Given the following data of the tension connection shown:
Using A36 steel , Fy =248 MPa
Fu = 400 MPa
Deadload PD = 450 KN.
Bolt diameter = 22 mm
Determine the Live load that can be carried by the member.
NSCP 2001 useful related code:
504.2.1 The allowable stress Ft shall not exceed o.06 Fy on the
Gross area nor 0.50 Fu on the effective net area. In addition, pin
Connected members shall meet the requirements of section
Section 504.4.1 at the pinhole.
502.3.2 The width of a bolt or rivets holes shall be taken
As 1.6 mm greater than nominal dimension.
Using NSCP 2001:
Gross area: Ag =350(22) =7,700 mm2
P = 0.60 FyAg = 0.60(248)(7700) = 1145.76 KN
Considering Net (effective area)
An = ⟮350 – 3(24+1.6)⟯(22) = 6,010.4 mm2
Ae = UAn but for this problem the total load is
uniformly distributed such that U=1
P = 0.50 FuAe = 0.50(400)(6010.4) = 1202.08 KN
For Block shear NSCP 2001:
510.5 Allowable shear Rupture
510.5.1 at a beam end connections where the top flange is
coped, and in similar situation where failure might occur by
shear along a plane through the fasteners, or by a combination
of shear along a fasteners plus tension along a perpendicular
plane:
Fp = 0.30 Fu (Acting on the net shear area Av)
Fp = 0.50 Fu (Acting on the net tension area A t)
1st possible failure:
At =⟮200-2(24+1.6)⟯ (22) = 3273.6 mm2
Av =⟮190-2.5(24+1.6)⟯ (22)(2) =5544mm2
P=0.50 FuAt + 0.30 FuAv
P=0.50 (400)(3273.6) + 0.30(400)(5544)
P = 1,320 KN
2nd possible failure:
At =⟮275-2.5(24+1.6)⟯ (22) = 4642 mm2
Av =⟮190-2.5(24+1.6)⟯ (22) =2772mm2
P=0.50 FuAt + 0.30 FuAv
P=0.50 (400)(4642) + 0.30(400)(2772)
P = 1,261.04 KN
Safe P = 1145.76 KN ← from gross area
P = PD + PL ≤1145.76 (note PD= 450 KN)
PL ≤ 695.76 KN
Using NSCP 2015 : Using LRFD
Tensile yielding : Ф = 0.90
Gross area: Ag =350(22) =7,700 mm2
Rn = FyAg =(248)(7700) = 1909.6 KN
Ф Rn = 0.90 (1909.6 KN) =718.64 KN (designed strength)
Tensile rupture: Rn = FuAe (Ф = 0.75)
In computing the net area for tension and shear the width of the bolt holes shall be taken 2 mm greater than
the nominal dimension of the hole. and Ae ≤ 0.85 Ag
Ae = UAn but for this problem the total load is
uniformly distributed such that U=1 ; therefore, Ae= 5984 mm2 ≤ 0.85 Ag ok.
Rn = FuAe =(400)(5984) = 2393.6 KN
Ф Rn = 0.75 (2393.6 KN) = 1795.2 KN
BLOCK SHEAR STRENGTH:
Rn = 0.60FuAnv + UbsFuAnt ≤ 0.60FyAgv + UbsFuAnt
Agv = gross area subject to shear
Ant = net area subject to tension
Anv = net area subject to shear
1st possible failure:
Where tension stress is uniform Ubs = 1 otherwise , It’s 0.50
Ant = (200 – 2(24+2))(22) = 3256 mm2
Anv = ⟮190 – 2.5(24+2)⟯(22)(2) = 5500 mm2
Agv = 190 (22)(2) = 8360 mm2
Rn1 = 0.60FuAnv + UbsFuAnt
Rn1 = 0.60(400)(5500) + (1)(400)(3256) = 2622.4 KN
Rn2 = 0.60FyAgv + UbsFuAnt
Rn2 = 0.60(248)(8360) + (1)(400)(3256) = 2546.37 KN
2nd possible failure:
Where tension stress is uniform Ubs = 1 otherwise , It’s 0.50
Ant = (275 – 2.5(24+2))(22) = 4620 mm2
Anv = ⟮190 – 2.5(24+2)⟯(22) = 2750 mm2
Agv = 190 (22) = 4180mm2
Rn1 = 0.60FuAnv + UbsFuAnt
Rn1 = 0.60(400)(2750) + (1)(400)(4620) = 2508 KN
Rn2 = 0.60FyAgv + UbsFuAnt
Rn2 = 0.60(248)(4180) + (1)(400)(4620) = 2470 KN (governing result)
Ф Rn = 0.75 (2470 KN) = 1852.5 KN
Summary :
Tension Yielding = 1718.64 KN (governing result)
Tension rupture = 1795.20 KN
Tension due to block shear = 1852.5 KN
Ru =1.2 PD + 1.6 PL ≤ Ф Rn
1.2 (450) + 1.6 PL ≤ 1718.64
PL =736.65 KN
2) The W150 x 24 tension member is connected as shown.
All steel is A36 steel. The bolts are A325 bolts with threaded
From shear planes (Fnv=467 MPa)
Given:
Bolt diameter, db =16 mm
Hole diameter, dh = 18 mm
Gusset plate thickness , tg= 12 mm
Fy =248 MPa
; Fu= 400 MPa
s1=40 mm ; s2= 80 mm ; s3= 30 mm
Properties of W150 x 24:
A= 3060 mm2
d=160 mm
bf =102 mm
tf= 10.3 mm
tw= 6.6 mm
Compute the design and allowable strength of the member
based on all possible mode of failure.
504.1 Slenderness Limitation
There is no maximum slenderness limit for design of members in tension.
User Note: For members designed on the basis of tension the slenderness ratio L/r preferable should not exceed
300.This not apply to rods or hanger in tension.
504.2 Tensile strength;
𝑃
The design tensile strength ФPn and allowable tensile strength 𝑛 of tension member shall be lower value
𝛺
Obtained according to the limit states of tensile yielding in the cross-section and the tensile rupture in the net
section.
Tensile yielding in the cross-section
Pn = Fy Ag
LRFD: Ф = 0.90
,
ASD: 𝛺 = 1.67
Tensile rupture in the net section
Pn = FuAe
LRFD: Ф = 0.75
,
ASD: 𝛺 = 2.00
Tensile yielding in the cross-section
Pn = Fy Ag ;
Ag = A= 3060 mm2
Pn = (248) (3060) = 758.88 KN
Design strength; ФPn = 0.90 (758.88 KN) = 682.992 KN (LRFD)
Pu = 1.2PD + 1.6PL where Pu ≤ ФPn
Allowable strength ;
𝑃𝑛
𝛺
P = PD + PL where P ≤
=
758.88
1.67
= 454.419 KN (ASD)
𝑃𝑛
𝛺
Net area rupture of W-section :
An = 3060 – 4(18+2)(10.3) = 2236 mm2
Determine the value of shear lag factor, from the table consider case 7:
2/3 of d = 2/3(160)=106.667>bf
Therefore use U= 0.85
Ae = 0.85 (2236) = 1900.6 mm2
Pn = FuAe = 400(1900.6) = 760.24 KN
Design strength; ФPn = 0.75 (760.24 KN) = 570.18 KN (LRFD)
Allowable strength ;
𝑃𝑛
𝛺
=
760.24
2.00
= 380.12 KN (ASD)
For Bolt Shear strength:
Fnv = 457 MPa
LRFD: Ф = 0.75
, ASD: 𝛺 = 2.00
𝜋
Area of bolt , Ab= (16)2 = 201.062 mm2
4
Rn = (457)(201.062) 12 = 1102.624 KN ←nominal capacity of the bolts
Design strength; ФPn = 0.75 (1102.624 KN) = 826.968 KN (LRFD)
Allowable strength ;
𝑃𝑛
𝛺
=
1102.624
2.00
= 551.312 KN (ASD)
Based on Bearing at bolt holes in the flange of W-section.
Assuming that the deformation at the bolt hole at service load
Is not a design consideration.
a.)..When the deformation of bolt hole service is a design consideration.
Rn = 1.2LctFu ≤ 2.4dbtFu
b.)..When the deformation of bolt hole service is not a design consideration.
Rn = 1.5LctFu ≤ 3.0dbtFu
db = nominal bolt diameter
Lc ← clear distance, in the direction of load ,between the
edge of the hole and edge of adjacent hole or edge of material.
t = thickness of the connected part.
Lc1 = S1 – dh/2 = 31 mm
Lc2 = S2 – dh = 62 mm
t= tf =10.3 mm
1.5Lc1tFu = 1.5(31)(10.3)(400) = 191.58 KN ≤ 3.0dbtFu = 3.0(16)(10.3)(400) = 197.76 KN
1.5Lc2tFu = 1.5(62)(10.3)(400) = 383.16 KN > 3.0dbtFu therefore use 197.76 KN
Rn =1.5LctFu + 3.0dbtFu
Rn =⟮1.5LctFu + 2(3.0dbtFu)⟯4 = ⟮(191.58) + 2( 197.76)⟯4 = 2,348.4 KN
Design strength; ФPn = 0.75 (2348.4 KN) = 1761.3 KN (LRFD)
Allowable strength ;
𝑃𝑛
𝛺
=
2348.4
2.00
= 1174.2 KN (ASD)
For block shear of W-section:
Rn = 0.60FuAnv + UbsFuAnt ≤ 0.60FyAgv + UbsFuAnt
possible failure:
Where tension stress is uniform Ubs = 1 otherwise , It’s 0.50
Ant = (S3 – 0.5de)(tf)(4)
Ant = (30 – 0.5(20))(10.3)(4) = 824 mm2
Agv = (40+2(80))(10.3) (4) = 8240 mm2
Anv = Agv – 2.5det(4)
Anv = 8240 – 2.5(20)(10.3)(4) = 6180 mm2.
Rn1 = 0.60FuAnv + UbsFuAnt
Rn1 = 0.60(400)(6180) + (1)(400)(824) = 1812.8 KN
Rn2 = 0.60FyAgv + UbsFuAnt
Rn2 = 0.60(248)(8240) + (1)(400)(824) = 1555.712 KN
Design strength; ФPn = 0.75 (1555.712 KN) = 1166.784 KN (LRFD)
𝑃
1555.712
Allowable strength ; 𝛺𝑛 = 2.00
= 777.856 KN (ASD)
For block shear of Gusset plate:
Rn = 0.60FuAnv + UbsFuAnt ≤ 0.60FyAgv + UbsFuAnt
Where tension stress is uniform Ubs = 1 otherwise , It’s 0.50
Ant = (S4 – de)(tf)(2)
Ant = (102-2(30) – (20))(12)(2) = 528 mm2
Agv = (40+2(80))(12) (4) = 9600 mm2
Anv = Agv – 2.5det(4)
Anv = 9600 – 2.5(20)(12)(4) = 7200 mm2.
Rn1 = 0.60FuAnv + UbsFuAnt
Rn1 = 0.60(400)(7200) + (1)(400)(528) = 1,939.2 KN
Rn2 = 0.60FyAgv + UbsFuAnt
Rn2 = 0.60(248)(9600) + (1)(400)(528) = 1,639.68 KN
Design strength; ФPn = 0.75 (1639.68 KN) = 1229.76 KN (LRFD)
𝑃
1639.68
Allowable strength ; 𝛺𝑛 = 2.00 = 819.84 KN (ASD)
Ae= CtAn or UAn
Effective Net Area
Type of members
Minimum
Number of
Fasteners
Per line
Special requirement
Effective
Net Area
Ae
(a) full length tension
members having all cross-sectional elements connected to transmit
the tensile force
1
None
An
(b) Short tension member fittings, such as splice plates, gusset plates, or
beam-to column fittings
1
None
(c ) W , M , or rolled
shapes
3
flangewidth
2

sec tiondepth 3
Connection is to flange
or flanges
An
But not to
exceeding
0.85Ag
0;90 An
(d) structural tees cut from sections meeting requirements of (c) above
(e) W , M,or
S
shapes
not meeting
the
conditions of
(c ) , and
other
shapes,
including
built-up
sections
,
having
unconnected
segments
not in the
plane of the
loading
(f) All shapes in (c) ,(d) or (e)
3
Connection is to
flange
or flanges
0;90 An
3
none
0;85 An
2
none
0;75 An