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50
Case Study in Steel
Building description
The building is a one-story steel structure, typical of an office building. The figure shows that it has three 30 ft. bays in the short direction and a large number of bays in the long direction. Some options for the structural system include fully restrained with rigid connections and fixed column bases, simple framing with “pinned” connections and column bases requiring bracing against sideway, and simple framing with continuous beams and shear connections, pinned column bases and bracing against sidesway. This last situation is the one we’ll evaluate as shown in Figure 2.5(c).
Loads
Live Loads: Snow on Roof: 30 lb/ft^2 (1.44 kPa)
Wind: 20 lb/ft^2 (0.96 kPa)
Dead Loads: Roofing: 8 lb/ft^2 (0.38 kPa) Estimated decking: 3 lb/ft^2 (0.14 kPa) Ceiling: 7 lb/ft^2 (0.34 kPa) Total: 18 lb/ft^2 (0.86 kPa)
Materials
A36 steel for the connection angles and plates (Fy = 36 ksi, Fu = 58 ksi) and A992 or Grade 50 steel for the beams and columns (Fy = 50 ksi, Fu = 65 ksi), K series open web joists and roof decking.
Decking:
Decking selection is typically allowable stress design. Tables will give allowable total uniform load (taking self weight into account) based on stresses and deflection criteria for typical spans and how many spans are supported. The table (and description) for a Vulcraft 1.0 E deck is provided.
21’
21’
8’
50
Areas in gray are governed by live load roof deflection. The total load with snow and roofing = 30 psf + 8 psf = 38 psf.
Open Web Joists:
Open web joist selection is either based on allowable stress design or LRFD resistance for flexure ( not for deflection ). The total factored distributed load for joists at 6 ft on center will be:
wtotal = (1.218lb/ft^2 + 1.630 lb/ft^2 )(6 ft) + 1.2 (8 lb/ft estimated) = 427.2 lb/ft (with 1.2 D + 1.6( L,or Lr, or S, or R ) by catalogue) wlive = 30 lb/ft^2 (6 ft) = 180 lb/ft
For the beams, we know that the maximum unbraced length is 6 ft. For the middle 6 feet of the end span, the moment is nearly uniform, so Cb = 1 is acceptable (Cb = 1.08 for constant moment). For the interior span, Cb is nearly 1 as well.
Choosing a W18x35 (Mu = 229 k-ft) for the end beams, and a W12x30 (Mu = 158 k-ft) for the interior beam, the self weight can be included in the total weight. The diagrams change to:
y x
34.67134.67145.
45.04534.
1 2 3 4 5
Check beam shear:
Exterior Vu = 34.67 k 1.0(0.6)(50 ksi)(17.1 in.)(0.3 in.) = 153.9 k OK W18x35: d = 17.7 in., tw = 0.3 in., Ix = 510 in.^4 Interior Vu = 45.05 k 1.0(0.6)(50 ksi)(12.3 in.)(0.26 in.) = 95.94 k OK W12x30: d = 12.3 in., tw = 0.26 in., Ix = 238 in.^4
Vu vVn 1. 0 ( 0. 6 FywAw)
y
0 x
00 155.925155.
155.925155.
00
(^1 2 3 4 )
Check deflection (NO LOAD FACTORS) for total and live load (gravity and snow).
Exterior Beam: worst deflection is from no live load on the center span: y
x
Maximum total = 2.20 in. Is (^) total L/240 = 360 in./240 = 1.5 in.? NO GOOD We need an I (2.20in./1.5in.)(510 in.^4 ) = 748 in.^4
Maximum live = 1.86 in. Is live L/360 = 360 in./360 = 1.0 in.? NO GOOD
We need an I (1.86in./1.0in.)(510 in.^4 ) = 949 in.^4
The W21x48 looks promising, but it has a note that it exceeds the compact limit for flexure.
Choose a W21 x 50 (Ix = 984 in.^4 ) (because the W21x48 would require extra work!)
Now, live = 1.07 in., which is reasonable close.
Interior Beam: worst deflection is from load on all spans: y x
Maximum total (at midspan) = 1.31 in. Is (^) total L/240 = 360 in./240 = 1.5 in.? OK
Maximum live (at midspan) = 0.94 in. Is live L/360 = 360 in./360 = 1.0 in.? OK
For ¾ in. diameter A325-N bolts and standard holes without a concern for deformation of the holes, the capacity per bolt is:
shear: : Ru vRn 0. 75 , Rn FnAb , where Fn = 54 ksi
]
(. in) k n(. )( ksi)[ 4
^2
so n 1.96. Use 2 bolts (1@3 in. + [email protected] 5.5 in. < 10.125 in.)
bearing for 2 rows of bolts: depends on thickness of thinnest web (t=0.26 in.) and the connected material Ru Rn 0. 75 , Rn 1. 5 LctFu 3. 0 dtFu
Lc = 1.75 in. from the vertical edge of the beam to the edge of a hole
35 k 2 bolts^ [ 0_._ 75 ( 1_._ 5 )( 1_._ 75 in)( 0_._ 26 in)( 65 ksi) = 38.0 k 2 bolts^ [ 0_._ 75 ( 3 )( 0_._ 75 in)( 0_._ 26 in)( 65 ksi) = 57.0 k OK
If the spacing between the holes across the splice is 4 in., the eccentricity, ex, is 2 inches. We need to find C, which represents the number of bolts that are effective in resisting the eccentric shear force.
rn is the nominal shear per bolt:
(. in)^2_._
. ( ksi)
k Cmin
(which we found as n )
C off the table is 2.54 bolts which is more than the minimum of 1.95 (which is why we have 2). OK.
If the plate is 3/8 in. thick x 8 in. wide x 9 in. tall, check bolt bearing on plate:
Rn 2. 4 dtFu (per bolt)
2 bolts[2.4(0.75 in.)(0.375 in.)(58 ksi) = 78.3 k > 35 k OK
Check flexure of the plate:
design moment:
R e
M^ u u =
35 k 4 in
= 70.0 k-in
yielding capacity: Mn FySx 0. 9 (5.5 in. tall section, 3/8 in. thick)
. in(. in )^2 .( ksi) = 61.25 k-in > 70.0 k-in NOT OK
with 6 in. tall, Mn 72.9 k-in
rupture Mn FuSnet 0. 75
c S Inet net ^ and can be looked up or calculated = 1.74 in
3
0_._ 75 ( 58 ksi)( 1_._ 74 in^3 )^ =^ 75.7 k-in > 70.0 k-in^ OK
Check shear yielding of the plate: Ru Rn 1. 00 Rn 0. 6 FyAg
(1.00)[0.6(36 ksi)(6 in.)(0.375 in.)] = 48.6 k > 35 k OK
Check shear rupture of the plate: Ru Rn 0. 75 Rn 0. 6 FuAnv
for ¾” diameter bolts, the effective hole width is (0.75 + 1/8) = 0.875 in.: (0.75)[0.6(58 ksi)(6 in. – 2 x 0.875 in.)(0.375 in.)] = 41.6 k > 35 k OK
Check block shear rupture of the plate: R u Rn 0. 75
Rn 0_._ 6 FuAnv UbsFuAnt 0_._ 6 FyAgv UbsFuAnt
with Ubs = 0.5 when the tensile stress is non-uniform. (The tensile stress switches direction across the splice.) (and assuming 2 in. of width to the center of the bolt hole)
Rn = 0_._ 60 ( 58 ksi)( 0_._ 375 in)[ 1_._ 5 in 3 in 1_._ 5 holes^ ( 0_._ 875 )] 0_._ 5 ( 58 ksi)( 0_._ 375 in)( 2 in ^0_.^875 in 2 ) 58._ 6_._ 9 k
0_._ 6 ( 36 ksi)( 0_._ 375 in)( 1_._ 5 in 3 in) 0_._ 5 ( 36 ksi)( 0_._ 375 in)( 2 in ^0_.^875 in 2 ) 47._ 0 k
35 k < 0.75(47.0 k) = 35.2 k OK
Column Base Plate:
Column base plates are designed for bearing on the concrete (concrete capacity) and plastic hinge development from flexure because the column “punches” down the plate and it could bend upward near the edges of the column (shown as 0.8bf and 0.95d ). The plate dimensions are B and N. The concrete has a compressive strength, f’c = 3 ksi.
