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
t_w = 0.38 in
b_f = 6.53 in
t_f = 0.535 in
Z = 110 in³
L = 10 ft

Weld Properties : E90XX

F_y,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 )
F_y = 36 ksi

Capacity check at web hole assuming an unreinforced opening:

Loading :

V_u = 34.46 kip
M_u = 2,510.7 kip-inch

Tee Properties :

s_t = 5.415 in
s_b = 5.415 in

Calculations

Check for local buckling of compression flange :

width to thickness ratio $F_{1} = \frac{b_{f}}{2 t_{f}} = 6.1028$

limiting ratio $= \frac{65}{\sqrt{F_{y}}} = 86.667 > F_{1}$ Hence O.K

Check for web buckling :

width to thickness ratio of web $W_{1} = \frac{(d - {2 t}_{f})}{t_{w}} = 52$

limiting ratio $= \frac{520}{\sqrt{F_{y}}} = 86.667 > W_{1}$ Hence O.K

Check for opening dimensions to prevent web buckling :

Since $W_{1} \leq \frac{420}{\sqrt{F_{y}}} = 70$ , a₀ / h₀ = 2 < 3.0

h₀ / d = 0.4801 < 0.7

Opening parameter $p_{0} = \frac{a_{0}}{h_{0}} + \frac{6 h_{0}}{d} = 4.8805 < 5.6$ Hence OK

Check for Tee dimension :

s_{t} = \frac{d}{2} - (\frac{h_{0}}{2} + e) = 5.415 in. \geq 0.15 d = 3.1245 in.

s_{b} = \frac{d}{2} - (\frac{h_{0}}{2} - e) = 5.415 in. \geq 0.15 d = 3.1245 in.

Aspect Ratio:

$υ_{t} = \frac{a_{0}}{s_{t}} = 3.6934 < 12.0$ ,

$υ_{b} = \frac{a_{0}}{s_{b}} = 3.6934 < 12.0$ Hence OK

Calculation of Maximum Moment Capacity :

For unperforated section M_p = F_yZ = 3,960 kip-inch

ΔA_s = h₀t_w = 3.8 in²

$M_{m} = M_{p} [1 - \frac{Δ A_{s} (h_{0} / 4 + e)}{Z}] = 3, 618 kip-in \leq M_{p}$ Hence OK

Calculation of Maximum Shear Capacity

V_{p b} = \frac{F_{y} t_{w} s_{b}}{\sqrt{3}} = 42.768 kips

V_{p t} = \frac{F_{y} t_{w} s_{t}}{\sqrt{3}} = 42.768 kips

For unreinforced opening :

μ_b = 0, μ_t = 0

Ratio of nominal shear capacity of tees :

α_{v b} = \frac{\sqrt{6} + μ_{b}}{υ_{b} + \sqrt{3}} = 0.4515 \leq 1.0

α_{v t} = \frac{\sqrt{6} + μ_{t}}{υ_{t} + \sqrt{3}} = 0.4515 \leq 1.0

Hence OK

V_mb = V_pbα_vb = 19.309 kips

Vm = V_mb+V_mb = 38.618 kips Hence OK

Check against Maximum Shear capacity :

V_{p} = \frac{F_{y} t_{w} d}{\sqrt{3}} = 164.52 kips

Since $W_{1} \leq \frac{420}{\sqrt{F_{y}}}$ , $V_{m} \leq \frac{2}{3} V_{p} = 109.68 kips$ Hence OK

Check against Moment Shear Interaction:

R_{1} = \frac{V_{u}}{ϕ V_{m}} = 0.8923 \leq 1.0

R_{2} = \frac{M_{u}}{ϕ M_{m}} = 0.6939 \leq 1.0

R = \sqrt[3]{R_{1}^{3} + R_{2}^{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 t_r = 0.1875 in
Width of Reinforcement b_r = 0.25 in

Check for local buckling of compression flange :

width to thickness ratio of web reinforcement $F_{2} = \frac{b_{r}}{t_{r}} = 1.3333$

limiting ratio $= \frac{65}{\sqrt{F_{y}}} = 10.833 > F_{2}$ Hence O.K

Area of Reinforcement A_r = t_rb_r = 0.0469 in²

Calculation of Maximum Moment Capacity :

For unperforated section M_p = F_yZ = 3,960 kip-inch

ΔA _s = h₀t_w - 2A_r = 3.7063 in²

Since t_we = 0 < A_r

$M_{m} = M_{p} [1 - \frac{t_{w} (h_{0}^{2} / 4 + h_{0} e - e^{2}) - A_{r} h_{0}}{Z}] = 3, 634.88 kip-in \leq M_{p}$ Hence OK

Calculation of Maximum Shear Capacity :

V_{p b} = \frac{F_{y} t_{w} s_{b}}{\sqrt{3}} = 42.768 kips

V_{p t} = \frac{F_{y} t_{w} s_{t}}{\sqrt{3}} = 42.768 kips

s_{t 1} = s_{t} - \frac{A_{r}}{2 b_{f}} = 5.4114 in

s_{b 1} = s_{b} - \frac{A_{r}}{2 b_{f}} = 5.4114 in

$P_{r} = F_{y} A_{r} = 1.6875 \leq \frac{F_{y} t_{w} a_{0}}{2 \sqrt{3}} = 78.982 kips$ Hence Ok

d_{r t} = s_{t} - \frac{1}{2} t_{r} = 5.3213

d_{r b} = s_{b} - \frac{1}{2} t_{r} = 5.3213

$υ_{t} = \frac{a_{0}}{s_{t 1}} = 3.6959 < 12.0$ , $μ_{t} = \frac{2 P_{r} d_{r t}}{V_{p t} s_{t}} = 0.0775$

$υ_{b} = \frac{a_{0}}{s_{b 1}} = 3.6959 < 12.0$ , $μ_{b} = \frac{2 P_{r} d_{r b}}{V_{p b} s_{b}} = 0.0775$

$α_{v b} = \frac{\sqrt{6} + μ_{b}}{υ_{b} + \sqrt{3}} = 0.4656 \leq 1.0$

α_{v t} = \frac{\sqrt{6} + μ_{t}}{υ_{t} + \sqrt{3}} = 0.4656 \leq 1.0

Hence OK

V_mb = V_pbα_vb = 19.911 kips

V_mt = V_ptα_vt = 19.911 kips

V_m = V_mb + V_mt = 39.823 kips

Check against Maximum Shear capacity :

V_{p} = \frac{F_{y} t_{w} d}{\sqrt{3}} = 164.52 kips

Since $W_{1} \leq \frac{420}{\sqrt{F_{y}}}$ , $V_{m} \leq \frac{2}{3} V_{p} = 109.68 kips$ kips Hence OK

Check against Moment Shear Interaction :

$R_{1} = \frac{V_{u}}{ϕ V_{m}} = 0.8653 \leq 1.0$

$R_{2} = \frac{M_{u}}{ϕ M_{m}} = 0.6907 \leq 1.0$

$R = \sqrt[3]{R_{1}^{3} + R_{2}^{3}} = 0.9924 < 1.0$ Hence OK

Calculation of length of Fillet Weld :

A_f = b_ft_f = 3.4936 in²

For reinforcing bars on one side of the web :

$A_{r} \leq \frac{A_{f}}{3} = 1.1645 {in}^{2}$ Hence OK

a₀ / h₀ = 2 ≤ 2.5 Hence OK

$V_{1} = \frac{s_{t}}{t_{w}} = 14.25$ , $V_{2} = \frac{s_{b}}{t_{w}} = 14.25$

$V_{1} and V_{2} \leq \frac{140}{\sqrt{F_{y}}} = 23.333 kips$ Hence OK

$\frac{M_{u}}{V_{u} d} = 3.4977 \leq 20$ Hence OK

$R_{w 1} = ϕ 2 P_{r} = 3.375 kips$ (strength of weld within the length of the opening)

$L_{1} = \max (\frac{a_{0}}{4}, \frac{A_{r} \sqrt{3}}{2 t_{w}}) = 5 in$ (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 R_wr = max(R_wr1, R_wr2) = 3.375 kips

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

Corner Radii :

Minimum Radii $= \max (2 t_{w}, \frac{5}{8}) = 0.76 in$

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

STAAD Input

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
LOAD 1 DEAD AND LIVE LOAD
SELFWEIGHT X 1
SELFWEIGHT Y -1
JOINT LOAD
4 5 FY -15
11 FY -35
MEMBER LOAD
8 TO 13 UNI Y -0.9
6 UNI GY -1.2
CALCULATE RAYLEIGH FREQUENCY
LOAD 2 WIND FROM LEFT
MEMBER LOAD
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
LOAD LIST 4
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 Output

                         STAAD.Pro CODE CHECKING - (AISC 9TH EDITION)   v1.0
                         ***********************
 ALL UNITS ARE - KIP  INCH (UNLESS OTHERWISE Noted)
 MEMBER     TABLE       RESULT/   CRITICAL COND/     RATIO/     LOADING/
                          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
   ----------------------
   SECTION    LOAD        MZ/           FY/                IR
                      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 ========================