peecc aisc lrfd calculation 19-02-13

7/30/2019 Peecc Aisc Lrfd Calculation 19-02-13

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Title: Calculation of base frame flexural member as per AISC for PEECC

Report No.:

Rev. No.

Multifacet Technical Center Pvt. Ltd. (MTCPL)

#F810, Vivekadhama, Bharathnagar

Off Magadi road, Bangalore 91

Ph: 080-32946374 fax: 080-23588044

Email:[email protected] ,[email protected] & [email protected]

Web: www.multifacet.net

Report No. Date:

14.02.2013

Prepared by KEM/GNA -

Approved By BVN -

Document Change Note: Total pages: 11Any previous copies of this

document issued to the

customer, must be

destroyed if not of this

current revision.

- - - - -

Rev. No By Description Appvd Date

Page 1 of 24

DESIGN OF BASE FRAME FLEXURAL

MEMBERS

CALCULATION AS PER AISC FOR PEECC

Cust.: M/s. Power gear Pvt. Ltd. MEPZ Chennai
mailto:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]://www.multifacet.net/mailto:[email protected]:[email protected]:[email protected]://www.multifacet.net/


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Report No.:

Rev. No.

Page 2 of 24

Disclaimer:

The information contained is specific to the technical data mentioned in

this report.

No document will be provided to customer in any editable format.

MTCPL is not responsible for any changes made to the report by any

means and any consequences arising as a result of manipulation of

data.

Interpretation of the information contained herein should be based oncomplete report and not on parts thereof.

This report is a property of MTCPL and should not be copied either

directly or indirectly without a written consent from the company.

Disputes, if any, will be within the Jurisdiction of Bangalore only.


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Report No.:

Rev. No.

DESIGN CALCULATION OF BASE FRAME FLEXURAL MEMBERS:

Nomenclature:

E= Modulus of Elasticity of steel in MPa

Fy = Yield stress of the steel in MPa for W 14X34 Designation (36 ksi or 250 MPa)

L= Length or Spam of beam in mm

Lnz = Length neutral zone in mm

Mu = Required flexural strength in N-mm

Mn = Nominal or Design flexural strength in N-mm

Mm = Maximum nominal flexural strength at the location of an opening under pure

bending in N-mm

R = Loading rate

S = Clear space between opening in mm

Vu = Required shear strength in N

Vn = Nominal shear strength in N

Vm = Maximum nominal shear strength at the location of an opening under pure

shearing in N

ao = Length of square or rectangular opening in mm

bf= Flange width in mm

d = Overall depth of steel section in mm

ho = Depth of rectangular opening in mm

wu = Uniformly distributed load on the simply supported beam in N/mm

tf= Flange thickness in mm

tw = Web thickness in mm

h= Clear distance between flanges less the corner radius

zo = Distance from left support to opening center line in mm

b= Resistance factor for flexure 0.90 for steel

b= Resistance factor for shear 0.90 for steel

Load details

1) Dead load, D = 32346.54 kgf or 317319.6 N

2) Floor live load , Lf = 17250.58 kgf or 169228.18 N

3) Roof live load, Lr = 5037.7 kgf or 49419.78 N

4) Earthquake, E = 18955.11 kgf or 185949.63 N

Load acting on base frame, 1.0E0.5Lf0.5Lr1.2DP +++=P= (1.2x317319.6) + (0.5x49419.78) + (0.5x169228.18) + (1.0x185949.63)

Page 3 of 24


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Report No.:

Rev. No.

P= 6.7625 x 105 N (Uniformly distributed load in N)

Uniformly distributed load per mm, wn =Ltotal

P

Where,

Load acting on the base frame, P= 6.7625 x 105 N

Total length, Ltotal of base frame is 69466 mm (From drawing)

Uniformly distributed load per mm, wn =Ltotal

P=

69466

10x6.7625 5= 9.7349

mm

N

The maximum bending moment for simply supported beam, loaded with a uniformlydistributed load at mid span

Figure 1 I-beam with rectangular openings

Figure 2 I-beam sectional view

Mu=

8

2Lw

n

(Ref-1)

Where,

Page 4 of 24


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Report No.:

Rev. No.

Uniformly distributed load per mm, wn = 9.7349 N/mm

Span or maximum length, L of base frame is 7238 mm (From drawing)

Mu=8

2Lwn = ( )

8

72389.73492

= 63.749x106 mm-N

Range of application to satisfy the design for W-shape section:

For UB 356 steel beam (Ref-2)

Where,

bf= Flange width is 171.5 mm

d = Overall depth of steel section is 355 mm

h= Clear distance between flanges less the corner radius

355-((11.5x2) + (10.2x2)) = 311.6 mm

tf= Flange thickness is 11.5 mm

tw = Web thickness is 7.4 mm

E= Modulus of Elasticity of steel is 210x103 MPa

Fy = Yield stress of the steel is 250 MPa

(Ref-3)

fbd >1.20

5.171355 >1.20 2.069 >1.20

yw F

E

t

h76.3

250

1021076.3

4.7

6.311

3 42.108 108.975

Check for compactness:

yf

f

F

E

t

b38.0

2

250

1021038.0

5.112

5.171

3

7.456 11.013

Since the beam flange is compact and laterally supported

Nominal or Design flexural strength:

Mn= xy ZF mm-N (Ref-4)

Where,

Mn orMp is Nominal or Design flexural strength in N-mm

Page 5 of 24


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Report No.:

Rev. No.

Fy = Yield stress of the steel in MPa for W 14X34 Designation (36 ksi or 250 MPa)

Zx = Plastic section modulus is 881.473x103 mm3 or 53.79 ksi

Mn= xy ZF =310473.881250 = 220.368x10 6 mm-N

Mn is limited to 1.5Mn to avoid excessive working-load deformation and

xyZF 1.5 xySF or

x

x

S

Z 1.5

Where,

SX= Elastic section modulus for I-shape about the strong axisx

x

S

Zwill be always be 1.5

x

x

S

Z 1.5

3

3

10665.784

10473.881

1.5 1.12 1.5

The flexural design strength of compact beams, laterally supported is given by:

nbM = Xyb ZF Xyb SF5.1

310473.8812509.0 310665.7842505.19.0

106331.198 610824.264

Maximum bending stress at mid span with uniformly distributed load for plastic region:

u =X

u

Z

M u =

3

6

10473.881

10749.63

= 72.32 MPa < 250 MPa

Maximum bending stress at mid span with uniformly distributed load for Elastic region:

u =X

u

S

M u =

3

6

10665.784

10749.63

= 81.243 MPa < 136.66 MPa

The required flexural and shear strengths are calculated in several intermediary sections

(rectangular openings), equally spaced along midspan L/2

Required flexural strength at intermediary section:

( )ozuM = ( )2

2oo

uzLz

w (1) (Ref-3)

Where,

Mu (zo) = Required flexural strength at opening centerline in N-mm

Page 6 of 24


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Report No.:

Rev. No.

wn =Uniformly distributed load per mm = 9.7349 N/mm

zo = Distance from left support to opening center line is 2998.8 mm (From drawing)

L = Span or maximum length of base frame is 7238 mm (From drawing)

( )ozuM = ( )28.299872388.2998

2

7349.9 =61.873x106 N-mm

Required shear strength at intermediary section:

( )ozuV =

ouz

Lw2

(2) (Ref-3)

Where,

Vu (zo) = Required shear strength at opening centerline in N-mm

wn =Uniformly distributed load per mm = 9.7349 N/mm

zo = Distance from left support to opening center line is 2998.8 mm (From drawing)

L = Span or maximum length of base frame is 7238 mm (From drawing)

( )ozuV =

8.2998

2

72387349.9 = 6037.584 N-mm

Shear design for beams:

For unstiffened webs with 260wt

hthe design shear strength is VV and Vn is given as:

No web instability: Foryw F

E

t

h45.2

250

1021045.2

4.7

6.311 3 0.71108.42

nV = wyAF6.0

Where,

Vn= Design shear strength at opening centerline for no web instability in NFy = Yield stress of the steel in MPa for W 14X34 Designation (36 ksi or 250 MPa)

Aw = Web area of the steel is d x tw = 355 X 7.4 = 2627 mm2

nV = 26272506.0 = 394.05X103 N

The design shear strength shall be larger than required shear strength load, applicable to

all beams with unstiffened webs, with 260wt

h 260

4.7

6.311 260108.42

unV VV 584.60371005.3949.03 584.603710645.354 3

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Report No.:

Rev. No.

The neutral zone length for each studied beam length, where the moment-shear interaction

is not considered a critical

1)(

3

)(

+

nv

zou

nb

zou

V

V

M

M

(Ref-3)

Where,

Mu (zo) = Required flexural strength at opening centerline is 61.873x106 N-mm

Mn = Design flexural strength is 220.368x106 mm-N

Vu (zo) = Required shear strength at opening centerline is 6037.584 N

b and v= Resistance factor for flexural and shear strength is 0.90 for steel

11005.3949.0

584.6037

10368.2209.0

10873.613

3

3

6

6

+

110934.4030.0 6 + 103.0

The value of wu can vary from near to zero to a maximum that can be obtained taking into

account the design flexural strength, nbM and Design shear strength, nvV of

unperforated section.

2

8

L

Mw

nb

u

2

6

7238

10368.2209.087349.9

286.307349.9 (3)

(Ref-3)

L

V

wnV

u

2

72381005.3949.02

7349.9

3 995.977349.9 (4)

(Ref-3)

Loading rate: (Ref-3)

Is the quotient between the required strength determined from factored loads and the

design strength of the beam without openings. R shall be obtained from the following

conditions Equations 3 and 4 for wu and Equations. 1 and 2 for Mu and Vu

Page 8 of 24


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Report No.:

Rev. No.

nb

zou

M

M

)(

6

6

10368.2209.0

10876.61

311.0

R R R

nV

zou

V

V

)(

31005.3949.0

584.6037

017.0

Load rating is the larger of both, R=0.311

The neutral zone must always be considered centered in relation to the depyh of the steel

beam. The chart in Figure-8 locate the neutral zone for beams with rectangular openings

according to Table 1. For in chart the input data are in the ratio between the beam span

and steel depth, L/d and loading rate, R. The output data ia the parameter k, which shall be

multiplied by the beam span, L to supply in two ends of the beam.

In Table-1 for rectangular opening, the depth of neutral zone is ho d/3 Figure-4 and

ho d/2 for Figure-8 (Ref-3)

388.20355

7238 ==d

L

Case-1

ho d/3 (Figure-4)

Where,

ho = Depth of rectangular opening is 150 mm (from drawing)

150 355/3 150 118.33 (Not satisfied)

Case-2

ho d/2 (Figure-8)Where,

ho = Depth of rectangular opening is 150 mm (from drawing)

150 355/2 150 177.5 (satisfied)

From Figure-8 for L/d and Load rating, R where k=0.05

When the beam has more than one opening, the minimum spacing between edges of two

adjacent openings, S must be in accordance with the following criterion to avoid interaction

between openings.

Page 9 of 24


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Report No.:

Rev. No.

For Square and Rectangular openings (Ref-3)

Where,

ao = Width of rectangular opening is 300 mm (from drawing)

oh 150 150

S S S

)()(

zounV

zou

oVV

Va

584.60371005.3949.0584.6037

3003

5.195

Maintain spacing between rectangular openings are S2dWhere S2d, S=2x355=710 which is 710150

Figure 3 I-beam Rectangular openings with centroid (Concentric opening)

We have to maintain the above parameters for I beam with rectangular openings

355=d mm

723805.0 =kL = 362 mm

)2( kLLLnz = = )7247238( =6514 mm (Neutral zone length)

dS = 2 = 3552 = 710 mm (Spacing between rectangular opening)

LIFTING PAD CALCULATIONS Case-1:

BOLT STRENGTH CALCULATIONS:

Total weight of structure, W = 32346.54 kgf or 317.319x103 N

For 4 numbers of lifting pad, W4 = 4

317.319x103= 79.329x103 N or 8086.544 kgf

Page 10 of 24


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Report No.:

Rev. No.

Considering impact load factor of 2

Design load, F1 = Impact factor x W4

F1= 79.328x103 x 2 = 158.659x103 N or 16.173 x 103 kgf (Lifting load per pad)

Bolt are subjected to direct shear (For 4 No. Bolts):

Fs= n

F1 = (Ref-5)

Where, n = 4 Number of bolts

Fs= 4

158.659x10 3=39.664x103 N (For each bolt)

Since the load F will try to tilt the bracket in the anticlockwise direction about the lower

edge, therefore the bolts subjected to tensile load due to turning moment. The maximum

loaded bolts are 3 and 4 (See Figure below), because they lie at the greatest distance from

the tilting edge A-A (i.e. lower edge)

Maximum tensile load carried by 3 and 4

L1=50 mm

L2=282 mm

L=150 mm

F1=16173.27 kgf or 158.659x103 N

Ft=

( ) ( )[ ]

L2

F

2

2

2

1

21

L

xLxL

+

=( ) ( )

[ ]

282502

28215010659.15822

3

+

=40.910x103 N (Ref-5)

Since bolts are subjected to shear load as well as tensile load, therefore equivalent tensile

load

Page 11 of 24


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Report No.:

Rev. No.

Fte=( ) ( )

2

422

++

sttFFF

=( ) ( )

2

1039.66441040.9101040.91023233

++

Fte = 65.083x103 N

Since bolts are subjected to shear load as well as tensile load, therefore equivalent shear

load

Fte=( ) ( )

2

422

+

stFF

=( ) ( )

2

1039.66441040.9102323

+

= 44.628x103 N

Material properties of bolt:Size M24x3P C.S.8.8 Grade (Ref-1)

UTS=800 MPa

YS= 640 MPa

Allowable stress, allow =Min (YS0.2, O.7UTS)

YS=640, 0.7x800=560, therefore allow =560 MPa

allow =0.4xY=0.4x640

allow =256 MPa

We know that the equivalent shear load (Fse)

allow = areac/s

Fte

Where,

A=c/s area of bolt= 4xd

2

c

dc =Core diameter of the bolt

Therefore, dc= x

4F

allow

se

x

= 605

444.628x10

3

=10.07 mm < 24 mm

(Hence we have chosen M24 bolt, then design is safe for 4 bolts)

Page 12 of 24


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Report No.:

Rev. No.

We know that the equivalent tensile load (Fte)

allow = areac/s

Fte

Where,


2

c


Therefore, dc= x

4F

allow

te

x

= 605

41065.0833

=12.16 mm < 24 mm


Checking direct Shear stress on bolt:

= A

Fr

= 353

1039.664 3= 112.365 N/mm2 < 256 MPa (For each bolt)

Checking Equivalent Shear stress bolt:

= A

Fr

= 353

44.628x103

= 126.424 N/mm2 < 256 MPa (For each bolt)

Plate strength calculations:

Material properties of plate: (Ref-6)

M.S. IS: 2062 Gr. B

UTS=410 MPa

YS= 250 MPa

Allowable stress, allow =Min (1/3UTS, 2/3YS)

1/3xUTS= 0.333x410 = 136.66 MPa

2/3xYs = 0.666x250 = 166.66 MPa

allow =136.66 MPa

allow =0.4xYS=0.4x250=100 MPa

Page 13 of 24


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Report No.:

Rev. No.

Bending stress, b

=Z

M

Bending moment of the plate, M=F x Perpendicular distance

M=158.659x103 x 130 = 20.625x106 N-mm

Where,

Z = Section modulus at c/s A-A = 97.584x106 mm3

t = Thickness of the plate is 32 mm

b = Depth of plate is 332 mm

W = width of plate is 200 mm

Bending stress, b = ZM

b = 97.584x10

20.625x106

6

= 0.211 MPa < 136.66 MPa

Direct Shear stress, d = 1

A

F

Where,

A = C/s area of plate is 32 * 161= 5152 mm2

Page 14 of 24


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Report No.:

Rev. No.

d = 5152

158.659x103= 30.795 MPa < 100 MPa

Checking tensile tear-out strength on plate:

t = 1 tl

Fr

=

3253

158.659x103

= 93.548 N/mm2 < 136.66 MPa

Since tensile tear-out stress value less than the allowable value for ligament, l1=53 mm,

than the ligament 144 (vertical) is safe. No need to check, it is less than allowable

Checking shear tear-out strength on plate:

s = 2 tl

Fr

=

32752

158.659x103

= 33.053 N/mm2 < 100 MPa

Since shear tear-out stress value less than the allowable value for ligament, l2=75 mm,


WELD STRENGTH CALCULATIONS:

Page 15 of 24


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Report No.:

Rev. No.

Material properties of weld: (Ref-6)

E60xx number

UTS=427 MPa

YS= 345 MPa

Allowable stress for bending, b =0.6Ys = 0.6x345 = 207 MPa

Allowable stress for shear, s =0.4Ys = 0.4x345 = 138 MPa


Considering impact load factor 2

Design load, F1 = Impact factor x W


Area of throat, A = 1.414xhxd

Where,

h = Throat size is 16 mm

d = Depth of weld is 332 mm

Page 16 of 24


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Report No.:

Rev. No.

A = 1.414x16x332 = 7511.168 mm2

Shear stress, fs= A

F1

fs=

MPa492.847511.168

634.638x103=

< 138

Bending stress, fb= Z

M

Bending moment=Force x distance

M=634.638x103 x130=82.502x106 N-mm

Section modulus, Z= throatsize3d 2

Z= 613

323 2

= 587.861x103 mm3

Bending stress, fb=Z

M

fb= 587.861x10

82.502x103

6

=140.342 MPa < 207

Combined bending stress & shear stress

Fte = 22 fbfs +

Fte = 22 342.140492.84 +

Fte = 163.813 MPa < 207 MPa

LIFTING PAD CALCULATIONS Case-2:

BOLT STRENGTH CALCULATIONS:


For 4 numbers of lifting pad, W4 = 4

317.319x10

3

= 79.329x103 N or 8086.544 kgf

Considering impact load factor of 2

Page 17 of 24


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Report No.:

Rev. No.

Design load, F1 = Impact factor x W4


Bolt are subjected to direct shear (For 4 No. Bolts):

Fs= n

F1 = (Ref-5)

Where, n = 4 Number of bolts

Fs= 4

158.659x10 3=39.664x103 N (For each bolt)

Since the load F will try to tilt the bracket in the anticlockwise direction about the lower

edge, therefore the bolts subjected to tensile load due to turning moment. The maximum

loaded bolts are 3 and 4 (See Figure below), because they lie at the greatest distance from

the tilting edge A-A (i.e. lower edge)

Maximum tensile load carried by 3 and 4

L1 = 50 mm

L2 = 276 mm

L = 199 mm

F1 = 16173.27 kgf or 158.659x103 N

Ft =( ) ( )[ ]L2F

2

2

2

1

21

L

xLxL

+=

( ) ( )[ ] 276502 27619910659.158 223

+ = 55.380x103 N (Ref-5)

Page 18 of 24


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Report No.:

Rev. No.

Since bolts are subjected to shear load as well as tensile load, therefore equivalent tensile

load

Fte=( ) ( )

2

422

++

sttFFF

=( ) ( )

2

1039.66441055.3801055.38023233

++

Fte = 76.063x103 N

Since bolts are subjected to shear load as well as tensile load, therefore equivalent shear

load

Fse=( ) ( )

2

422

+

stFF

=( ) ( )

2

1039.66441055.3802323

+= 48.373x103 N

Material properties of bolt:

Size M24x3P C.S.8.8 Grade (Ref-1)

UTS=800 MPa

YS= 640 MPa

Allowable stress, allow =Min (YS0.2, O.7UTS)

YS=640, 0.7x800=560, therefore allow =560 MPa

allow =0.4xY=0.4x640

allow =256 MPa

We know that the equivalent shear load (Fse)

allow = areac/s

Fte

Where,


2

c


Therefore, dc= x

4F

allow

se

x=

605

448.373x103

=10.487 mm < 24 mm

Page 19 of 24


20/24


Report No.:

Rev. No.


We know that the equivalent tensile load (Fte)

allow = areac/s

Fte

Where,


2

c


Therefore, dc= x

4F

allow

te

x

= 605

410063.67 3

=13.15 mm < 24 mm


Checking direct Shear stress on bolt:

= A

Fr

= 353

1039.664 3= 112.365 N/mm2 < 256 MPa (For each bolt)

Checking Equivalent Shear stress bolt:

= A

Fse=

353

48.373x103= 137.034 N/mm2 < 256 MPa (For each bolt)

Plate strength calculations:

Material properties of plate: (Ref-6)

M.S. IS: 2062 Gr. B

UTS=410 MPa

YS= 250 MPa

Allowable stress, allow = Min (1/3UTS, 2/3YS)

1/3xUTS= 0.333x410 = 136.66 MPa

2/3xYs = 0.666x250 = 166.66 MPa

allow =136.66 MPa

allow =0.4xYS=0.4x250=100 MPa

Page 20 of 24


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Report No.:

Rev. No.

Bending stress, b

=Z

M

Bending moment of the plate, M=F x Perpendicular distance

M=158.659x103 x 130 = 20.625x106 N-mm

Where,

Z = Section modulus at c/s A-A = 97.584x106 mm3

t = Thickness of the plate is 32 mm

b = Depth of plate is 332 mm

W = width of plate is 200 mm

Bending stress, b = ZM

b = 97.584x10

20.625x106

6

= 0.211 MPa < 136.66 MPa

Direct Shear stress, d = 1

A

F

Where,

A = C/s area of plate is 32 * 161= 5152 mm2

Page 21 of 24


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Report No.:

Rev. No.

d = 5152

158.659x103= 30.795 MPa < 100 MPa

Checking tensile tear-out strength on plate:

t = 1 tl

Fr

=

3253

158.659x103

= 93.548 N/mm2 < 136.66 MPa

Since tensile tear-out stress value less than the allowable value for ligament, l1=53 mm,


Checking shear tear-out strength on plate:

s = 2 tl

Fr

=

32752

158.659x103

= 33.053 N/mm2 < 100 MPa

Since shear tear-out stress value less than the allowable value for ligament, l2=75 mm,


WELD STRENGTH CALCULATIONS:

Page 22 of 24


23/24


Report No.:

Rev. No.

Material properties of weld: (Ref-6)

E60xx number

UTS=427 MPa

YS= 345 MPa

Allowable stress for bending, b = 0.6Ys = 0.6x345 = 207 MPa

Allowable stress for shear, s =0.4Ys = 0.4x345 = 138 MPa


Considering impact load factor 2

Design load, F1 = Impact factor x W


Area of throat, A = 1.414xhxd

Where,

h = Throat size is 16 mm

d = Depth of weld is 332 mm

Page 23 of 24


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Report No.:

Rev. No.

A = 1.414x16x332 = 7511.168 mm2

Shear stress, fs= A

F1

fs=

MPa492.847511.168

634.638x103=

< 138

Bending stress, fb= Z

M

Bending moment=Force x distance

M=634.638x103 x130=82.502x106 N-mm

Section modulus, Z= throatsize3d 2

Z= 613

323 2

= 587.861x103 mm3

Bending stress, fb=Z

M

fb= 587.861x10

82.502x103

6

=140.342 MPa < 207

Combined bending stress & shear stress

Fte = 22 fbfs +

Fte = 22 342.140492.84 +

Fte = 163.813 MPa < 207 MPa

List of Reference:

1. CMTI-Machine tool design handbook

2. Jindal steel & power, Annexure-A

3. Design aids for unreinforced web openings in steel and composite beams with W-shapes

4. Design of beams and others Flexural Members ASIC LRFD 3rd Edition (2001)

5. Machine design by R.S. Khurmi and J.K.Gupta

6. Mechanical engineering design ninth edition by Richard G. Budynas and J.Keith Nisbett

peecc aisc lrfd calculation 19-02-13

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