PSD 323 Module 3

DON HONORIO VENTURA STATE UNIVERSITY COLLEGE OF ENGINEERING
Cabambangan, Villa de Bacolor 2001, Pampanga, Philippines AND ARCHITECTURE

Tel. No. (6345) 458 0021; Fax (6345) 458 0021 Local 211 DHVSU Main Campus, Villa de Bacolor, Pampanga E-
URL: http://dhvsu.edu.ph
Mail Address: ceadhvtsu@gmail.com
PSD 323: Principles of Steel Design
Module 3: Analysis of Tension Members (ASD and LRFD)
I. Course Objectives:
1. The students will be able to familiarize the general consideration of code designs
and introduce the concept of designing tension members.
II. Course Outline:

3.1 Introduction
3.2 Nominal Strengths of Tension Members
3.3 Net Areas
3.4 Effective Net Areas
3.5 Sample Problems
3.6 Block Shear
3.7 Sample Problems
III. Learning Content:
3.1 INTRODUCTION:
Tension members are encountered in most steel structures. They are found in bridge and
roof trusses, towers, and bracing systems, and in situations where they are used as tie rods.
The selection of a section to be used as a tension member is one of the simplest problems
encountered in design. They frequently appear as tie rods to stiffen a trussed floor system
or to provide intermediate support for a wall girt system. Tension members may consist of
a single structural shape or they may be built up from a number of structural shapes. The
cross-sections of some typical tension members are shown below.
3.2 NOMINAL STRENGTHS OF TENSION MEMBERS

NSCP 2015 – Sec. 504.2 (Tensile Strength)
The design tensile strength, ∅𝒕 𝑷𝒏 , and the allowable tensile strength, 𝑷𝒏 /𝛀𝒕 , of tension
members, shall be the lower value obtained according to the limit states of tensile yielding
in the gross section and tensile rupture in the net section.
1. For tensile yielding in the gross section:
𝑷𝒏 = 𝑭𝒚 𝑨𝒈
∅𝒕 = 𝟎. 𝟗𝟎 (𝑳𝑹𝑭𝑫) 𝛀𝒕 = 𝟏. 𝟔𝟕 (𝑨𝑺𝑫)
2. For tensile rupture in the net section:
𝑷𝒏 = 𝑭𝒖 𝑨𝒆
∅𝒕 = 𝟎. 𝟕𝟓 (𝑳𝑹𝑭𝑫) 𝛀𝒕 = 𝟐. 𝟎𝟎 (𝑨𝑺𝑫)
Where:
𝑨𝒆 = effective net area, mm2
𝑨𝒈 = gross area of member, mm2
𝑭𝒚 = specified minimum yield stress of the type of steel being used, MPa
𝑭𝒖 = specified minimum tensile strength of the type of steel being used, MPa
NSCP 2001 – Sec. 504.2 (Allowable Stress)
1. 𝑭𝒕 = 𝟎. 𝟔𝟎𝑭𝒚 on the gross area
𝑭𝒕 = 𝟎. 𝟓𝟎𝑭𝒖 on the effective net area
2. For pin-connected members,
𝑭𝒕 = 𝟎. 𝟒𝟓𝑭𝒚 on the net area of the pinhole
For tension members consisting of rolled steel shapes, there actually is a third limit state,
block shear.
3.3 NET AREA
Whenever a tension member is to be fastened by means of bolts or rivets, holes must be
provided at the connection. As a result, the member cross-sectional area at the connection is
reduced and the strength of the member may also be reduced depending on the size and
location of the holes.
The net area, 𝑨𝒏 , of a member is the sum of the products of the thickness and the net width
of each element computed as follows:
In computing net area for tension and shear,

NSCP 2001 (Sec.502.3.2) NSCP 2015 (Sec.504.3.2)
The width of a bolt or rivet The width of a bolt hole shall
hole shall be taken as 1.6 mm be taken 2 mm greater than
greater than the nominal the nominal dimension of the
dimension of the hole hole.
dh = dn + 1.6 mm dh = dn + 2 mm
Where: dh = diameter of the hole in computing the net area for tension and shear
dn = nominal dimension of the hole
In considering the area of such items, it is important to realize that it is usually necessary to
subtract an area a little larger than the actual hole.
(Table 510.3.3) Nominal Hole Dimensions, mm

Bolt Short-Slot Long-Slot
Standard Oversize
Diameter (Width x (Width x
(Dia) (Dia)
Length) Length)
M16 18 20 18 x 22 18 x 40
M20 22 24 22 x 26 22 x 50
M22 24 28 24 x 30 24 x 55
M24 27(a) 30 27 x 32 27 x 60
M27 30 35 30 x 37 30 x 67
M30 33 38 33 x 40 33 x 75
≥M36 d+3 d+8 (d+3) x (d+10) (d +3) x 2.5 d
(a) Clearance provided allows the use of a M30 bolt if desirable.
For a chain of holes extending across a part in any diagonal or zigzag line, the net width of
the part shall be obtained by deducting from the gross width the sum of the diameters or slot
dimensions as provided from the table above, of all holes in the chain, and adding, for each
gage space in the chain, the quantity,
𝒔𝟐 /𝟒𝒈
𝒔𝟐
𝑨𝒏 = 𝑨𝒈 − ∑ 𝒅𝒉 (𝒕) + ∑ 𝟒𝒈(t)
𝒔𝟐
𝒘𝒏 = 𝒘𝒈 − ∑ 𝒅𝒉 + ∑ 𝟒𝒈
Where: s = longitudinal center-to-center spacing (pitch) of any two consecutive

holes, mm
g = transverse center-to-center spacing (gage) between fastener gage lines,
mm
For angles, the gage for holes in opposite adjacent legs shall be the sum of the gages from
the back of the angles less the thickness.
The critical net area, 𝐴𝑛 , of the part is obtained from that chain which gives the least net
width.
3.4 EFFECTIVE NET AREAS
The net area computed gives the reduced section that resists tension but still may not
accurately reflect the strength. That is when the tension member has a profile consisting of
elements not in a common plane and where the tensile load is transmitted at the end of the
member by connection to some but not all of the elements. For such cases, the tensile force
is not uniformly distributed over the net area.
When a tensile load is applied eccentrically to a wide plate, the stress distribution across the
width of the plate is non-uniform. This nonuniformity of stress is referred to as “shear lag”.
The shear lag factor (U) accounts for the effects of eccentricity and shear lag on tension
members connected through only part of their cross-sectional elements.
The effective area of tension members shall be determined as follows: (NSCP 2015 - Sec. 504.3.3)
𝑨𝒆 = 𝑼 𝑨𝒏
Where: 𝑨𝒆 = effective net area of the member
𝑨𝒏 = net area of the member
𝑼 = shear lag factor
Sec. 510.4.1.(2) limits 𝑨𝒏 to a maximum of 0.85𝑨𝒈 connection design for splice plates with
holes.
𝑨𝒆 = 𝑨𝒏 ≤ 𝟎. 𝟖𝟓𝑨𝒈
NSCP 2001 (Sec. 502.4.4) – Reduction Coefficients, U

Condition U
W, M, or S shapes with flange widths not less than 2/3 the depth, and structural
tees cut from these shapes, provided the connection is to the flanges. Bolted or
0.9
riveted connections shall have no fewer than three fasteners per line in the
direction of stress
W, M, or S shapes not meeting the conditions mentioned above, structural tees cut
from these shapes and all other shapes including built-up cross sections. Bolted or
0.85
riveted connections shall have no fewer than three fasteners per line in the
direction of stress
All members with bolted or riveted connections having only two fasteners per line
0.75
in the direction of stress
NSCP 2015 (Table 504.3.1) – Shear Lag Factors for Connections to Tension Members
NSCP 2001 (Sec.502.4.1) NSCP 2015 (Table 504.3.1)

When the load is transmitted All tension members where
directly to each of the cross- the tension load is
sectional elements by transmitted directly to each of
connectors, the effective net cross-sectional elements by
area 𝐴𝑒 . is equal to the net fasteners or welds (except as
area 𝐴𝑛 . in Cases 3, 4, 5, and 6)
Ae = An
When the load is transmitted directly to each of the cross-sectional elements by connectors,
the effective net area 𝐴𝑒 . is equal to the net area 𝐴𝑛 . (NSCP 2001 Sec. 502.4.1)
Also, under NSCP 2015 – Table 504.3.1 Case 1 where all tension members have their
tension load transmitted directly to each of cross-sectional elements by fastener
3.5 SAMPLE PROBLEMS
1. A 130 mm x 12 mm plate is used for tension member. It is connected to a gusset plate
with 4-16 mm diameter bolt as shown. Assume Ae = An. Fy = 248 MPa and Fu = 345 MPa.
Using NSCP 2015 provisions,
a. Determine the design strength for LRFD
b. Determine the allowable strength for ASD
 Based on gross area (yielding):
Ag = 130(12) = 1560 mm2
Pn = 248(1560)
Pn = 386.88 kN (on gross area)
 Based on net area (rupture):
Bolt diameter = 16 mm
Nominal hole dimension = 18 mm
Note: dh = dn + 2 mm
An = [130 − 2(18 + 2)](12) = 1,080 mm2 = Ae

0.85Ag = 0.85(130)(12) = 1,326 mm2
Ae < 0.85Ag ; ∴ Ok
Pn = 345(1,080) = 372.60 kN (on net area)
Based on gross area, Pn = 386.88 kN

Based on net area, Pn = 372.60 kN
LRFD (on gross area): ASD (on gross area):
∅t Pn = (0.90)386.88 Pn /Ωt = 386.88/1.67
∅𝐭 𝐏𝐧 = 𝟑𝟒𝟖. 𝟏𝟗 𝐤𝐍 𝐏𝐧 /𝛀𝐭 = 𝟐𝟑𝟏. 𝟔𝟔 𝐤𝐍
LRFD (on net area): ASD (on net area):
∅t Pn = (0.75)372.60 Pn /Ωt = 372.60/2.00
∅𝐭 𝐏𝐧 = 𝟐𝟕𝟗. 𝟒𝟓 𝐤𝐍 (design strength) 𝐏𝐧 /𝛀𝐭 = 𝟏𝟖𝟔. 𝟑𝟎 𝐤𝐍 (allowable strength)
2. Given the following data of the slip-critical tension connection shown with 22-mm
diameter bolts in standard holes with threads excluded from shear planes. Considering the
plates only, compute the design tensile strength (LRFD) and the allowable tensile strength
(ASD) using NSCP 2015. (Fy = 248 MPa, Fu = 400 MPa)

Side Plate (350 mm x 12 mm):

Ag = 350(12) = 4200 mm2
Pn
= 248(4200)
2
Pn = 2083.2 kN
Middle Plate (350 mm x 22 mm):
Ag = 350(22) = 7700 mm2
Pn = 248(7700) = 1,909.6 kN
Use the lower value. Therefore,
Pn = 1,909.6 kN (based on gross area, middle plate governs)

Side Plate (350 mm x 12 mm):
An = [350 − 3(24 + 2)](12) = 3,264 mm2 = Ae
0.85Ag = 0.85(350)(12) = 3,570 mm2
Ae < 0.85Ag ; ∴ Ok
Pn
= 400(3,264)
2
Pn = 2,611.2 kN
Middle Plate (350 mm x 22 mm):
An = [350 − 3(24 + 2)](22) = 5,984 mm2 = Ae
0.85Ag = 0.85(350)(22) = 6,545 mm2
Ae < 0.85Ag ; ∴ Ok
Pn = 400(5,984)
Pn = 2,393.6 kN
Use the lower value. Therefore, Pn = 2,393.6 kN (based on net area, middle plate governs)
Based on gross area, Pn = 1,909.6 kN

Based on net area, Pn = 2,393.6 kN
LRFD (on gross area): ASD (on gross area):
∅t Pn = (0.90)1,909.6 Pn /Ωt = 1,909.6/1.67
∅𝐭 𝐏𝐧 = 𝟏, 𝟕𝟏𝟖. 𝟔𝟒 𝐤𝐍 (design strength) 𝐏𝐧 /𝛀𝐭 = 𝟏, 𝟏𝟒𝟑. 𝟒𝟕 𝐤𝐍 (allowable strength)
LRFD (on net area): ASD (on net area):
∅t Pn = (0.75)2,393.6 Pn /Ωt = 2,393.6/2.00
∅𝐭 𝐏𝐧 = 𝟏, 𝟕𝟗𝟓. 𝟐𝟎 𝐤𝐍 𝐏𝐧 /𝛀𝐭 = 𝟏, 𝟏𝟗𝟔. 𝟖𝟎 𝐤𝐍
3. Compute the critical net area of the riveted connection shown if the thickness of the
plate is 12 mm. Diameter of rivets = 20 mm.
Nominal hole diameter for 20 mm rivets = 22 mm (Refer to Table 510.3.3)

dh = dn + 2 mm = 22 + 2
dh = 24 mm
Considering route ABCD: (2 holes only)
(50)2
Net width = 300 − 2(24) + 4(175)
Net width = 255.5714 mm

Considering route ABECD: (3 holes)
(75)2 (25)2
Net width = 300 − 3(24) + 4(100) + 4(75)
Net width = 244.1458 mm

Use smaller net width, 𝑤𝑛 = 244.1458 mm
Net area, 𝐴𝑛 = 𝑤𝑛 𝑡
An = 244.1458 (12)
𝐀 𝐧 = 𝟐, 𝟗𝟐𝟗. 𝟕𝟓 𝐦𝐦𝟐 < 0.85𝐴𝑔 = 0.85(300)(12) = 3,060 mm2
4. Determine the LRFD design tensile strength and the ASD allowable design tensile
strength for a W 10 x 45 with two lines of 20-mm diameter bolts in each flange using
A572 Grade 50 steel, with Fy = 345 MPa and Fu = 448 MPa and the NSCP 2015 provisions.
There are assumed to be at least three bolts in each line 100 mm on center and the bolts
are not staggered with respect to each other.
Properties of W 10 x 45:
Ag = 8,581 mm2
d = 256.50 mm
bf = 203.70 mm
t f = 15.70 mm
Sol’n:
Pn = (345)(8,581)
𝐏𝐧 = 𝟐, 𝟗𝟔𝟎. 𝟒𝟓 𝐤𝐍 (𝐛𝐚𝐬𝐞𝐝 𝐨𝐧 𝐠𝐫𝐨𝐬𝐬 𝐲𝐢𝐞𝐥𝐝𝐢𝐧𝐠)
An = 8,581 mm2 − 4(22 + 2)(15.70) = 7,073.8 mm2
𝑥̅
From Table 504.3.1 (Case 2), 𝑈 = 1 − 𝐿
Referring to tables in in Manual for one-half of a W10x45 (or, that is, a WT5x22.5), we find
that
x̅ = 23.04 mm (y̅ from ASEP Steel Handbook)
Length of connection, L = 2(100 mm) = 200 mm
23.04
𝑈 = 1− = 0.8848 (Case 2)
200
2 2
But bf = 203.70 mm > d = (256.50) = 171 mm
3 3
∴ U from Table 504.3.1 (Case 7) is 0.90 (The larger value of U is permitted to be used)
Ae = UAn = 0.90(7,073.8 mm2 ) = 6,366.42 mm2
Pn = Fu Ae = (448)(6,366.42)
𝐏𝐧 = 𝟐, 𝟖𝟓𝟐. 𝟏𝟔 𝐤𝐍 (based on tensile rupture)
Based on gross section yielding, Pn = 2,960.45 kN

Based tensile rupture strength, Pn = 2,852.16 kN
LRFD (on gross section yielding): ASD (on gross section yielding):
∅t Pn = (0.90)2960.45 Pn /Ωt = 2,960.45/1.67
∅𝐭 𝐏𝐧 = 𝟐, 𝟔𝟔𝟒. 𝟒𝟏 𝐤𝐍 𝐏𝐧 /𝛀𝐭 = 𝟏, 𝟕𝟕𝟐. 𝟕𝟐 𝐤𝐍
LRFD (on tensile rupture): ASD (on tensile rupture):
∅t Pn = (0.75)2,852.16 Pn /Ωt = 2,852.16/2.00
∅𝐭 𝐏𝐧 = 𝟐, 𝟏𝟑𝟗. 𝟏𝟐 𝐤𝐍 (design strength) 𝐏𝐧 /𝛀𝐭 = 𝟏, 𝟒𝟐𝟔. 𝟎𝟖 𝐤𝐍 (allowable strength)
3.6 BLOCK SHEAR

The failure of a member may occur along a path involving tension in one plane and shear on
a perpendicular plane as shown in Fig. TM-01, where several possible block shear failures
are illustrated. For these situations, it is possible for a “block” of steel to tear out.
Fig. TM-01 – Block shear.

NSCP 2015 – Sec. 510.4.3 (Block Shear Strength)
The available strength for the limit state of block shear rupture along a shear failure path
or path(s) and a perpendicular tension failure path shall be taken as
𝑷𝒏 = 𝟎. 𝟔𝑭𝒖 𝑨𝒏𝒗 + 𝑼𝒃𝒔 𝑭𝒖 𝑨𝒏𝒕
𝑷𝒏(𝒎𝒂𝒙) = 𝟎. 𝟔𝑭𝒚 𝑨𝒈𝒗 + 𝑼𝒃𝒔 𝑭𝒖 𝑨𝒏𝒕
∅𝒕 = 𝟎. 𝟕𝟓 (𝑳𝑹𝑭𝑫) 𝛀𝒕 = 𝟐. 𝟎𝟎 (𝑨𝑺𝑫)
Where:
𝐴𝑔𝑣 = gross area subject to shear, mm2
𝐴𝑛𝑡 = net area subject to tension, mm2
𝐴𝑛𝑣 = net area subject to shear, mm2
Where the tension stress is uniform, 𝑼𝒃𝒔 = 𝟏; where the tension stress is nonuniform,
𝑼𝒃𝒔 = 𝟎. 𝟓.
The tensile stress is generally considered to be uniform for angles, gusset (or connection)
plates, and for coped beams with one line of bolts. The connections of part (a) of Fig. TM-02
fall into this class.
Tensile stress in coped beams with two lines of bolts tends to be nonuniform such as
illustrated in part (b) of the figure. The stress there is nonuniform because the row of bolts
nearer the end of the beam picks up the largest proportion of the shear load.
3.7 SAMPLE PROBLEMS

5. The A572 Grade 50 (Fu = 448 MPa) tension member shown is connected with three 18-
mm bolts. For the given member, determine the following:
a. LRFD block shear rupture strength
b. ASD allowable block-shear rupture strength
c. LRFD design tensile strength
d. ASD allowable tensile strength
Properties of L 150 x 90 x 12:
Ag = 2,751.45 mm2
𝑥 = 21.16 𝑚𝑚
Sol’n:
(a) and (b) For Block shear strength:

Anv = [250 − 2.5(20 + 2)](12) = 2,340 mm2
Ant = [62.5 − 0.5(20 + 2)](12) = 618 mm2
Ubs = 1.0
Pn = 0.6(448)(2,340) + 1.0(448)(618)
𝐏𝐧 = 𝟗𝟎𝟓. 𝟖𝟔 𝐤𝐍
Agv = 250(12) = 3,000 mm2
Pn(max) = 0.6(345)(3,000) + 1.0(448)(618)
𝐏𝐧(𝐦𝐚𝐱) = 𝟖𝟗𝟕. 𝟖𝟔 𝐤𝐍
𝐏𝐧 > 𝐏𝐧(𝐦𝐚𝐱)
∴ Use 𝐏𝐧(𝐦𝐚𝐱) = 𝟖𝟗𝟕. 𝟖𝟔 𝐤𝐍 (for block shear rupture strength)
a. LRFD block shear rupture strength

∅t Pn = 0.75(897.86 kN) = 𝟔𝟕𝟑. 𝟒𝟎 𝐤𝐍
b. ASD allowable block-shear rupture strength

Pn /Ω𝑡 = (897.86 kN)/2.00 = 𝟒𝟒𝟖. 𝟗𝟑 𝐤𝐍
For gross section yielding:
𝑷𝒏 = 𝑭𝒚 𝑨𝒈 = (345)(2,751.45) = 𝟗𝟒𝟗. 𝟐𝟓 𝐤𝐍
For tensile rupture strength:

An = 2,751.45 − (20 + 2)(12) = 2,487.45 mm2
𝑥̅
From Table 504.3.1 (Case 2), 𝑈 = 1 −
𝐿
𝑥̅ = 21.16 mm
L = 100 + 100 = 200 mm
21.16
𝑈 =1− = 0.8942
200
From Table 504.3.1 (Case 8), with 2 or 3 fasteners in the direction of loading, 𝑈 = 0.60. Use
calculated 𝑈 = 0.8942.
Ae = U An = 0.8942(2,487.45 mm2 ) = 2,224.28 mm2
Pn = Fu Ae
Pn = (448)(2,224.28) = 𝟗𝟗𝟔. 𝟒𝟖 𝐤𝐍
Based on gross section yielding, Pn = 949.25 kN
Based tensile rupture strength, Pn = 996.48 kN
LRFD (on gross section yielding): ASD (on gross section yielding):
∅t Pn = (0.90)949.25 Pn /Ωt = 949.25/1.67
∅𝐭 𝐏𝐧 = 𝟖𝟓𝟒. 𝟑𝟑 𝐤𝐍 𝐏𝐧 /𝛀𝐭 = 𝟓𝟔𝟖. 𝟒𝟏 𝐤𝐍
LRFD (on tensile rupture): ASD (on tensile rupture):
∅t Pn = (0.75)949.25 Pn /Ωt = 996.48 /2.00
∅𝐭 𝐏𝐧 = 𝟕𝟏𝟏. 𝟗𝟒 𝐤𝐍 𝐏𝐧 /𝛀𝐭 = 𝟒𝟗𝟖. 𝟐𝟒 𝐤𝐍
LRFD (on block shear strength): ASD (on block shear strength):
∅t Pn = 𝟔𝟕𝟑. 𝟒𝟎 𝐤𝐍 (design strength) 𝐏𝐧 /𝛀𝐭 = 𝟒𝟒𝟖. 𝟗𝟑 𝐤𝐍 (allowable strength)
Block shear governs in both LRFD and ASD.
6. Using the example from Problem 2, determine its block shear strength in LRFD and ASD.
Fy = 248 MPa, Fu = 400 MPa. 22-mm diameter bolts in standard holes.
Since the combined thickness of the side plates is larger than that of the middle plate (given
that they have the same width), the middle plate has a smaller cross-sectional area.
Therefore, the middle plate will govern.
Case 1:

𝐴𝑛𝑣 = [190 − 2.5(24 + 2)](22)(2) = 5,500 𝑚𝑚2

𝐴𝑛𝑡 = [200 − 2(24 + 2)](22) = 3,256 𝑚𝑚2
𝐴𝑔𝑣 = 190(22)(2) = 8,360 𝑚𝑚2
𝑃𝑛 = 0.6(400)(5,500) + 1.0(400)(3,256) = 2,622.4 𝑘𝑁

𝑃𝑛(𝑚𝑎𝑥) = 0.6(248)(8,360) + 1.0(400)(3,256) = 2,546.37 𝑘𝑁
Use 𝐏𝐧 = 𝟐, 𝟓𝟒𝟔. 𝟑𝟕 𝐤𝐍 (𝐂𝐚𝐬𝐞 𝟏)

Case 2:
𝐴𝑛𝑣 = [190 − 2.5(24 + 2)](22) = 2,750 𝑚𝑚2

𝐴𝑛𝑡 = [275 − 2.5(24 + 2)](22) = 4,620 𝑚𝑚2
𝐴𝑔𝑣 = 190(22) = 4,180 𝑚𝑚2
𝑃𝑛 = 0.6(400)(2,750) + 1.0(400)(4,620) = 2,508 𝑘𝑁

𝑃𝑛(𝑚𝑎𝑥) = 0.6(248)(4,180) + 1.0(400)(4,620) = 2,469.98 𝑘𝑁
Use 𝐏𝐧 = 𝟐, 𝟒𝟔𝟗. 𝟗𝟖 𝐤𝐍 (𝐂𝐚𝐬𝐞 𝟐)

∴ 𝐀𝐝𝐚𝐩𝐭 𝐏𝐧 = 𝟐, 𝟒𝟔𝟗. 𝟗𝟖 𝐤𝐍
LRFD: ASD:
𝑃𝑢 = ∅𝑡 𝑃𝑛 = 0.75(2,469.98) 𝑃𝑛 2,469.98
𝑃𝑎 = =
Ω𝑡 2.00
𝐏𝐮 = 𝟏, 𝟖𝟓𝟐. 𝟒𝟗 𝐤𝐍
𝐏𝐚 = 𝟏, 𝟐𝟑𝟒. 𝟗𝟗 𝐤𝐍
IV. REFERENCES
1. Steel Structures Design and Behaviour by Charles G. Salmon & John E. Johnson
2. Structural Steel Design 5th Ed. By Jack C. McCormac & Stephen F. Csernak
3. National Structural Code of the Philippines (Chapter 5: Steel Structures) 2015 (Buildings,
Towers & other Vertical Structures).
4. American Institute of Steel Construction (AISC) specifications.

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DON HONORIO VENTURA STATE UNIVERSITY COLLEGE OF ENGINEERING

Cabambangan, Villa de Bacolor 2001, Pampanga, Philippines AND ARCHITECTURE

II. Course Outline:

3.2 NOMINAL STRENGTHS OF TENSION MEMBERS

In computing net area for tension and shear,

(Table 510.3.3) Nominal Hole Dimensions, mm

Where: s = longitudinal center-to-center spacing (pitch) of any two consecutive

NSCP 2001 (Sec. 502.4.4) – Reduction Coefficients, U

NSCP 2001 (Sec.502.4.1) NSCP 2015 (Table 504.3.1)

Ag = 130(12) = 1560 mm2

An = [130 − 2(18 + 2)](12) = 1,080 mm2 = Ae

Pn = 345(1,080) = 372.60 kN (on net area)

Based on gross area, Pn = 386.88 kN

 Based on gross area (yielding):

Side Plate (350 mm x 12 mm):

 Based on net area (rupture):

Based on gross area, Pn = 1,909.6 kN

Nominal hole diameter for 20 mm rivets = 22 mm (Refer to Table 510.3.3)

Net width = 255.5714 mm

Net width = 244.1458 mm

Based on gross section yielding, Pn = 2,960.45 kN

3.6 BLOCK SHEAR

Fig. TM-01 – Block shear.

3.7 SAMPLE PROBLEMS

(a) and (b) For Block shear strength:

Agv = 250(12) = 3,000 mm2

Pn(max) = 0.6(345)(3,000) + 1.0(448)(618)

∴ Use 𝐏𝐧(𝐦𝐚𝐱) = 𝟖𝟗𝟕. 𝟖𝟔 𝐤𝐍 (for block shear rupture strength)

a. LRFD block shear rupture strength

b. ASD allowable block-shear rupture strength

For tensile rupture strength:

𝑷𝒏 = 𝟎. 𝟔𝑭𝒖 𝑨𝒏𝒗 + 𝑼𝒃𝒔 𝑭𝒖 𝑨𝒏𝒕

𝐴𝑛𝑣 = [190 − 2.5(24 + 2)](22)(2) = 5,500 𝑚𝑚2

𝑃𝑛 = 0.6(400)(5,500) + 1.0(400)(3,256) = 2,622.4 𝑘𝑁

Use 𝐏𝐧 = 𝟐, 𝟓𝟒𝟔. 𝟑𝟕 𝐤𝐍 (𝐂𝐚𝐬𝐞 𝟏)

𝐴𝑛𝑣 = [190 − 2.5(24 + 2)](22) = 2,750 𝑚𝑚2

𝐴𝑔𝑣 = 190(22) = 4,180 𝑚𝑚2

𝑃𝑛 = 0.6(400)(2,750) + 1.0(400)(4,620) = 2,508 𝑘𝑁

Use 𝐏𝐧 = 𝟐, 𝟒𝟔𝟗. 𝟗𝟖 𝐤𝐍 (𝐂𝐚𝐬𝐞 𝟐)

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