 # V. Design of Steel Beam with Web Opening

This example illustrates the design procedure applied at the location of the web opening. The design at locations where openings are not present is not shown in this example.

## Given

Design Method : ASD

Beam No : 5

Beam Section : W21X50

• d = 20.83 in
• tw = 0.38 in
• bf = 6.53 in
• tf = 0.535 in
• Z = 110 in3
• L = 10 ft

Weld Properties : E90XX

Fy,weld = 90 ksi

Weld Stress = 54 ksi

Opening Type: Rectangular

Number of openings : 1

• Section location of Opening : 0.6
• a0 = 20 in
• h0 = 10 in
• e = 0 in ( concentric opening )
• Fy = 36 ksi

Capacity check at web hole assuming an unreinforced opening:

• Vu = 34.46 kip
• Mu = 2,510.7 kip-inch

Tee Properties :

• st = 5.415 in
• sb = 5.415 in

## Calculations

Check for local buckling of compression flange :

width to thickness ratio $F 1 = b f 2 t f = 6.1028$

limiting ratio $= 65 F y = 86.667 > F 1$ Hence O.K

Check for web buckling :

width to thickness ratio of web $W 1 = ( d − 2 t f ) t w = 52$

limiting ratio $= 520 F y = 86.667 ⁢ > W 1$ Hence O.K

Check for opening dimensions to prevent web buckling :

Since $W 1 ≤ 420 F y = 70$ , a0 / h0 = 2 < 3.0

h0 / d = 0.4801 < 0.7

Opening parameter $p 0 = a 0 h 0 + 6 h 0 d = 4.8805 < 5.6$ Hence OK

Check for Tee dimension :

$s t = d 2 − ( h 0 2 + e ) = 5.415 in. ≥ 0.15 d = 3.1245 in.$
$s b = d 2 − ( h 0 2 − e ) = 5.415 in. ≥ 0.15 d = 3.1245 in.$

Aspect Ratio:

$υ t = a 0 s t = 3.6934 < 12.0$,

$υ b = a 0 s b = 3.6934 < 12.0$ Hence OK

Calculation of Maximum Moment Capacity :

For unperforated section Mp = FyZ = 3,960 kip-inch

ΔAs = h0tw = 3.8 in2

$M m = M p [ 1 − Δ A s ( h 0 / 4 + e ) Z ] = 3 , 618 kip-in ≤ M p$ Hence OK

Calculation of Maximum Shear Capacity

$V p b = F y t w s b 3 = 42.768 kips$
$V p t = F y t w s t 3 = 42.768 kips$

For unreinforced opening :

μb = 0, μt = 0

Ratio of nominal shear capacity of tees :

$α v b = 6 + μ b υ b + 3 = 0.4515 ≤ 1.0$
$α v t = 6 + μ t υ t + 3 = 0.4515 ≤ 1.0$

Hence OK

Vmb = Vpbαvb = 19.309 kips

Vmb = Vpbαvb = 19.309 kips

Vm = Vmb+Vmb = 38.618 kips Hence OK

Check against Maximum Shear capacity :

$V p = F y t w d 3 = 164.52 ⁢ kips$

Since $W 1 ≤ 420 F y$ , $V m ≤ 2 3 V p = 109.68 kips$ Hence OK

Check against Moment Shear Interaction:

 $R 1 = V u ϕ V m = 0.8923 ≤ 1.0$

 $R 2 = M u ϕ M m = 0.6939 ≤ 1.0$

 $R = R 1 3 + R 2 3 3 = 1.0147 > 1.0$

Not OK… Try with a Reinforced web opening .

Capacity check at web hole assuming a reinforced opening:

Reinforcement should be selected to reduce R to 1.0

Let us assume,

• Thickness of Reinforcement tr = 0.1875 in
• Width of Reinforcement br = 0.25 in

Check for local buckling of compression flange :

width to thickness ratio of web reinforcement $F 2 = b r t r = 1.3333$

limiting ratio $= 65 F y = 10.833 > F 2$ Hence O.K

Area of Reinforcement Ar = trbr = 0.0469 in2

Calculation of Maximum Moment Capacity :

For unperforated section Mp = FyZ = 3,960 kip-inch

ΔA s = h0tw - 2Ar = 3.7063 in2

Since twe = 0 < Ar

$M m = M p [ 1 − t w ( h 0 2 / 4 + h 0 e − e 2 ) − A r h 0 Z ] = 3 , 634.88 kip-in ≤ M p$ Hence OK

Calculation of Maximum Shear Capacity :

$V p b = F y t w s b 3 = 42.768 kips$
$V p t = F y t w s t 3 = 42.768 kips$
$s t 1 = s t − A r 2 b f = 5.4114 in$
$s b 1 = s b − A r 2 b f = 5.4114 in$

$P r = F y A r = 1.6875 ≤ F y t w a 0 2 3 = 78.982 ⁢ kips$ Hence Ok

$d r t = s t − 1 2 t r = 5.3213$
$d r b = s b − 1 2 t r = 5.3213$

$υ t = a 0 s t 1 = 3.6959 < 12.0$, $μ t = 2 P r d r t V p t s t = 0.0775$

$υ b = a 0 s b 1 = 3.6959 < 12.0$, $μ b = 2 P r d r b V p b s b = 0.0775$

$α v b = 6 + μ b υ b + 3 = 0.4656 ≤ 1.0$

$α v t = 6 + μ t υ t + 3 = 0.4656 ≤ 1.0$

Hence OK

Vmb = Vpbαvb = 19.911 kips

Vmt = Vptαvt = 19.911 kips

Vm = Vmb + Vmt = 39.823 kips

Check against Maximum Shear capacity :

$V p = F y t w d 3 = 164.52 kips$

Since $W 1 ≤ 420 F y$, $V m ≤ 2 3 V p = 109.68 kips$kips Hence OK

Check against Moment Shear Interaction :

$R 1 = V u ϕ V m = 0.8653 ⁢ ≤ 1.0$

$R 2 = M u ϕ M m = 0.6907 ≤ 1.0$

$R = R 1 3 + R 2 3 3 = 0.9924 < 1.0$ Hence OK

Calculation of length of Fillet Weld :

Af = bftf = 3.4936 in2

For reinforcing bars on one side of the web :

$A r ≤ A f 3 = 1.1645 in 2$ Hence OK

a0 / h0 = 2 ≤ 2.5 Hence OK

$V 1 = s t t w = 14.25$, $V 2 = s b t w = 14.25$

$V 1 and V 2 ≤ 140 F y = 23.333 kips$ Hence OK

$M u V u d = 3.4977 ≤ 20$ Hence OK

$R w 1 = ϕ 2 P r = 3.375 ⁢ kips$ (strength of weld within the length of the opening)

$L 1 = max ⁡ ( a 0 4 , A r 3 2 t w ) = 5 in$ (length extended on each side of the opening)

Thus, Length of bar = a0 + 2L1 = 30 in

$R w 2 = ϕ F y A r = 1.6875 kips$ (strength of weld for extension on each side of opening)

Strength of Weld Rwr = max(Rwr1, Rwr2) = 3.375 kips

Fillet Weld Size = 0.0044 in (rounded to nearest weld size of 0.0625 in = 1/16 in)

Minimum Radii $= max ⁡ ( 2 t w , 5 8 ) = 0.76 in$

## Comparison

Table 1. Web opening design for member no. 5
Result Type Hand Calculations STAAD.Pro Difference
Interaction Ratio, R 0.99 1.00 negligible
Reinforcement bar (required at web opening) Length, in 30 30 none
Width, in 0.25 0.25 none
Thickness, in 0.1875 0.1875 none
Fillet Weld Size, in 1/16 0.0625 (1/16") none

The file C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\Samples\ is typically installed with the program.

STAAD PLANE Design of Steel Beam with Web Opening
START JOB INFORMATION
ENGINEER DATE 18-May-05
END JOB INFORMATION
UNIT FEET KIP
JOINT COORDINATES
1 0 0 0; 2 30 0 0; 3 0 20 0; 4 10 20 0; 5 20 20 0; 6 30 20 0; 7 0 35 0;
8 30 35 0; 9 7.5 35 0; 10 22.5 35 0; 11 15 35 0; 12 5 38 0; 13 25 38 0;
14 10 41 0; 15 20 41 0; 16 15 44 0;
MEMBER INCIDENCES
1 1 3; 2 3 7; 3 2 6; 4 6 8; 5 3 4; 6 4 5; 7 5 6; 8 7 12; 9 12 14;
10 14 16; 11 15 16; 12 13 15; 13 8 13; 14 9 12; 15 9 14; 16 11 14;
17 11 15; 18 10 15; 19 10 13; 20 7 9; 21 9 11; 22 10 11; 23 8 10;
MEMBER PROPERTY AMERICAN
1 3 4 TABLE ST W14X90
2 TABLE ST W10X49
5 6 7 TABLE ST W21X50
8 TO 13 TABLE ST W18X35
14 TO 23 TABLE ST L40404
MEMBER TRUSS
14 TO 23
MEMBER RELEASE
5 START MZ
UNIT INCHES KIP
DEFINE MATERIAL START
ISOTROPIC MATERIAL1
E 29000
POISSON 0.3
DENSITY 0.000283
ISOTROPIC STEEL
E 29732.7
POISSON 0.3
DENSITY 0.000283
ALPHA 1.2e-005
DAMP 0.03
END DEFINE MATERIAL
CONSTANTS
BETA 90 MEMB 3 4
MATERIAL MATERIAL1 MEMB 1 TO 4 6 TO 23
MATERIAL STEEL MEMB 5
UNIT FEET KIP
SUPPORTS
1 FIXED
2 PINNED
PRINT MEMBER INFORMATION LIST 1 5 14
PRINT MEMBER PROPERTIES LIST 1 2 5 8 14
SELFWEIGHT X 1
SELFWEIGHT Y -1
4 5 FY -15
11 FY -35
8 TO 13 UNI Y -0.9
6 UNI GY -1.2
CALCULATE RAYLEIGH FREQUENCY
1 2 UNI GX 0.6
8 TO 10 UNI Y -1
* 1/3 RD INCREASE IS ACCOMPLISHED BY 75% LOAD
LOAD COMB 3 75 PERCENT DL LL WL
1 0.75 2 0.75
LOAD COMB 4 75 PERCENT DL LL WL
1 2.75 2 2.75
PERFORM ANALYSIS
UNIT INCHES KIP
PARAMETER
CODE AISC
*WEB OPENINGS
*********************
RHOLE 0.6 MEMB 5
RDIM 20.0 10.0 MRMB 5
electrode 3
*********************
CHECK CODE MEMB 5
FINISH


                         STAAD.Pro CODE CHECKING - (AISC 9TH EDITION)   v1.0
***********************
ALL UNITS ARE - KIP  INCH (UNLESS OTHERWISE Noted)
FX            MY             MZ       LOCATION
=======================================================================
*    5  ST   W21X50                   (AISC SECTIONS)
FAIL     AISC- H2-1         2.268         4
50.64 T          0.00       -4151.62      120.00
OUTPUT FOR WEB OPENING
----------------------