Transcript
Page 1: Connections-I Bolted Connections

Structural Connections – I Bolted Connections

A.K.Mandal

Connections form an important part of any structure and are designed more conservatively than members. This is because, connections are more complex than members to analyse and the discrepancy between analysis and actual behavior is large. Moreover, in case of overloading, we prefer the failure confined to an individual member rather than in connection, which could affect many members. Connections account for more than half the cost of structural steelwork and so their design and detailing are of primary importance for economy of the structure.

Just as members are classified as bending members or axially loaded members depending on the dominant force/moment resisted, connections are also classified into idealized types while designing. But the actual behaviour of the connection may be different and this point should always be kept in mind so that the connection designed does not differ from the intended type. Take for example, the connection of an axially loaded truss members at a joint. If the truss is assumed to be pin jointed, then the members should ideally be connected by means of a single pin or bolt. However, in practice, if the pin or bolt diameter works out to be larger than that possible, more than one bolt is used. The truss can then be considered pin-jointed only if the bending due to self-weight or other superimposed loads between joints is negligible.

Connection elements consist of components such as cleats, gusset plates, brackets, connecting plates and connectors such as rivets, bolts, pins and weld. The connections provided in steel structures can be classified as 1) riveted, 2) bolted and 3) welded connections. Riveted connections were once very popular and are still used in some cases, but will gradually be replaced by bolted connections. This is due to the low strength of rivets, higher installation cost and inherent inefficiency of the connection. Welded connections have the advantage that no holes need to be drilled in the member and consequently have higher efficiencies. However, welding in the field may be difficult, costly and time consuming. Welded connections are also susceptible to failure by cracking under repeated cyclic loads due to fatigue which may be due to working loads such as trains passing over a bridge (high-cycle fatigue) or earthquakes (low-cycle fatigue). A special type of bolted connections using High Strength Friction Grip (HSFG) bolts has been found to perform better under such conditions than the conventional black bolts used to resist predominantly static loading. Bolted connections are also easy to inspect and replace. The choice of using a particular type of connection is entirely that of the designer.

Riveted Connections:

A rivet is made up of round ductile steel bar (mild or high tensile) called shank, with a head at one end. The head can be of different shapes. The usual form of rivet head used in structural steel construction is the snap head. The snap heads and pan heads form a projection beyond the plate face, whereas countersunk head remains flashed with plate face.

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2 Connections-I: Bolted Connections

The diameter of the shank is called the nominal diameter. A hole slightly greater than the nominal diameter is drilled in the parts to be connected. The rivet is inserted and the head is formed at the other end. The rivets may be formed in cold state or may be heated to red hot state prior to driving. The rivets may be placed in a variety of patterns. The most common types of rivet patterns are chain riveting and diamond riveting. Staggered patterns can also be provided.

The design of riveted connections is same as that of bolted connections with ordinary bolts with the following differences –

1. The diameter of rivet to be used in calculation is the diameter of the hole, whereas for bolted connections it is the nominal diameter of the bolt.

2. The design stresses are different, lesser for bolts. Bolted Connections:

Classification of connections

a) Classification based on the type of resultant force transferred: The bolted connections are referred to as concentric connections (force transfer in tension and compression), eccentric connections (in reaction transferring brackets) or moment resisting connections (in beam to column connections in frames)

Ideal concentric connections should have only one bolt passing through all the members meeting at a joint [Fig 2 (a)]. However, in practice, this is not usually possible and so it is only ensured that the centroidal axes of the members meet at one point [Fig 2 (b)].

The Moment connections are more complex to analyse compared to the above two types and are shown in Fig 3 (a) and Fig 3 (b). The connection in Fig 3 (a) is also known as bracket connection and the resistance is only through shear in the bolts.

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Connections-I: Bolted Connections 3

The connection shown in Fig 3 (b) is often found in moment resisting frames where the beam moment is transferred to the column. The connection is also used at the base of a column where a base plate is connected to the foundation by means of anchor bolts. In this connection, the bolts are subjected to a combination of shear and axial tension.

b) Classifications based on the types of force experienced by the bolts: The bolted connections can also be classified based on geometry and loading conditions into three types namely, shear connections, tension connections and combined shear and tension connections.

Typical shear connections occur as a lap or a butt joint used in the tension members [Fig 4]. While the lap joint has a tendency to bend so that the forces tend to become collinear, the butt joint requires cover plates. Since the load acts in the plane of the plates, load transmission at the joint will ultimately be through shearing forces in the bolts.

In case of a lap joint or a single cover plate butt joint, there is only one shearing plane and so

the bolts are said to be in single shear. In case of double cover butt joint, there are two shearing planes and so the bolts will be in double shear. It should be noted that the single cover type butt joint is nothing but lap joints in series and also bends so that the centre of cover plate becomes collinear with the forces. In case of single cover plate (lap) joint, the thickness of the cover plate is chosen to be equal to or greater than the connected plates. While in double cover plate (butt) joint, the combined thickness of the cover plates should be equal to or greater than the connected plates.

A hanger connection is shown in Fig 5 (a). In this connection, load transmission is by pure tension in the bolts. In the connection shown in Fig 5 (b), bolts are subjected to both tension and shear.

c) Classification based on force transfer mechanism by bolts: The bolted connections are classified as bearing type (bolts bear against the holes to transfer the force) or friction type (force transfer between the plates due to the clamping force generated by the pre-tensioning of the bolts).

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4 Connections-I: Bolted Connections Bolts and bolting

Bolts used in steel structures are of three types- a) Unfinished bolts or black bolts (also called ordinary, common or rough bolt) b) Turned and fitted bolts c) High strength friction grip (HSFG) bolts.

Bolts are designated by grade x.y. In this nomenclature x indicates one-hundredth of the nominal ultimate tensile strength of the bolt in N/mm2 and the second number y indicates one-tenth of the ratio of yield stress to ultimate stress, expressed as percentage. Thus, grade 4.6 will have the ultimate tensile strength of 400 N/mm2 and yield strength of 0.6 x 400 i.e., 240 N/mm2.

Black bolts are unfinished and are made from mild steel rods with square or hexagonal head and nuts and usually of grade 4.6. Black bolts have adequate strength and ductility when used properly, but when tightening the nut snug tight (“Snug tight” is defined as the tightness that exists when all plies in a joint are in firm contact) will twist off easily if tightened too much. These bolts are used for light structures subjected to static loads and for secondary members such as purlins, bracings etc., and for roof trusses. They are not recommended for connections subjected to impact load, vibration and fatigue. These bolts are available from 5 to 36 mm diameter and designated as M5 to M36. However, in structural steel work, the most common ones are M16, M20, and M24 to M30.

Turned and fitted bolts are similar to unfinished bolts, with the difference that the shanks of these bolts are formed from hexagonal rods. The surface of these bolts are prepared and machined carefully to fit in the hole. Tolerances allowed are about 0.15 mm to 0.5 mm. These bolts are used when no slippage is permitted between the connected parts and where accurate alignment of components is required. They are mainly used in special jobs (in some machines and where there are dynamic loads). These bolts are available from grade 4.6 to 8.8.

High strength friction grip bolts (HSFG) provide extremely efficient connection and perform well under fluctuating/fatigue load condition. They are made from quenched and tempered alloy steel with grades from 8.8 to 10.9. The most common are general grade of 8.8 and have medium carbon content which makes them less ductile. The 10.9 grade have much higher tensile strength, but lower ductility. The size of the bolts available are from M16 to M36, but bolts of sizes M16, M20, M24 and M30 are commonly used. HSFG bolts require hardened washer to distribute the load under bolt heads. The tension in the bolt ensures that no slip takes place under working conditions and so the load transmission from plate to bolt is through friction and not by bearing.

The tightening of HSFG bolts can be done by either of the following methods (IS 4000):

1. Turn-of-nut tightening method: In this method the bolts are first made snug tight and then turned by specific amounts (usually either half or three-fourth turns) to induce tension equal to proof load.

2. Calibrated wrench tightening method: In this method the bolts are tightened by a wrench, calibrated to produce the required tension.

3. Alternate design bolt installation: In this method special bolts are used which indicate the bolt tension. Presently such bolts are not available in India.

4. Direct tension indicator method: In this method special washers with protrusions are used [Fig 6]. As the bolt is tightened, these protrusions are compressed and the gap produced by them gets reduced in proportion to the load. This gap is measured by means of a feeler gauge, consisting of small bits of steel plates of varying thickness, which can be inserted into the gap.

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Since HSFG bolts under working loads, do not rely on resistance from bearing, holes larger than

usual can be provided to ease erection and take care of lack-of-fit. Typical hole types that can be used are standard, extra large and short or long slotted. These are shown in Fig 7. However, the type of hole will govern the strength of the connection.

Spacing and Edge Distance of bolt holes

Pitch (p): The centre-to-centre distance between individual fasteners (i.e. bolts or rivets) in a line, in the direction of load/stress is called pitch. The distance between any two consecutive fasteners in a zigzag pattern of bolts, measured parallel to the direction of load/stress is called a staggered pitch. For wide plates, pitch may be defined as the c/c distance of bolts, measured along the transverse direction of load. A minimum spacing of 2.5 times the nominal diameter of the fastener is specified in the code (cl. 10.2.2) to ensure that there is sufficient space to tighten the bolts, prevent overlapping of washers and provide adequate resistance to tear-out of bolts. Similarly, the code also specified the maximum pitch value (cl. 10.2.3).

Fig 7 – Hole types for HSFG bolts

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6 Connections-I: Bolted Connections Gauge distance (g): The distance between adjacent parallel line of bolts, transverse to the direction of load/stress, or the distance between the back of the rolled section and the first bolt line or centre to centre distance between two consecutive bolts measured along the width of the member is called gauge distance. IS Handbook SP-1 provides gauge distance for various rolled sections.

Edge distance (e): The distance from the centre of a bolt hole to the edge of a member, measured at right angles to the direction of load is called the edge distance. The distance in the direction of stress from the centre of a hole to the end of the member is called end distance.

IS 800 specifies spacing and edge distance for bolt holes.

Types of Bolted Connections

a) Lap joint b) Butt joint.

Load transfer mechanism of bolts

Depending upon the types of bolts, ordinary or high-strength bolts, the load transfer from one connected part to another may be by shear and bearing or by friction; the former being called bearing-type connection and the latter slip-resistance (or slip-critical) connection.

A) Bearing-type connections

In this type of connections, it is assumed that the load to be transferred is larger than the frictional resistance caused by tightening the bolts. Consequently members slip a little over each other and the bolts bear on the plates, placing the bolts in shear and bearing. Fig 9 shows the free-body diagram of the shear force transfer in bearing-type bolted connection. It is seen that tension in one plate is equilibrated by the bearing stress between the bolt and the hole in the plate. Since there is a clearance between the bolt and the hole in which it is fitted, the bearing stress is mobilized only after the plates slip relative to one another and start bearing on the bolt. The section x-x in the bolt is critical section for shear. Since it is a lap joint, there is only one critical section in shear (single shear) in the bolt. In case of butt joint there would be two critical sections in shear in the bolt (double shear), corresponding to two cover plates.

Assumptions in design of bearing bolts

a) Friction between the plates is negligible b) Shear is uniform over the cross-section of the bolt c) Distribution of stress on the plates between the bolt holes is uniform d) Bolts in a group subjected to direct loads share the load equally e) Bending stresses developed in the bolts is neglected.

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Connections-I: Bolted Connections 7 Design of bearing bolts

The failure of a joint can be in bolt or in plate. Thus the strength of a joint will be the minimum of the strength of joint based on (a) strength of bolts in the joint and (b) the net tensile strength of the plate. Since the bolt can fail either in shear or in bearing, the strength of bolts should be taken as the minimum strength of bolts in shear and in bearing.

1.0 Shearing strength of bearing-type bolts

The resistance of a bolt to shear is called nominal capacity of bolt in shear and is denoted by Vnsb. It depends on the ultimate tensile strength of bolt fub, the number of shear planes n, the nominal shank area of bolt Asb and net tensile stress area of bolt through the threads Anb. If threads do not lie in the shear plane, nominal shank area Asb is used as shear area, whereas if threads lie in the shear plane, net tensile stress area Anb is used as shear area. Net tensile stress area is specified in the code and is usually about 0.78 times the shank area.

A bolt subjected to a factored shear force (Vsb) shall satisfy Vsb ≤ Vdsb where, Vsb = factored shear force Vdsb = design shear strength of bolt

Design shear strength of bolt (Vdsb) is given by Vdsb = Vnsb / γmb where, Vnsb = nominal shear capacity of a bolt γmb = partial safety factor for the material of bolt = 1.25 [Refer Table 5 of IS 800:2007]

Nominal shear capacity of a bolt (Vnsb) is given by

Vnsb = fub

√3 ( nn Anb + ns Asb ) [cl. 10.3.3]

where, fub = ultimate tensile strength of a bolt nn = number of shear planes with threads intercepting the shear plane ns = number of shear planes without threads intercepting the shear plane Asb = nominal plain shank area of bolt

Anb = net tensile area at thread, may be taken as area corresponding to root diameter at thread

For bolts in single shear, either nn or ns is one and the other is zero. For bolts in double shear the sum of nn and ns is two.

Then design strength of bolt in shear is given by

Vdsb = fub

√3 γmb ( nn Anb + ns Asb )

Reduction factors for nominal shear capacity of bolt

In long joints, the bolts farther away from the centre of the joint will carry more loads than the bolts located close to the centre. Therefore, for joints having more than two bolts on either side of the connection with distance between the first and last bolt exceeding 15d in the direction of load, the shear capacity of the joint shall be reduced by the factor βlj, given by cl. 10.3.3.1 as

βlj = 1.075 lj

200 d but 0.75 ≤ βlj ≤ 1.0

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8 Connections-I: Bolted Connections where, d = nominal diameter of bolt lj = length of joint (equals to distance between first & last row of bolts measured in the

direction of load transfer.

Similarly, if the grip length exceeds five times the nominal diameter, the strength is reduced by a factor βlg as per cl. 10.3.3.2. The strength is also reduced if thickness of packing plate exceeds 6 mm by a factor βpk as per cl. 10.3.3.3.

So the design shear capacity of a bolt for long joint, long grip length and with packing plates (if provided) will be modified as below.

Vdsb = fub

√3 γmb ( nn Anb + ns Asb ) βlj βlg βpk

2.0 Bearing strength of bolt

If the connected plates are made of high strength steel, then failure of bolt can take place by bearing of the plates on the bolt. If the plate material is weaker than the bolt material, then failure will occur by bearing of the bolt on the plate and the hole will elongate.

The design bearing strength of a bolt on any plate Vdpb as governed by bearing is given by Vdpb = Vnpb / γmb where, Vnpb = nominal bearing strength of a bolt & γmb = 1.25

Nominal bearing strength of a bolt is given by

Vnpb = 2.5 kb d t fu

where,

kb is smaller of e

3 d0 ,

p3 d0

0.25 , fubfu

and 1.0

e, p = end and pitch distances of the fastener along bearing direction d0 = diameter of hole fub, fu = ultimate tensile stress of the bolt and the ultimate tensile strength of the plate

respectively d = nominal diameter of the bolt t = summation of thicknesses of the connected plates experiencing bearing stress in

the same direction. In case the bolts are countersunk, then it should be taken as the thickness of the plate minus half of the depth of countersinking.

Then design strength of a bolt in bearing is given by

Vdpb = 2.5 kb d t fuγmb

NOTE: In case of butt joints, thickness of cover plate should not be less than ⅝ t, where t is the thickness of main plate.

3.0 Tensile strength of bolt When bolts are subjected to tensile load (see Fig 5), the bolts are designed as tension member.

As per cl. 10.3.5, a bolt subjected to a factored tension force (Tb) shall satisfy

Tb ≤ Tdb

where, Tdb = design tension capacity of bolt

= Tnb / γmb where, Tnb = nominal tensile capacity of bolt & γmb = 1.25

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The nominal tensile capacity of bolt (Tnb) is calculated as

Tnb = 0.90 fub An < fyb Asb (γmb / γm0) γmb = 1.25 & γm0 = 1.10

where, fub = ultimate tensile strength of bolt fyb = yield stress of bolt An = net tensile stress area of bolt Asb = shank area of bolt.

Tension capacity of plate

The plate in a joint may fail in tension through the weakest section due to holes. The holes may be arranged in the longitudinal direction of plate, so that the number of holes is equal in all rows across the width [Fig 10 (a)] or staggered so that the number of holes across the width is reduced [Fig 10 (b)]. In the first case, the plate will fail across the weakest section, whereas in the second case the failure is along a zigzag pattern. The design tension capacity (Tdn) of plate is expressed as

Tdn = 0.9 fuγm 1

An [cl. 6.3.1]

where, fu = ultimate stress of material An = net effective area of the plate γm1 = partial safety factor for failure at ultimate stress = 1.25

The net effective area of the plate (An) depends on arrangement of bolt holes. If holes are not staggered, then An = (b n dh) t

For staggered holes, An =

where, b = width of plate n = number of holes dh = diameter of hole pi = staggered pitch gi = gauge distance

Efficiency of the joint

Efficiency of the bolted joint η = Strength of bolted joint per pitch lengthstrength of solid plate per pitch length

x 100

Combined shear and tension

A bolt required to resist both shear and tension at the same time shall satisfy

� VsbVdsb

�2

+ � TbTdb

�2

≤ 1.0

where, Vsb = factored shear force on bolt Vd = design shear capacity Tb = factored tensile force on bolt Tdb = design tension capacity

�b − n dh + �pi2 /(4 gi)

m

i=1

� t

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Connection with HSFG bolts

Free body diagram of an HSFG connection is shown in Fig 11. It can be seen that the pretension in the bolt causes clamping forces between the plates even before the external load is applied. When the external load is applied, the tendency of the two plates to slip against one another is resisted by the friction between the plates. Until the externally applied force exceeds this frictional resistance, the relative slip between the plates is prevented. The HSFG connections are designed such that under service load the force does not exceed the frictional resistance so that the relative slip is avoided during service. When the external force exceeds the frictional resistance the plates slip until bolts come in contact with the plate and start bearing against the hole. Beyond this point the external force is resisted by the combined action of frictional resistance and bearing resistance.

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Connections-I: Bolted Connections 11 1.0 Shear strength of HSFG bolts

HSFG bolts will come into bearing only after slip takes place. So if slip is critical (i.e., if slip cannot be allowed), then slip resistance is to be calculated which will govern the design. However, if slip is not critical, and limit state method is used, then bearing failure can occur at the limit state of collapse and needs to be checked. Even in the limit state method, since HSFG bolts are designed to withstand working load without slipping, slip resistance has to be checked anyway as a serviceability limit state.

a) Slip reistance: Design for friction type bolting in which slip is required to be limited, a bolt subjected only to a factored design shear force Vsf in the interface of connections shall satisfy the following [cl. 10.4.3]-

Vsf ≤ Vnsf / γmf

The design slip resistance or nominal shear capacity of a bolt as governed by slip for friction type connection is calculated as

Vnsf = μf ne Kh Fo

where, μf = coefficient of friction (slip factor) [as specified in Table 20, page 77 of IS 800] ne = number of effective interfaces offering frictional resistance to slip Kh = 1.0 for fasteners in clearance holes = 0.85 for fasteners in oversized & short slotted holes and for fasteners in long

slotted holes loaded perpendicular to the slot =0.7 for fasteners in long slotted holes loaded parallel to the slot γmf = 1.10 (if slip resistance is designed at service load) = 1.25 (if slip resistance is designed at ultimate load) Fo = minimum bolt tension (proof load) at installation (= Anb fo) Anb = net area of bolt at threads fo = proof stress = 0.70 fub fub = ultimate tensile stress of bolt

b) Bearing strength of HSFG bolts: HSFG bolts will come into bearing only after slip takes place. So if slip is not critical and limit state method is used, bearing is to be checked at ultimate limit state. The design bearing capacity of bolt can be determined using the same formula used for bearing-type i.e., ordinary (or black) bolts

Vdpb = Vnpb / γmb and Vnpb = 2.5 kb d t fu

c) Tensile strength of HSFG bolts: The design tensile strength of HSFG bolt is similar to that of bearing-type i.e., ordinary (or black) bolts and is given by

Tnf = 0.90 fub An < fyb Asb (γm1 / γm0) where, fu is the ultimate tensile stress of the bolt and An is the net tensile area of the bolt and may be taken as the area at the root of the threads. Asb is shank area of the bolt, γm1 is the partial safety factor for material resistance governed by ultimate stress = 1.25, γm0 is the partial safety factor for material resistance governed by yielding = 1.10, and fyb is the yield stress of the bolt. As per IS 800, HSFG bolt subjected to a factored tension force Tf should satisfy

Tf ≤ Tnf / γmb where, γmb = partial safety factor for material of the bolt = 1.25.

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d) Combined shear and tension in HSFG bolt: The following equation should be satisfied -

� VsfVdsf

�2

+ � TfTdf�

2 ≤ 1.0

where, Vsf is applied factored shear at design load, Vdsf is design shear strength, Tf is externally applied factored tension at design load and Tdf is design tensile strength.

Examples:

P-1: Find the efficiency of the lap joint shown in the figure. Given: M20 bolts of grade 4.6 and Fe 410 plates are used.

Ans:

For M20 bolt of grade 4.6, Diameter of bolt, d = 20 mm Diameter of bolt hole, d0 = 22 mm [refer table 19 of IS 800] Ultimate strength, fub = 400 N/mm2 Partial safety factor, γmb = 1.25 [refer table 5 of IS 800]

For Fe 410 plate Ultimate strength, fu = 410 N/mm2 [refer table 1 of IS 800] Partial safety factor, γm1 = 1.25 [refer table 5 of IS 800]

Strength of bolts

(i) Design strength in shear Bolts will be in single shear. Considering shear plane at threads, nn = 1 and ns = 0 Anb = 0.78 x π

4 d2 = 0.78 x 𝜋𝜋

4 x 202 = 245.04 mm2

Strength of a single bolt in shear

Vdsb = fub

√3 γmb ( nn Anb + ns Asb ) =

400√3 x 1.25

( 1 x 245.04 + 0 ) = 45,272 N = 45.272 kN

Design strength of joint in shear = number of bolts x Vdsb = 6 x 45.272 = 271.632 kN . . . . . (1)

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Connections-I: Bolted Connections 13 (ii) Design strength in bearing

Vdpb = 2.5 kb d t fuγmb

kb is least of

a) e3 d0

= 303 x 22

= 0.4545

b) p3 d0

− 0.25 = 603 x 22

− 0.25 = 0.6591

c) fubfu

= 400410

= 0.9756

d) 1.0 ∴ kb = 0.4545 Design strength of a single bolt in bearing

Vdpb = 2.5 kb d t fuγmb

= 2.5 x 0.4545 x 20 x 20 x 4101.25

= 149,076 N = 149.076 kN

Design strength of joint in bearing = number of bolts x Vdpb = 6 x 149.076 = 894.456 kN . . . . . (2) Hence strength of bolts = lesser of (1) and (2) = 271.632 kN . . . . . (3)

Strength of plates

Net tensile strength of plate is

Tdn = 0.9 fuγm 1

An = 0.9 x 4101.25

x (180 3 x 22) x 20

= 673,056 N = 673.056 kN . . . . . (4)

Hence strength of the given bolted joint = lesser of (3) and (4) = 271.632 kN

Strength of solid plate = 0.9 fuγm 1

A = 0.9 x 4101.25

x (180 x 20)

= 1062,720 N = 1062.72 kN

Efficiency of the bolted joint η = Strength of bolted joint strength of solid plate

x 100

= 271.632

1062.720 x 100 = 25.56 %

P-2: Find the efficiency of the joint, if in the above example (P-1) instead of lap, butt joint is made using two cover plates each of 12 mm thick and 6 numbers of bolts on each side.

Thickness of main plate is less than sum of thicknesses of cover plates. So, in this case strength of plates and strength of bolts in bearing are same as in example P-1. However strength of bolts in shear will be different.

Strength of bolts in shear Bolts are in double shear. Consider one shear plane at root of threads and the other at shank of bolt.

Asb = π4

x d2 = π4

x 202 = 314.16 mm2

Anb = 0.78 x π4 x d2 = 0.78 x π4 x 202 = 245 mm2

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nn = 1 and ns = 1

Vdsb = fub

√3 γmb ( nn Anb + ns Asb ) =

400√3 x 1.25

( 1 x 245 + 1 x 314.16)

= 103,306 N = 103.306 kN

Hence design strength of bolts in shear = 6 x 103.306 = 619.836 kN . . . . . (1) Design strength of bolts in bearing = 894.456 kN [from example P-1] . . . . . (2) Hence design strength of bolts = lesser of (1) and (2) = 619.836 kN . . . . . (3) Net tensile strength of plate = 673.056 kN [from example P-1] . . . . . (4) ∴ Design strength of joint = lesser of (3) and (4) = 619.836 kN

Design strength of solid plate = 1062.720 kN [from example P-1]

∴ Efficiency η = Strength of bolted joint strength of so lid plate

x 100 = 619.836

1062.720 x 100

= 58.33 %

P-3: Two plates 10 mm and 18 mm thick are to be jointed by a double cover butt joint. Assuming cover plates of 8 mm thickness, design the joint to transmit a factored load of 500 kN. Assume Fe 410 plate and 4.6 grade of bolts.

Ans:

Assume diameter of bolt, d = 20 mm Diameter of bolt hole, d0 = 22 mm For Fe 410 grade of steel, fu = 410 N/mm2 and γm1 = 1.25 For bolts of grade 4.6, fub = 400 N/mm2 and γmb = 1.25 Thickness of plate from bearing consideration will be least of 10 mm, 18 mm and 16 mm (2 x8) and hence t = 10 mm. The bolts are in double shear and bearing. Since two plates to be joined are of 10 mm and 18 mm thick, packing plate of 18 – 10 = 8 mm thick will be required. Since thickness of packing plate is more than 6 mm, shear strength of bolts has to be reduced by a factor βpkg whose value is βpkg = 1 0.0125 x tpkg = 1 0.0125 x 8 = 0.9 Now, bolt strength in double shear per pitch length will be

Vdsb = fub

√3 γmb ( nn Anb + ns Asb ) x βpkg

Considering one shear plane at root of threads and the other at shank of bolt,

Asb = π4

x d2 = π4

x 202 = 314.16 mm2

Anb = 0.78 x π4 x d2 = 0.78 x π4 x 202 = 245 mm2

nn = 1 and ns = 1

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Connections-I: Bolted Connections 15

∴ Vdsb = fub

√3 γmb ( nn Anb + ns Asb ) x βpkg =

400√3 x 1.25

( 1 x 245 + 1 x 314.16) x 0.9

= 92,980 N = 92.98 kN

Since pitch of bolt is not known, the factor kp and thus bearing strength of bolt can not be determined now. However, since in general shear controls the strength of bolts, not the bearing, we can calculate number of bolts on the basis of strength in shear of the bolt [ in other words, assuming strength of bolt in bearing to be more than that in shear, no of bolts can be calculated].

So, number of bolts required = 50092.90

= 5.38

Provide 6 bolts, 3 in a line. Since there will be 2 bolts per pitch length, strength of joint per pitch length on the basis of shearing strength of bolts = 2 x 92.98 = 185.96 kN.

Now equating strength of bolt in shear per pitch length to the net tensile strength of plate per pitch length, we find pitch (p) of the bolts.

Tdn = 0.9 fuγm 1

(p n x d0) x t = 185.96 x 103

or, 0.9 x 4101.25

x (p 1 x 22) x 10 = 185.96 x 103

or, p = 84.99 mm Provide pitch of 80 mm > 50 mm ( = 2.5 d)

Consider end distance, e = 1.7 x d0 = 1.7 x 22 = 37.4 mm [cl. 10.2.4.2] Provide e = 40 mm.

kb is least of

a) e3 d0

= 403 x 22

= 0.61

b) p3 d0

− 0.25 = 803 x 22

− 0.25 = 0.96

c) fubfu

= 400410

= 0.9756

d) 1.0 ∴ kb = 0.61

Design strength of bolt in bearing per pitch length

Vdpb = 2.5 kb d t fuγmb

= 2.5 x 0.61 x 20 x 10 x 4101.25

= 100,040 N

= 100.04 kN > 92.98 kN

Hence there is no need to increase end distance. Provide additional two bolts of 20 mm dia on the packing plate as shown.

NOTE: End distance causes value of kb to be the least among other values of kb. So if Vdpb comes out to be less than Vdsb, value of e can be increased and consequently the bearing strength.

Page 16: Connections-I Bolted Connections

16 Connections-I: Bolted Connections

P-4: A member of a truss consists of two angles ISA 75 x 75 x 6 placed back to back. It carries an ultimate tensile load of 150 kN and is connected to a gusset plate 8 mm thick placed in between the two connected legs. Determine the number of 16 mm diameter 4.6 grade ordinary bolts required for the joint. Assume fu of plate as 410 MPa.

Ans:

For M16 bolt of grade 4.6, Diameter of bolt, d = 16 mm Diameter of bolt hole, d0 = 18 mm [refer table 19 of IS 800] Ultimate strength, fub = 400 N/mm2 Partial safety factor, γmb = 1.25 [refer table 5 of IS 800]

For gusset plate Ultimate strength, fu = 410 N/mm2 [refer table 1 of IS 800] Partial safety factor, γm1 = 1.25 [refer table 5 of IS 800]

Strength of bolts

(i) strength of bolt in shear Bolts are in double shear. Consider one shear plane at root of threads and the other at shank of bolt.

Asb = π4

x d2 = π4

x 162 = 201.06 mm2

Anb = 0.78 x π4 x d2

= 0.78 x π4 x 162 = 156.83 mm2

nn = 1 and ns = 1

Vdsb = fub

√3 γmb ( nn Anb + ns Asb )

= 400

√3 x 1.25 ( 1 x 156.83 + 1 x 201.06)

= 66,121 N = 66.121 kN

(ii) strength of bolt in bearing Assume pitch, p = 2.5 d = 2.5 x 16 = 40 mm [cl. no. 10.2.2] end distance, e = 1.5 d0 = 1.5 x 18 = 27 mm [ cl. no. 10.2.4.2]

≈ 30 mm Thickness of plate from bearing consideration will be minimum of 8 mm and 2 x 6 = 12 mm. So t = 8 mm.

kb is least of

a) 𝑒𝑒3 𝑑𝑑0

= 303 𝑥𝑥 18

= 0.56

b) p3 d0

− 0.25 = 403 x 18

− 0.25 = 0.49

c) fubfu

= 400410

= 0.9756

d) 1.0 ∴ kb = 0.49

Page 17: Connections-I Bolted Connections

Connections-I: Bolted Connections 17

Design strength of bolt in bearing per pitch length

Vdpb = 2.5 kb d t fuγmb

= 2.5 x 0.49 x 16 x 8 x 4101.25

= 51,430 N = 51.43 kN

Hence strength of bolt = minimum of strength in shear and in bearing = 51.43 kN ∴ No of bolts required = 150 / 51.43 = 2.92 So provide 3 nos of bolts as shown in the fig.

NOTE: Also study Example 5.8, page 203 of SK Duggal or Example 5.13, page 405 of N Subramanian.

P-5: Design a single bolted double cover butt joint to connect boiler plates of thickness 12 mm for maximum efficiency. Use M16 bolts of grade 4.6. Boiler plates are of Fe 410 grade. Also find the efficiency of the joint.

Ans:

Strength of bolts

Design strength in shear d = 16 mm d0 = 18 mm Asb = π

4 d2 = 𝜋𝜋

4 x 162 = 201.06 mm2

Anb = 0.78 x π4

d2 = 0.78 x 𝜋𝜋4

x 182 = 156.83 mm2

Bolts will be in double shear. Considering one shear plane at threads and the other at shank, nn = 1 and ns = 1 Strength of a single bolt in shear

Vdsb = fub

√3 γmb ( nn Anb + ns Asb ) =

400√3 x 1.25

( 1 x 156.83 + 1 x 201.06 )

= 66,121 N = 66.121 kN Assuming strength of bolts in bearing will be more than strength of bolts in shear, Design strength of bolts per pitch length = 66.121 kN For maximum efficiency of the joint, net tensile strength of the plate per pitch length must be equal to strength of bolts per pitch length. Thickness of cover plates = ⅝ t = ⅝ x 12 = 7.5 mm Provide 2 cover plates of 8 mm thick. Then sum of thicknesses of cover plates (2 x 8 =16 mm) > thickness of plate (=12 mm). Hence O.K. Net tensile strength of plate per pitch length is

Tdn = 0.9 fuγm 1

An = 0.9 x 4101.25

x (p 1 x 18) x 12

where p = pitch of bolts Equating it with strength of bolts,

0.9 x 4101.25

x (p 1 x 18) x 12 = 66.121 x 103 or, p = 36.67 mm Provide pitch of 40 mm ≥ 2 d (2.5 x 16 = 40 mm) Now, we have to check the strength of bolts in bearing. Let e = 30 mm ( > 1.5 d0 = 1.5 x 18 = 27 mm)

Page 18: Connections-I Bolted Connections

18 Connections-I: Bolted Connections

kb is least of

a) e3 d0

= 30403 x 18

= 0.5556 b) p3 d0

− 0.25 = 403 x 18

− 0.25 = 0.4907

c) fubfu

= 400410

= 0.9756 d) 1.0

∴ kb = 0.4907

Design strength of bolt in bearing per pitch length

Vdpb = 2.5 kb d t fuγmb

= 2.5 x 0.4907 x 16 x 12 x 4101.25

= 77,256 N

= 77.256 kN > 66.121 kN Hence our assumption of bearing strength is more than shear strength is ok and so there is no need to revise the design. Draw the sketch of connection. Now, we shall check the efficiency of the joint.

Design strength of solid plate per pitch length = 0.9 fuγm 1

A = 0.9 x 4101.25

x (40 x 12)

= 141,696 N = 141.696 kN

Efficiency of the bolted joint η = Strength of bolted joint per pitch length strength of solid plate per pitch length

x 100

= 66.121

141.696 x 100 = 46.66 %

P-6: An ISA 100 x 100 x 10 mm carries a factored tensile force of 100 kN. It is to be jointed with a 12 mm thick gusset plate. Design a high strength bolted joint when (a) no slip is permitted, (b) when slip is permitted. Steel of grade Fe 410.

Ans:

Let us provide HSFG bolts of grade 8.8S and diameter 16 mm. For 8.8S grade bolts: fub = 800 N/mm2 Anb = 0.78 x π

4 x 162 = 157 mm2

For Fe 410 grade of steel: fu = 410 N/mm2 When slip is not permitted, the joint will be a slip-critical connection and when the sections

jointed are allowed to slip, the joint will be a bearing-type connection. (a) Slip-critical connection:

Proof load (i.e. minimum bolt tension at installation) = Anb x 0.7 fub = 157 x 0.7 x 800 x 10-3 = 87.9 kN Slip resistance or design shear capacity of bolt, Vdsf = μf ne Kh Fo/γmf Here, μf = 0.5 (assuming) ne = number of effective interfaces offering frictional resistance to slip = 1 γmf = 1.25 at ultimate load kh = 1.0 (assuming bolt in clearance holes)

Slip resistance of bolt = 0.5 x 1 x 1 x 87.91.25

= 35.16 kN

Number of bolts required = 10035.16

= 2.84 ≈ 3

So, provide 3 nos 16 mm diameter 8.8S grade HSFG bolts.

Page 19: Connections-I Bolted Connections

Connections-I: Bolted Connections 19 (b) Bearing-type connection

Strength of bolt in single shear, assuming shear plane in threaded area (nn = 1, ns = 0)

Vdsb = fub

√3 γmb ( nn Anb + ns Asb ) =

400√3 x 1.25

x 1 x 157 x 10-3 = 58 kN

Strength of bolt in bearing,

Vdpb = 2.5 kb d t fuγmb

= 2.5 x 1 x 16 x 10 x 4101.25

x 10-3 = 131.2 kN [assuming kb = 1]

Hence strength of bolt = 58 kN

Number of bolts required = 10058

= 1.72 ≈ 2

Provide 2 nos 16 mm diameter HSFG bolts of grade 8.8S.

P-7: The connection shown in the figure uses 20 mm dia 10.9S grade bolts with threads in the shear plane to connect an ISF 150 x 12 mm with gusset plate. Determine the strength of the joint, if

(a) Slip is not permitted (b) Slip is permitted.

Block shear strength of the joint need not be considered.

Ans:

For Fe 410 grade of steel: fu = 410 N/mm2 For 10.9S grade bolts: fub = 1040 N/mm2 Anb = 0.78 x π

4 x 202 = 245 mm2

(a) Slip-critical connection (slip is not permitted)

Proof load = F0 = Anb x 0.7 fub = 245 x 0.7 x 1040 x 10-3 = 178.36 kN Slip resistance of bolt, Vdsf = μf ne Kh Fo/γmf Assuming, μf = 0.5, Kh = 1, n = 1 & γmf = 1.25 (at ultimate load) Slip resistance of bolt = 0.5 x 1 x 1 x 178.36/1.25 = 71.34 kN Hence, strength of connection = no of bolts x slip resistance = 4 x 71.34 = 285.36 kN

(b) Bearing-type connection (slip is permitted)

Strength of bolt in single shear (nn =1, ns = 0),

Vdsb = fub

√3 γmb ( nn Anb + ns Asb ) =

1040√3 x 1.25

x 1 x 245 x 10-3 = 117.68 kN

Strength of bolt in bearing,

Vdpb = 2.5 kb d t fuγmb

= 2.5 x 1 x 20 x 12 x 4101.25

x 10-3 = 196.8 kN [assuming kb = 1]

So, strength of bolts = 117.68 kN [minimum of strength of bolt in shear and bearing] Hence, strength of connection = no of bolts x strength of bolt = 4 x 117.68 = 470.72 kN.

Prying Forces

In the design of HSFG bolts subjected to tensile forces, an additional force, called prying force is to be considered. This additional force is mainly due to flexibility of connected plates. Consider the connection of a T-section to a plate as shown in the Fig 12, subjected to tensile force 2Te. As tensile force acts, the flange of T-section bends in the middle portion and presses connecting plate near

Page 20: Connections-I Bolted Connections

20 Connections-I: Bolted Connections bolts. It gives rise to additional contact forces known as prying forces. IS 800 – 2007 has given the following expression for the prying force Q (cl. no. 10.4.7 of code) -

Q = lv2le

�Te − β η f0 be t4

27 le lv2 � . . . . (1)

where, lv = distance from the bolt centerline to the toe of fillet weld or to half the root radius for a rolled section

le = distance between prying force and bolt centerline and is minimum of either end distance or the value given by

le = 1.1 t �β f0fy

β = 2 for non pre-tensioned bolt and 1 for pre-tensioned bolt η = 1.5 be = effective width of flange per pair of bolts f0 = proof stress t = thickness of end plate.

Neglecting second term of eqn (1), which is relatively small, we get

Q = Te lv2 le

The thickness of end plate to avoid yielding of plate is obtained by equating the moment in the plate at bolt centre and at a distance lv from it to the plastic moment capacity of plate Mp. Thus, we get

MA = Q le and MC = Te lv - Q le MA = MC = Te lv / 2 = Mp

Taking Mp = (fy/1.1) (be t2/4), the minimum thickness of end plate is obtained as

tmin = �4.4 Mp /�fy be�

P-8: Design a hanger joint as shown to carry a factored load of 300 kN. Use an end plate of size 250 mm x 150 mm and appropriate thickness, M24 HSFG bolts (2 nos) and Fe 410 steel for end plate (fy = 250 MPa).

Ans:

We shall consider grade of HSFG bolt as 8.8 [Note: grade of 8.8 is commonly used]

Fig 12 Prying action

Page 21: Connections-I Bolted Connections

Connections-I: Bolted Connections 21 For Fe 410 grade of steel fu = 410 N/mm2, fy = 250 N/mm2 For bolt of grade 8.8, fub = 800 N/mm2 For 24 mm diameter bolt, Asb = π/4 x 242 = 452 mm2, Anb = 0.78 x π/4 x 242 = 353 mm2 do = 30 + 3 = 33 mm Proof stress, fo = 0.7 fub = 0.7 x 800 = 560 N/mm2 = 0 .56 kN/mm2 For pre-tensioned bolt, β = 1 Assuming an 8-mm fillet weld between hanger and end plate and an end distance of 40 mm, the distance from centre line of bolt to the toe of the fillet weld is lv = (250/2) – 10 – 8 – 40 = 67 mm For minimum thickness, Mp = Telv / 2 = (300/2) x 67/2 = 5025 kNmm

and tmin = �4.4 Mp/�fy be�

= �[(4.4 x 5025 x 103)/(250 x 150)] = 24.28 mm ≈ 25 mm

le = 1.1 t �β f0fy

= 1.1 x 25 x �1 x 560250

= 41.158 mm

le is lesser of 41.158 mm and 40 mm (end distance). So le = 40 mm

Prying force, Q = lv2le

�Te − β η f0 be t4

27 le lv2 � = 67

2 x 40 �150− 1 x 1.5 x 0.56 x 150 x 254

27 x 40 x 672 � = 117 kN

Total load on bolt = Te + Q = 150 + 117 = 267 kN Tension capacity of bolt, Tdb = Tnb / γmb Tnb = 0.9 Anb fub = 0.9 x 353 x 800 x 10-3 = 254 kN ≯ Asb fyb γmb / γm0 = 452 x 640 x 1.25 / 1.10 x 10-3 = 329 kN Hence, Tdb = 254 / 1.25 = 203 kN < 267 kN. So, it is unsafe. To reduce prying force, let us provide a thicker plate. Assume 45 mm thick plate. Tensile force in bolt = 150 kN Tension capacity of bolt = 203 kN So, prying force that can be allowed = 203 – 150 = 53 kN Moment at toe of weld = Te lv – Q le = 150 x 67 - 53 x 40 = 7930 kNmm

Moment capacity = fy

1.1 x be t2

4 = 250

1.1 x 150 x 452

4 x 10-3 = 17,258 kNmm > 7930 kNmm

So, ok.

Prying force, Q = lv2le

�Te − β η f0 be t4

27 le lv2 � = 67

2 x 40 �150− 1 x 1.5 x 0.56 x 150 x 454

27 x 40 x 672 � = 36.37 kN < 53 kN

Hence design is O.K.

* * * * *


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