## 3

## Tension Members

#### 3.1 Introduction

A member carrying a tension is known as a tension member. It is also known as a tie (Figure 3.1). In practice, the axial force may not be acting through the centroid of the cross section. There may be a small eccentricity of the load which causes a non-uniform stress distribution in the member. But, this is usually neglected if the eccentricity is small. Tension members are used in a variety of applications, shown in Figures 1.1 to 1.5, as members of a truss (in roofs, bridges and towers), as bracing members in buildings, as suspenders in suspension bridges, as cables in cable stayed bridges.

Wires/cables, rods/bars, single/double structural sections and built-up sections may be used as tension members. A cable is a flexible tension member with one or more groups of wires or strands or ropes. Wire ropes with strand cores (Figure 3.2(a)) or independent wire rope cores are used in structural applications such as hoists, derricks, rigging slings, guy wires and hangers for suspension bridges. A strand consists of wires placed helically about a centre wire. A rope consists of strands laid helically about a centre strand. The advantage of wires/cables is their flexibility and strength. Small tension members may be hot rolled square or round rods or flat bars (Figure 3.2(b), 3.2(c)). These are used in bracing systems or in very light structures such as towers. The disadvantage of rods and bars is their inadequate stiffness. If some amount of rigidity is required or reversal of load may occur, structural shapes – single (Figure 3.2(d)–3.2(f)) or double (Figure 3.2(g)) or built-up sections (Figure 3.2(h)–3.2(l)) are used. With the advent of welding, tubular sections are being extensively used in roof trusses and towers. Angles are also used in towers. Built-up sections are mainly used in truss girder bridges. In the built-up sections, dashed lines indicate a lateral system such as lacing or battening, which makes the different components act as one section.

#### 3.2 Net Area

If no holes are made in a tension member for its end connection, the gross sectional area is effective in resisting the tension. If the end connection is made with bolts, holes are made in the member. The presence of a hole in a plate subjected to tension produces a stress concentration at the hole edge in the elastic range but the stress distribution becomes uniform at yielding or at ultimate stress as shown in Figure 3.3. The effective area of the cross section of a plate at the hole is the net area.

In a plate or flat, if the holes are in a single line, the net sectional area is obtained by multiplying the net effective width *b _{e}* with the thickness

*t*. If there are

*N*holes in a line,

*b*=

_{e}*b*−

*nd*

_{0}, where

*d*

_{0}is the diameter of the hole. For example, in the Figure 3.4(a), there are 2 holes in a line (i.e.,

*n*= 2). Therefore,

*b*=

_{e}*b*–

*2d*

_{0}. The net sectional area of the flat,

*A*=

_{n}*b*.

_{e}t

If the holes are staggered at close pitch (Figure 3.4(b)), the net effective area is obtained by multiplying the net effective width *b _{e}* with the thickness

*t*of the plate or flat. The net effective width

*b*is obtained using the following expression.

_{e}

where *p* = staggered pitch

*g* = gauge

*n* = number of holes

*n*_{1} = number of inclined lines between the holes

*d*_{0} = diameter of the hole

For example, in Figure 3.2(b), *n* = 4 and *n*_{1} = 3.

Again, the net sectional area of the flat, *A _{n}* =

*b*.

_{e t}#### 3.3 Shear-lag

Consider a long rectangular flat subjected to tension along one edge of the flat as shown in Figure 3.5. The normal stress distribution and deformation in the flat near the ends where forces are applied is non-uniform with more normal stress or elongation along the loaded edge and less normal stress or elongation at the unloaded edge. This phenomenon which is due to shear stresses and shear deformation is known as shear-lag. The distribution of normal stress and deformation of the flat becomes uniform at a section away from the ends i.e., in the middle of the flat.

This phenomenon has a practical significance in the design of steel sections such as angles, channels and tees, as they are connected to a gusset through one leg or web or flange, when used as tension or compression members. The connecting elements of the section are more stressed than the unconnected elements of the section near the loaded ends. For example, in the case of a single angle, the connected leg is more stressed than the outstanding leg near the loaded ends. Suitable modifications are made in their design to account for this phenomenon.

#### 3.4 Design of Tension Members as per IS800:2007

The tension members can sustain loads up to the ultimate load where they fail by rupture at a weakest (net) section. However, the yielding of the gross section of the member may take place before the rupture at the weakest section. Flats and other rolled sections may also fail by block shear, especially at bolted end connections. The factored design tension in the member (*T*) should be less than or equal to the design strength of the member (*T _{d}*) which is the least of the design strength due to the yielding of the gross section

*T*, the design strength due to the rupture at the Net section

_{dg}*T*and the design strength due to the block shear

_{dn}*T*.

_{db}#### 3.4.1 Design Strength Due to Yielding

*T*

_{dg}=

*A*

_{g}

*f*

_{y}/γ

_{m0}(3.2)

where *f _{y}* = yield strength of the material

*A _{g}* = gross area of the cross section

*γ*_{m0} = partial safety factor for the yielding = 1.1

#### 3.4.2 Design Strength Due to Rupture

#### 3.4.2.1 Flats/Plates

*T*

_{dn}= 0.9

*A*

_{n}

*f*

_{u}/γ

_{m1}(3.3)

where *f _{u}* = ultimate strength of the material

*A _{n}* = netwidth of the outstanding leg area of the cross section

*γ*_{m1} = partial safety factor for the rupture = 1.25

#### 3.4.2.2 Single Angle

The rupture strength of a single angle connected through one leg and affected by shear lag is given by

where

in which *w* = width of the outstanding leg

*b _{s}* = shear lag width (as shown in Figure 3.6)

*L _{c}* = length of the end connection, i.e., distance between the outermost bolts or length of the weld along the direction of load

*A _{nc}* = net sectional area of the connected leg

*A _{go}* = gross area of the outstanding leg

*t* = thickness of the angle

#### 3.4.2.3 Other Sections

The rupture strength of double-angles, tee, channel, I section etc. are also affected by shear-lag and may be calculated using Equation 3.4. For the calculation of β, *b _{s}* should be taken as the distance from the farthest edge of the outstanding leg to the nearest bolt or weld line in the connecting leg of a section. Double angles should be tack welded or bolted along their length at a spacing not exceeding 1000 mm.

#### 3.4.3 Design Strength Due to Block Shear

The block shear strength of a bolted connection is the least of

or

where *A _{vg}*,

*A*are the minimum gross and net areas in shear along the bolt line parallel to the line of the action of force, respectively (along 1-2-3 in Figure 3.7);

_{vn}*A*,

_{tg}*A*are the minimum gross and net areas in tension from the bolt hole to the edge of a plate or between the bolt holes, perpendicular to the line of action of force, respectively (along 3-4-5 in Figures 3.7); ƒ

_{tn}_{u}, ƒ

_{y}are the ultimate and yield strengths of the material of the plates, respectively.

Equation 3.5 accounts for the shearing strength (*V _{bdg}*) on the gross area

*A*at yielding and the tensile strength due to rupture (

_{vg}*T*) on the net area

_{bdn}*A*. Equation 3.6 accounts for the shearing strength (

_{tn}*V*) on the net area

_{bdn}*A*and the tensile strength due to yielding (

_{vn}*T*) on the gross area

_{bdg}*A*.

_{tg}Deesign a tubular section to carry a tension of 150 kN. The end connection is done using fillet welds by flattening the section. Also, design the end connection.

Using Yst 250 grade steel tubes with the yield strength = 250 MPa

The required cross sectional area of the steel tube

From Appendix D, steel tube with the following specifications may be provided.

Nominal diameter = 80 mm

Outside diameter = 88.9 mm

Class: Medium

Area of the cross section = 1,070 mm^{2}

Thickness = 4.0 mm

The end connection is done using fillet welding with the ends of the tube flattened as shown in Figure 3.8.

Length of the weld = perimeter of the tube = π × 88.9 = 279 mm

Design shear strength of the weld, ƒ_{wd} = 189.4 MPa (From Example 2.1)

Size of the weld (*s*) is calculated from

*s*× 189.4 = 1.5 × 150 × 10

^{3}

*s*= 6.08 mm

**Example 3.2**

Re-do Example 3.1 using a square hollow section.

For the required sectional area of 990 mm^{2} (as in Example 3.1), 63.5 mm × 63.5 mm × 4.5 mm hollow section may be selected from Appendix E.

Its sectional area is 1,010 mm^{2}.

The length of the fillet weld = the outer perimeter of the section = 4 × 63.5 = 254 mm

∴ 254 × 0.7*s* ×189.4 = 1.5 × 150 × 10^{3}

or the size of the fillet weld, gap *s* = 6.7 mm

Determine the minimum net area of cross section of 300 × 12 mm flat shown in Figure 3.9. The holes are of diameter 17.5 mm.

Pitch (*p*) = 40 mm

Gauge (*g*) = 60 mm

Net effective width *b _{e}*along the net section 1-2-5-6

Net effective width *b _{e}*along the section 1-2-3-4-7

(from Equation (3.1))

Net effective width *b _{e}* along the net section 1-2-3-4-5-6

∴ Minimum net area of the section = 243.3 × 12 = 2,920 mm^{2}

**Example 3.4**

Find the ultimate design strength of ∠ 100 100 × 10 in tension which is connected to a gusset 12 mm thick through 100 mm leg using M20 bolts of property class 4.6 in a single line. Assume that the bolt threads are outside the shear plane. The yield and the ultimate strengths of the steel are 250MPa and 410MPa, respectively.

The diameter of the hole for M20 bolt = 20 + 2 = 22 mm

The gross sectional of ∠ 100 100 × 10, *A _{g}* = 1,900 mm

^{2}(from Appendix A)

The design strength of the angle due to yielding,

*T _{dg}* =

*A*

_{g}f_{y}/γ_{m0}

= 1,900 × 250/1.1 = 4,31,818 N

*Design of the end connection*

(for shear plane out of bolt threads)

*V _{dsb}* = 72,552/1.25 = 58,041.6N

∴ *k _{b}* = .0.606

*V _{npb}* = 2.5

*k*= 2.5 × 0.606 × 20 × 10 × 410 = 1,24,230 N

_{b}dtf_{u}*V _{dpb}* = 1,24,230/1.25 = 99,384 N

Number of bolts = 4,31,818/58,041.6 = 7.4

8 bolts may be provided as shown in Figure 3.10(b).

*Design strength due to rupture*

*b _{s}* = 60 + 100 − 10 = 150 mm

*L _{c}* = 7 × 60 = 420 mm

*Design strength due to the block shear*

*A _{vg}* = 460 × 10 = 4,600 mm

^{2}(along 1–9 in Figure 3.10(b))

*A _{vn}* = (460 − 7.5 × 22) × 10 = 2,950 mm

^{2}(along 1–9 in Figure 3.10(b))

*A _{tg}* = 40 × 10 = 400 mm

^{2}(along 9–10 in Figure 3.10(b))

*A _{tn}* = (40 − 0.5 × 22) × 10 = 290 mm

^{2}(along 9–10 in Figure 3.10(b))

=6,89,202N, or 5,93,689N

∴ T_{bd} =5,93,689N

The ultimate design strength of ∠ 10075 × 10 = Least of 4,31,811N,4,19,331N,5,93,689N

**Example 3.5**

Re-do Example 3.4 if the end connection is made using fillet welds. The yield and ultimate strengths of the steel are 250 MPa and 410 MPa, respectively.

Design the strength of the angle due to yielding = 4,31,818 N (as in Example 3.4)

The size of the weld (*s*) may be 5 mm which satisfies the maximum and minimum limits.

The design shear strength of the weld,

Providing two fillet welds on both the edges of connecting leg (Figure 3.11),

2 (0.7 × 5) × l × 189.4 = 4,31,818

Length of each weld, *l* = 326 mm

*b _{s}* =

*w*= 100 mm

*Design strength angle due to rupture*

The ultimate design strength of ∠ 10075 × 10 = Least of 4,31,818N,4,19,331N,5,52,485N

Find the ultimate load carrying capacity in the following cases:

- 2 ∠ 75 75 × 10 connected to the same side of the gusset,
- 2 ∠ 75 75 × 10 connected to both sides of the gusset.

Assume that in either case tacking bolts are provided and the end connection is done using M24 bolts of property class 4.6. The yield and ultimate strengths of the steel are 250 MPa and 410 MPa, respectively.

Diameter of the bolt hole = 24 + 2 = 26 mm

Gross area of the cross section of each angle = 1,400 mm^{2} (from Appendix A)

Design strength of the angles due to yielding, *T _{dg}* = 2 × 1400 × 250/1.1 = 6,36,363.6 N

*(i) 2 ∠ 75 75 × 10 connected to the same side of the gusset*

*Design of the end connection*

*V*= 1,04,474.8/1.25 = 8,35,780 N

_{dsb}*K*is least of

_{b}*K*= 0.64

_{b}*V*= 2.5 × 0.64 × 24 × 10 × 410 = 1,57,440 N

_{npb}*V*= 1,57,440/1.25 = 1,25,952 N

_{npb}

Number of bolts = 6,36,363.6/8,35,780 = 7.6

Four bolts in each angle may be provided as shown in Figure 3.12.

*Design strength due to rupture*

*b _{s}* = 40 + 75 − 10 = 105 mm

Length of the connection, *L _{c}* = 3 × 80 = 240 mm

For 2 ∠ 75 75 × 10, *T _{dn}* = 2 × 3,28,433.5 = 6,56,867 N

*Design strength due to block shear*

*A _{vg}* = 290 × 10 = 2,900 mm

^{2}(along 1-5 in Figure 3.12)

*A _{vn}* = (290 − 3.5 × 26) × 10 = 1,990 mm

^{2}(along 1-5 in Figure 3.12)

*A _{tg}* = 35 × 10 = 350 mm

^{2}(along 5-6 in Figure 3.12)

*A _{tn}* = (35 − 0.5 × 26) × 10 = 220 mm

^{2}(along 5-6 in Figure 3.12)

For 2 ∠ 75 75 × 10, *T _{bd}* = 2 × 4,18,709 = 8,37,418 N

The ultimate load carrying capacity of 2 ∠ 75 75 × 10

= least of 6,36,363.6 N, 6,56,867 N, 8,37,418 N

= 6,36,363.6 N

*(ii) 2 ∠ 75 75 × 10 connected to both sides side of gusset*

*Design of the end connection*

Let the thickness of the gusset = 12 mm

The bolts are in double shear. Assume that one shear plane is within and another outside the threads of the bolts.

*V _{dsb}* = 1,88,059/1.25 = 1,50,447 N

*k _{b}* = 0.64 as in case (i) and

*t*= the thickness of the gusset = 12 mm

*V _{npb}* = 2.5 × 0.64 × 24 × 12 × 410 = 1,88,928 N

*V _{dpb}* = 1,88,928/1.25 = 1,51,142 N

Number of bolts = 6,36,363.6 / 1,50,447 = 4.2

Five bolts may be provided as shown in Figure 3.13.

*Design strength due to rupture*

Length of the connection, *L _{c}* = 4 × 80 = 320 mm

For 2 ∠ 75 75 ×10, *T _{dn}* = 2 × 3,34,479 = 6,68,958 N

*Design strength due to the block shear*

*A _{vg}* = 370 × 10 = 3,700 mm

^{2}(along 1-6 in Figure 3.13)

*A _{vn}* = (370 − 4.5 × 26) × 10 = 2,530 mm

^{2}(along 1-6 in Figure 3.13)

*A _{tg}* = 35 × 10 = 350 mm

^{2}(along 6-7 in Figure 3.13)

*A _{tn}* = (35 − 0.5 × 26) × 10 = 220 mm

^{2}(along 6-7 in Figure 3.13)

For 2 ∠ 75 75 × 10, *T _{bd}* = 2 × 5,10,743 = 1,021,486 N

The ultimate load carrying capacity of 2 ∠ 75 75 × 10

= least of 6,36,363.6 N, 6,68,958 N, 10,21,486 N

= 6,36,363.6 N

Design a single angle to carry a tension of 100 kN. The end connection is to be done using M20 bolts of product Grade C and property Class 4.6. The yield and ultimate strengths of the steel are 250 MPa and 410 MPa, respectively.

Design strength of the angle due to yielding,

Factored design tension = 1.5 × 100 = 150 kN = 1,50,000 N

or *A _{g}* = 660 mm

^{2}

Try ∠ 60 60 × 6: A = 684 mm^{2} (from Appendix A)

*Design of the end connection*

*V _{dsb}* = 72,552/1.25 = 58,041.6 N

*k _{b}* = 0.606 as in Example 3.4

*V _{npb}* = 2.5 × 0.606 × 20 × 6 × 410 = 74,538 N

*V _{dpb}* = 74,538/1.25 = 59,630 N

Number of bolts = 1.5 × 100 × 103 / 58,041.6 = 2.58

Three bolts may be provided as shown in Figure 3.14.

*Design strength due to rupture*

Length of the connection, *L _{c}* = 2 × 60 = 120 mm

*b _{s}* = 35 + 60 − 6 = 89 mm

Not satisfactory. So, try ∠ 60 60 × 8:

b_{s} = 35 + 60 − 8 = 87 mm

OK

*Design strength due to the block shear*

*A _{vg}* = 160 × 8 = 1,280 mm

^{2}(along 1-4 in Figure 3.14)

*A _{vn}* = (160 − 2.5 × 22) × 8 = 840mm

^{2}(along 1-4 in Figure 3.14)

*A _{tg}* = 25 × 8 = 200 mm

^{2}(along 4-5 in Figure 3.14)

*A _{tn}* = (25 − 0.5 × 22) × 8 = 112 mm

^{2}(along 4-5 in Figure 3.14)

= 2,01,019 N or 1,88,619 N

= 1,88,619 N > 1,50,000 N

OK

So, ∠ 60 60 × 8 is alright.

**Example 3.8**

Re-do Example 3.7 if the end condition is done using fillet welds. The yield and ultimate strengths of the steel are 250 MPa and 410 MPa, respectively.

As the end connection is done by welding, ∠ 60 60 × 6 may be all right. 3 mm size of weld satisfies the maximum and minimum requirements. The design shear strength of the weld, ƒ_{wd} = 189.4 MPa (from Example 3.5). Providing two fillet welds on both the edges of the connecting leg (Figure 3.15),

2 (0.7 × 3) × *l* × 189.4 = 1,50,000.

Length of each weld, *l* = 189 mm

*b _{s}* =

*w*= 60 mm

*Design strength of angle due to rupture*

OK

**Example 3.9**

Design a double angle section to carry a tension of 300 kN. The end connection is to be made using M20 bolts of product Grade C and property class 5.6. Assume that the angles are provided on both sides of the gusset. The yield and ultimate strengths of the steel are 250 MPa and 410 MPa, respectively.

Diameter of the bolt hole = 20 + 2 = 22 mm

Design strength of the angle due to yielding,

Factored design tension = 1.5 × 300 = 450 kN = 4,50,000 N

or *A _{g}* = 1,980 mm

^{2}

Try 2 ∠ 75 75 × 8: *A* = 2,280 mm^{2} (from Appendix A)

*Design of the end connection*

As the bolts are in double shear, it is assumed that one shear plane interferes with the threads of the bolts.

*V _{dsb}* = 1,63,242/1.25 = 1,30,593.7 N

*k _{b}* = 0.606 (as in Example 3.4)

*V _{npb}* = 2.5 × 0.606 × 20 × 10 × 410 = 1,24,230 N

(Thickness of the gusset = 10 mm)

*V _{dpb}* = 1,24,230 / 1.25 = 99,384 N

Number of bolts 4.5

Five bolts may be provided as shown in Figure 3.16.

*Design strength due to rupture*

Length of the connection, *L _{c}* = 4 × 75 = 300 mm

*b _{s}* = 40 + 75 − 8 = 107 mm

For 2 ∠ 75 75 × 8, *T _{dn}* = 2 × 2,76,436.6 = 5,52,873 N > 4,50,000 N

OK

*Design strength due to block shear*

*A _{vg}* = 340 × 8 = 2,720 mm

^{2}(along 1–6 in Figure 3.16)

*A _{vn}* = (160 − 4.5 × 22) × 8 = 1,928 mm

^{2}(along 1–6 in Figure 3.16)

*A _{tg}* = 35 × 8= 280 mm

^{2}(along 6–7 in Figure 3.16)

*A _{tn}* = (35 − 0.5 × 22) × 8 = 192 mm

^{2}(along 6–7 in Figure 3.16)

For two angles, *T _{bd}* = 2 × 3,92,233 = 7,84,466 N > 4,50,000 N

OK

Tack bolts are to be provided at a spacing not greater than 1,000 mm.

**Example 3.10**

Re-do the Example 3.9 if the end connection is made using fillet welds.

As the end connection is done by welding, 2 ∠ 70 70 × 8 (A = 2 × 1,060 = 2,120 mm^{2}) may be all right.

Let the thickness of the gusset be 10 mm. A 5 mm size of weld may be provided which satisfies the maximum and minimum requirements.

The design shear strength of the weld, ƒ_{wd} = 189.4 MPa (from Example 3.5).

Providing two fillet welds on both the edges of a connecting leg as shown in Figure 3.17, 4 (0.7 × 5) × *l* × 189.4 = 1.5 × 300 × 10^{3}

∴ The length of each weld, *l* = 170 mm

*b _{s}* =

*w*= 70 mm

*Design strength angle due to rupture*

*= 2 × 3,3,03,826 = 6,07,651 N > 4,50,000 N*

_{dn}OK

#### 3.5 Lug Angles

A piece of angle used to reduce the length of the bolted end connection of a single angle or a channel type tension member with gusset is known as lug angle (Figure 3.18). In the case of a channel type tension member, lug angles are provided on either side of the flange. The provision of lug angles make the stress distribution in single angle or channel type tension members uniform and the shear-lag effect is nullified. This is because the unconnected leg of a single angle or the flanges of a channel are also connected to the gusset. Therefore, the whole net area of angle or channel section is effective including the lug angle. The reduction in the length of the end connection is achieved due to the less number of bolts provided in the connected leg of single angle or the web of channel type tension members. The following are the provisions of IS800: 2007 in respect to lug angles.

In the case of angle members, the lug angles and their connections to the gusset or other supporting member shall be capable of developing a strength not less than 20% in excess of the force in the outstanding leg of the angle and the attachment of the lug angle to the angle member shall be capable of developing 40% in excess of that force.

In the case of channel members, the lug angles and their connection to the gusset or other supporting member shall be capable of developing a strength of not less than 10% in excess of the force not accounted for by the direct connection of the member and the attachment of the lug angles to the members shall be capable of developing 20% in excess of that force.

At least two bolts should be used to attach the lug angle to the gusset or any other supporting member.

**Example 3.11**

Design the end connection for ∠ 100 100 × 10 using a lug angle for its full design strength. Use M20 bolts of product Grade C and property class 4.6. The grade of the steel is E 250.

Diameter of the bolt hole = 20 + 2 = 22 mm

Sectional area of ∠ 100 100 × 10 = 1,900 mm2 (from Appendix A)

Design strength of ∠ 100 100 × 10 due to yielding = 1,900 × 250/1.1 = 4,31,818 N

Net sectional area of this angle = 1,900 − 22 × 10 = 1,680 mm^{2}

Design strength of ∠ 100 100 × 10 due to rupture = 0.9 × 1,680 × 410/1.25

= 4,95,936 N

∴ The design strength of ∠ 100 100 × 10 = 4,31,818 N

The width of the outstanding leg = 100 − 10 = 90 mm

The load carried by the unconnected leg

The load carried by the connected leg = 4,31,818 − 2,27,273 = 2,04,545 N

The lug angle should be designed for 20% in excess of the load carried by the unconnected leg.

∴ The load to be carried by the lug angle = 1.2 × 2,27,273 = 2,72,727 N

The required gross area of the lug angle

Try ∠ 100 100 × 8 as the lug angle.

Gross sectional area of this angle = 1,540 mm^{2} (from Appendix A)

Net sectional area of this angle = 1,540 − 22 × 8 = 1,364 mm^{2}

Design strength of ∠ 100 100 × 8 due to rupture = 0.9 × 1,364 × 410/1.25 = 4,02,653 N

This is greater than 2,72,727 N. It is OK.

*Design of the end connection*

Assume that the thickness of the gusset is more than 10 mm

The bolts provided are subjected to single shear.

(for the shear plane out of the bolt threads)

*k _{b}* = 0.606 as in Example 3.4

*V _{npb}* = 2.5

*k*= 2.5 × 0.606 × 20 × 10 × 410 = 1,24,230 N (on 10 mm thickness)

_{b}dtf_{u}*V _{npb}* = 2.5

*k*= 2.5 × 0.606 × 20 × 8 × 410 = 99,384 N (on 8 mm thickness)

_{b}dtf_{u}*V _{dpb}* = 1,24,230/1.25 = 99,384 N (for 10 mm thick angle)

*V _{dpb}* = 99,384/1.25 = 79,507 N (for 8 mm thick angle)

∴ The design strength of the bold = 58,041.6 N

Number of bolts required to connect ∠ 100 100 × 10 with the gusset

Four bolts may be provided.

Number of bolts required to connect ∠ 100 100 × 10 with lug angle

Six bolts may be provided.

Number of bolts required to connect the lug angle with the gusset

Five bolts may be provided

The details are shown in Figure 3.19.

#### Problems

- Find the net width of the plate (Figure 3.20). The holes are of 24 mm in diameter.
- What is the ultimate design load carrying capacity of the flats in tension? (Figure 3.21) ƒ
= 250 MPa and ƒ_{y}= 410 MPa. Assume that the flats are connected with a lap._{u} - Design a single angle section to carry a tension of 210 kN. M22 bolts of the product Grade C and the property class 4.6 are provided as shown in Figure 3.22. ƒ
= 250 MPa and ƒ_{y}= 410 MPa._{u} - Design a double angle section (both sides of the gusset) to carry a tension of 400 kN. The end connection is done using high strength M24 bolts of the property class 8.8. Assume the connection as bearing type. ƒ
= 250 MPa and ƒ_{y}= 410 MPa._{u} - What is the ultimate design load carrying capacity in tension of MC 250 connected to a gusset as shown in Figure 3.23. Use M20 bolts of the product Grade C and the property class 5.6. ƒ
= 250 MPa and ƒ_{y}= 410 MPa._{u} - Determine the number of M20 bolts of the product Grade C and the property class 5.6 needed for the effective tensile capacity of the pair of angles placed on both sides of the gusset as shown in Figure 3.24. ƒ
= 250 MPa and ƒ_{y}= 410 MPa._{u} - Design an end connection for the strength of MC 200 in tension using lug angles. Use M16 bolts of the product Grade C and the property class 4.6. ƒ
= 250 MPa and ƒ_{y}= 410 MPa._{u}