# V. AISC ASD - Design of Steel Beam with Web Opening

Design the web reinforcement for a web opening in a steel beam.

## Details

Note: The design at locations where openings are not present is not shown in this example.

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

## Validation

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 $= max ⁡ ( 2 t w , 5 8 ) = 0.76 in$

## Results

Table 1. Web opening design for member no. 5
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
----------------------
MOM. CAP. (Mm)  SHR. CAP. (Vm) ((MZ/Mm)^3+(FY/Vm)^3)^0.33
=======================================================================
0.600       4        2510.65         34.46
3634.88         39.64             1.00
REINFORCING BARS : TO BE PLACED ON ONE SIDE OF WEB
BAR DIMENSION	: LENGTH = 30.00 WIDTH =  0.25 THICKNESS = .1875
FILLET WELD      : SIZE REQD = 0.0625 inch LENGTH REQD =  30.00
======================= END WEB OPENING OUTPUT ========================