = 50,000 × 104 mm4

Zsel ≥ 773.7 × 103 mm3

Selection Minimum weight section

of section d ≥ 364 mm

Imin = 50,000 × 104 mm4

From beam selection tables,

Trial section no.1: W360 × 44

d = 352 mm < dmin

Ix = 12,100 × 104 mm4

Moving upwards in the same tables,

Trial section no.2: W410 × 46.1

d = 403 mm > dmin

Ix = 15,600 × 104 mm4


d = 355 mm < dmin

Ix = 14,200 × 104 mm4


Ix = 21,200 × 104 mm4

Skipping W410 × 60 and W460 × 60,


Ix = 35,100 × 104 mm4


Ix = 41,000 × 104 mm4


Ix = 56,200 × 104 mm4

Trial Section: W610 × 82

After selection of the trial section, all the checks are

to be performed. This part of the exercise is left for

the reader to be completed.

End of Chapter 4

FLOW CHART FOR DESIGN OF BEAMS - WEC...

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