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
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

Validation

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

Results

Table 1. Web opening design for member no. 5
Result Type		Reference	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.Pro 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.Pro 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 ========================