Tension Member MCQ Quiz in मल्याळम - Objective Question with Answer for Tension Member - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Net sectional area (A_net) of the plate is given by

\({A_{net}} = \left( {B - n \times {d_h}} \right) \times t\)

where B = width of the plate, n = number of bolt hole at the section considered, d_h = bolt hole diameter, and t = thickness of the plate.

Given:

B = 200 mm,

n = 1,

t = 10 mm and

diameter of the bolt = 16 mm

Therefore, diameter of the bolt hole = 16 + 2 = 18 mm

Net sectional area, \({A_{net}} = \left( {200 - 1 \times 18} \right) \times 10 = 1820\;m{m^2}\)

Explanation:

As per IS: 800 - 2007

Calculation:

The net area of the tension member is given by

A_net = A₁ + kA₂

Area of rivet hole = 20 + 1.5 = 21.5 mm

\(\begin{array}{l} {{\rm{A}}_1} = \left( {100 - \frac{{10}}{2} - 21.5} \right) \times 10 = 735{\rm{\;}}m{m^2}\\ {{\rm{A}}_2} = \left( {100 - \frac{{10}}{2}} \right) \times 10 = 950{\rm{m}}{{\rm{m}}^2} \end{array}\)

∴ A_net = 735 + 0.7 × 950 = 1400 mm²

∴ P_safe = σ_at × A_net

= 150 × 1400 × 10^-3

Concept:

Specification for the design of splices:

1. Where the ends of the compression members are faced for complete bearing over the whole area there the splices are designed to hold the members accurately in position and to resist any tension where bending is also there.
2. A splice connection for a tension member is usually bolted connection and it is designed for a force of at least 0.3 times the member design capacity in tension or the design action, whichever is more.
3. Splices are designed as short columns.
4. A splice plate should be located at a place where a flexural moment in the column is zero i.e. at the location of the point of contra flexure.

Explanation

As per code IS 800:2007 

As per Clause 6.3 Design Strength Due to Rupture of Critical Section of IS 800: 2007, design strength is given for different members separately as follows.

\(T_{u}={α A_nf_u\over\gamma_{ml}}\)

When α = 1, above equation can be written as

\(T_{u}={A_nf_u\over\gamma_{ml}}\)

Where γml- partial safety factor for failure at ultimate stress = 1.25 (refer to Table 5 of IS 800: 2007)

fu - ultimate stress of the material

An - the net effective area of the member

\(A_n = {T_u \over {f_{u}/1.25}}\)

Important Points Net section rupture 

• When a tension member is connected using bolts, the cross-section reduces because of the holes present and this is referred to as net area. Holes in the members cause stress concentration at service loads. From the theory of elasticity, the tensile stress adjacent to a hole will be about two to three times the average stress on the net area.
• The ratio of maximum elastic stress to the average stress (fmax/favg) is called the stress concentration factor.
• Stress concentration is an important factor when a member is subjected to dynamic load where there is a possibility of brittle fracture or when the repeated application of load may lead to fatigue failure.
• In static loading of a tension member with a hole, the point adjacent to the hole reaches the field stress (fy) first. With further loading, the stress at that point remains constant at yield stress and each fiber away from the hole progressively reaches the yield stress.
• Deformations continue with increasing load until rupture/tension failure of the member occurs when the entire net cross-section of the member reaches the ultimate stress (fu).

Explanation

As per IS 800-1984 (Working Stress Method),

Shear lag factor (K) for

a) Single angle section connected to gusset plate by one leg only ⇒ K = \(\frac{{3{A_1}}}{{3{A_1} + {A_2}}}\)

b) Double angle connected back to back on the same side of gusset plate and tack riveted ⇒ K = \(\frac{{5{A_1}}}{{5{A_1} + {A_2}}}\)

Where,

A₁ = Net area of the connecting cross-section and A₂ = Area of the outstanding leg

∴ Net area of the section (A_net) = A₁ + k × A₂

Here A₁/A₂ = a/b

For a single angle in tension, the total area of the section is given by:

\(\rm{A=A_1+k\times A_2=a+(\frac{{3{a}}}{{3{a} + {b}}})\times b=a+(\frac{{1}}{{\frac{3{a}}{3a} +\frac{{b}}{3a}}})\times b=a+\frac{b}{1+0.33\times\frac{b}{a}}}\approx a+\frac{b}{1+0.35\times\frac{b}{a}}\)

∴ \(A = a+\frac{b}{1+0.35\times\frac{b}{a}}\)

Concept: -

As per IS 800:2007, cl. 8.2.1.5,

The shear lag effects in flanges may be disregarded when –

1. For outstands elements (supported along one edge),

b_o ≤ L_o / 20

2. For internal elements (supported along two edges),

b_i ≤ L_o / 10

Here,

L_o – length between the points of zero moment in the span

Concept:

The suitability of rolled steel sections for tension members is governed by shear lag effect.

Shear lag is the non-uniform straining of member due to tension and it occurs when action and reaction do not pass through the centre of gravity of cross section.

For sections other than plates like angles, I sections, T-sections, Channel-sections Shear Lag effect occurs which in turn reduces the tensile carrying capacity.  

Concept:

Modes of failure in steel tension member:

a) Gross section yielding: Generally a tension member without bolt holes can resist loads up to the ultimate load without failure. But such a member will deform in the longitudinal direction considerably (nearly 10%-15% of its original length) before fracture. At such a large deformation a structure becomes unserviceable.

b) Net section rupture Failure: When a tension member with a hole is loaded statically, the point adjacent to the hole reaches the yield stress first. On further loading, the stress at that point remains constant at yield stress and each fibre away from the hole progressively reaches the yield stress. Deformations continue with increasing load until finally rupture of the member occurs.

c) Block Shear Failure: The block shear failure becomes a possible mode of failure when the material bearing strength and bolt shear strength are higher.Block shear failure is the rupturing of the net tension plane(BC) and yielding on the gross shear plane(AB & CD),as shown in figure,which results in rupturing of shear plane as the connection length becomes shorter.

∴ Buckling takes place in compression members and not in tension members.

• Two channels placed back-to-back Fig(1) result in a small value of radius of gyration about y-y axis and are, therefore, seldom recommended.
• Two channels placed face to face Fig.(2) provide a larger value of radius of gyration as compared to channels back-to-back (Fig.(1)) and separated apart for the same spacing.
• Thus two channels face to face are the ideal section for compression members as these provide more rigidity.
• However, when a large radius of gyration is not the criteria for the type required, channels are placed back-to-back as the lacing is minimized.
• Also the exposed location of both ends of each rivet makes it easier to fabricate and further lower the cost. Channels placed face to face are also called placed toe-to-toe or placed with flanges turned inwards.

Additional Information In a built-up section carrying tensile force, the flanges of two channels are turned outward to have greater lateral rigidity.

Built-up Section- :

A built-up section is made by joining or combining two or more sections to have a section of required strength as per the design criteria. It can be combined by welding or bolting.

Here the flanges of the channel sections are kept outward to have a better lateral rigidity.

Sometimes rolled steel sections do not fulfill the design criteria. In such situations, built-up sections can be used.