Concept

As per IS 800:2007, the effective slenderness ratio of the laced built-up column shall be taken as 1.05 times maximum slenderness ratio of the column. This is done to account for the shear deformations.

∴ X = 1.05 and Y = to account for shear deformations.

Note:

The same value in the case of a battened built-up column system is 1.1

Concept:

The deflection of the given truss will be similar to its equivalent simply supported beam i.e. It will sag due to which top members of truss are subjected to pure compression and bottom members of truss are subjected to pure tension.

For this it can be concluded that Member AB is in compression and member CD is in tension.

Now, as per IS 800: 2007, Cl- 3.8, Slenderness ratio (SR) of steel sections is limited to avoid buckling in compression members  and its value is 180 for pure compression member and SR is limited to avoid excessive vibration or undesirable later movements and its value is 400 for pure tension member .

For Pure compression member, AB:

SR_max = 180

SR_actual­ = 200 > SR_max ⇒ Member AB will buckle

For Pure Tension member, CD:

SR_max = 400

SR_actual­ = 450 > SR_max

⇒ Although SR exceeded the limited value but member CD will not buckle as tension member does not buckle. However, member CD will behave like a cable.

Slenderness Limits for other conditions as per IS 800:2007

Concept:

A plate may buckle locally due to compressive stresses. This local buckling can be avoided before the limit state is reached, by limiting the width to thickness ratio (b/t) of each element of the cross-section. On this basis 4 classes of sections are defined as follows:

1) Plastic section (Class 1): Those sections which can develop plastic hinges and has the rotational capacity required for the failure of the structure by plastic mechanism formation.

2) Compact section (Class 2): Those sections which can develop plastic hinge but have inadequate rotational capacity for the formation of plastic mechanism due to local buckling.

3) Semi-compact section (class 3): Those section in which extreme fibers can reach yield stress (f_y) but can not develop the plastic moment of resistive due to local buckling.

4) Slender-section (Class 4): Those section in which local buckling occurs before yield stress is reached are called slender section.

Hence statement I is false and statement II is true for a semi-compact section.

Concept:

As per yura,

Bucking load of column, \(p_e\; =p + \frac{{2M}}{d}\)

Where, p = Factored axial load

M = Factored axial moment

d = depth of beam section

Calculation:

Given, p = 500 kN, M = 45 kNm and d = 200 mm = 0.2 m

Concept:

The slenderness ratio is the ratio of effective length and radius of gyration i.e.

\(SR = \;\frac{{{l_{eff}}}}{r}\)

For the Maximum Slenderness Ratio, the radius of gyration (r) has to be minimum.

The radius of gyration about any axis, r is given as

\(r = \sqrt {\frac{I}{A}} \)

Where I is the MOI about that axis, A is the total area of the system.

The radius of gyration about Z-axis and about Y-axis needs to be evaluated separately and the minimum value is to be taken to find the maximum SR.

In this problem, the built-up tension member involves two-channel section and one plate. This entire system is divided into three parts as shown in the figure.

Let Y̅ and Z̅  be the centroid of this built-up system in y-direction and z direction respectively.

Calculation:

Given, l_e = 10 m, C_y = 23.5 mm, Plate : (280 × 6) mm

A₁ = 4630 mm², A₂ = 4630 mm², A₃ = 280 × 6 = 1680 mm²

y̅₁ = 150 + 6 = 156 mm                     (measured from top)

y̅₂ = 150 + 6 = 156 mm                     (measured from top)

y̅₃ = 3 mm  

∴ \(\bar y = \frac{{{A_1}{{\bar y}_1} + {A_2}{{\bar y}_2} + {A_3}{{\bar y}_3}}}{{{A_1} + {A_2} + {A_3}}}\)  \( = \frac{{4630 \times 156 + 4630 + 156 + 1680 \times 3}}{{4630 + 4630 + 1680}} = 139.5\;mm\) 

Due to symmetry about y axis, z̅ be the midpoint of the spacing between channel section.

i.e. \(\bar z = \frac{{300}}{2} = 150\;mm\) 

Total area, A_t = 4630 + 4630 + 1680

A_t = 10940 mm²

\({I_{yy}} = \left[ {313 \times {{10}^4} + 4630 \times {{\left( {\frac{{300}}{2} - 235} \right)}^2}} \right] \times 2 + \frac{{{{\left( {280} \right)}^3} \times 6}}{{12}}\) 

∴ I_yy = 165.42 × 10⁶ mm

\({I_{zz}} = \left[ {6420 \times {{10}^4} + 4630 \times {{\left( {156 - 132.5} \right)}^2}} \right] \times 2 + \frac{{{6^3} \times 280}}{{12}} + 1680 \times \left( {132.5 - \frac{6}{2}} \right)\)  = 133.74 × 10⁶ mm⁴

\({r_z} = \sqrt {\frac{{{I_{zz}}}}{{{A_{total}}}}} = \sqrt {\frac{{133.74 \times {{10}^6}}}{{10940}}} = 110.56\;mm\) 

\({r_y} = \sqrt {\frac{{{I_{yy}}}}{{{A_{total}}}}} = \sqrt {\frac{{165.42 \times {{10}^6}}}{{10940}}} = 123\;mm\) 

r = min (r_y, r_z) = 110.56 mm

\(Slenderness\;Rati{o_{max}} = \frac{{10 \times {{10}^3}}}{{110.56}} = 90.45\) 

Concept:

The bending strength or the plastic moment capacity of a laterally unsupported beam is given by:

Md = β_b Z_p f_bd

Where

β_b = 1 → for plastic and concept section

\(= \frac{{{{\rm{z}}_{\rm{e}}}}}{{{{\rm{z}}_{\rm{p}}}}} \to {\rm{for\;semi}} - {\rm{compact\;section}}.\)

Z_e = Elastic section modulus

Z_p = Plastic section modulus

f_bd = design bending compressive strength

\(= {{\rm{X}}_{{\rm{LT}}}}\frac{{{{\rm{f}}_{\rm{y}}}}}{{{{\rm{y}}_{{\rm{mo}}}}}}\)

y_mo = Partial safety factor for material = 1.1

X_LT = bending stress reduction factor to account for lateral torsional buckling

X_LT ≤ 1.0

Calculation:

For a semi-compact section of a laterally unsupported steel bean plastic moment capacity

\({{\rm{M}}_{\rm{d}}} = \frac{{{\rm{ze}}}}{{{\rm{zp}}}} \times {\rm{zp}} \times {{\rm{X}}_{{\rm{LT}}}}\frac{{{\rm{fy}}}}{{{{\rm{y}}_{{\rm{mo}}}}}}{\rm{\;}}\)

X_LT = 1.0

γ_mo = 1.1

f_y = 250 Mpa for grade fe 410

\(\therefore {\rm{\;Md\;}} = {\rm{\;}}200{\rm{\;}} \times {\rm{\;}}{10^{ - 5}}{{\rm{m}}^3} \times {\rm{\;}}1{\rm{\;}} \times \frac{{250}}{{1.1}}{\rm{N}}/{\rm{m}}{{\rm{m}}^2}\)

Strut r_x = 1.51 cm (same … for the single angle)

\({I_y} = 2{I_y} + 2AC_y^2 = 2 \times 12.9 + 2 \times 5.68 \times {\left( {1.45} \right)^2}\)

= 49.68 cm⁴

\({r_y} = \sqrt {\frac{{{I_y}}}{A}} = \sqrt {\frac{{49.08}}{{2A}}}\)

= 2.04 cm                                ___________________(2)

r_min = 1.51 cm = Lesser of \(\left\{ {\begin{array}{*{20}{c}} {1.51\ cm}\\ {2.04\ cm} \end{array}} \right.\)

\({\lambda _{min}} = \frac{l}{{{r_{min}}}} = \frac{{1.75 \times 100}}{{1.52}} = 115.89\)

\({\sigma _x} = 72 - \frac{{\left( {72 - 64} \right)}}{{110 - 120}}\left( {110 - 115.89} \right)\)

= 67.288 MPa

P_…. = σ_a.. × A = 67.288 × (2× 5.68)× 100 = 76.44 kN

Concept

1. As per IS 800:2007, Cl: 10.12.2, in case of angle members the lug angle and their connection to gusset or any other supporting member should be capable of developing strength of not less than 20 %  in excess of the force in the outstanding leg of the angle and connection of lug angle to main angle member should be capable of developing 40% in excess of the force.

2. The force in each part of the angle member gets distributed in the ratio of the gross area of the connected and outstanding leg of the main angle member.

3. The bolted connection to the lug angle to the gusset plate and the outstanding leg of the main angle are in single shear.

Calculation:       

Given:

Main angle member → ISA 100 × 75 × 10

Gusset plate thickness, t_g = 10 mm

Design Tensile load, T = 250 kN

Long angle → ISA 150 × 75 × 8

Gross area of the connected leg of the main angle \({A_{gc}} = \left( {150 - \frac{{10}}{2}} \right) \times 10\) 

∴ A_gc = 1450 mm²

Gross area of outstanding leg of main angle, \({A_{g0}} = \left( {75 - \frac{{10}}{2}} \right) \times 10\) 

∴ A_g0 = 700 mm²

Load shared by the outstanding log of the main angle is P₀

\({P_0} = \frac{{250 \times 700}}{{\left( {700 + 1450} \right)}} = 81.4\;kN\) 

Load shared by the connected leg of Main angle, P_c

\({P_c} = \frac{{250 \times 1450}}{{\left( {700 + 1450} \right)}} = 168.60\;kN\) 

Connection of lug angle to gusset plate

The longer leg of the lug angle is connected to the gusset plate.

Load on lug angle, P_LC = 1.2 × 81.4 = 97.68 kN

No. of bolts required, \(X = \frac{{97.68}}{{50}} = 1.95\) 

∴ Minimum bolts required, X = 2

Connection of Lug angle to the outstanding leg of the main angle.

The lug angle is connected to an outstanding leg of the main angle.

Load on the outstanding leg of lug Angle, P_L0 = 1.4 × 81.4

P_L0 = 114 kN

No. of bolts required. \(Y = \frac{{114}}{{50}} = 2.28\) 

∴ Minimum Bolts required, Y = 3

Effective Length of column = 6 m

Area = 19.03 cm²

I_x = I_y = 177 cm⁴

Cx = Cy = 2.84 cm.

\(\begin{array}{l}\lambda = \frac{L}{{r\;min}}\\r\;min = \sqrt {\frac{{{I_{min}}}}{A}} \\{I_{min}} = 4 \times \left[ {177 \times {{10}^4} + 1903 \times {{\left( {200 - 28.4} \right)}^2}} \right]\;\end{array}\)

⇒ 23122.72 cm⁴

A = 4 × 19.03 = 76.12 cm²

\(\begin{array}{l}r\min = \sqrt {\frac{{23122.72}}{{76.12}}} = 17.43\;cm\\\lambda = \frac{{{l_{eff}}}}{{r\;min}} = \frac{{600}}{{17.43}} = 34.42\\{\sigma _{ac}}\left( {from\;table} \right) = 145 - \frac{6}{{10}} \times 4.42\end{array}\)

= 142.34

P_safe = 4 × 19.03 × 10² × 142.34

P_safe ⇒ 1083.5 KN

Concept:

As per IS 800:2007, the tensile strength based on rupture of the net section of the plate is given as:

\({T_{n\;}} = \frac{{0.9{f_u}{A_{net}}}}{\gamma }\)

Where,

f_u = Ultimate strength of the plate

A_net = Minimum net area of the section

γ = Factor of safety for plate

To calculate A_net, all possible failure paths (straight as well as zig-zag) are required to be considered and the corresponding net area is to be computed for each possible failure path and the minimum net area is to be selected.

\({A_{net}} = \left\{ {B - n{d_h} + \;{\rm{\Sigma }}\left( {\frac{{{s^2}}}{{4g}}} \right)} \right\}t\;\)

Where,

t is the thickness of the plate joined.

d_h­ is the diameter of the hole

s is staggered pitch , g is gauge distance, b is width of plate

Calculation:

Given: B = 210 mm, P = 50 mm, g = 30 mm, r = 45 mm

Path (1 - 2)

A_net = (210 – 2 × 18) × 12 = 2088 mm²

Path – (1 – 3 – 4 - 2) or (1 – 3 – 4 - 7)

\({A_{net}} = \left( {210 - 4 \times 18 + \frac{{{{45}^2}}}{{2 \times 30}} \times 2} \right) \times 12 = 2466\;m{m^2}\) 

Path (1 – 3 – 4 - 5) or (6 – 3 – 4 - 5)

\({A_{net}} = \left( {210 - 4 \times 18 + \frac{{{{45}^2}}}{{2 \times 30}}} \right) \times 12 = 2061\;m{m^2}\) 

Path (1 – 3 – 4 - 5) or (6 – 3 – 4 - 5)

\({A_{net}} = \left( {210 - 4 \times 18 + \frac{{{{45}^2}}}{{2 \times 30}}} \right) \times 12 = 2061\;m{m^2}\) 

Path (6 – 3 - 7)

\({A_{net}} = \left( {210 - 3 \times 18 + 2 \times \frac{{{{45}^2}}}{{2 \times 30}}} \right) \times 12 = 2682\;m{m^2}\) 

From above, the minimum net area is obtained for path either 1 – 3 – 4 – 5 or 6 – 3 – 4 – 5..

∴ A_net = 2061 mm²

Now,

\({T_n} = \frac{{0.9\;{f_u}}}{\gamma }{A_{net}}\)                      [γ  = 1.25 as per IS 800, and f_u = 410 MPa]

⇒ T_n \(= \left( {\frac{{0.9 \times 410 \times 2061}}{{1.25}}} \right)\) = 608.41 kN