As per IS 800: 1984, Clause 3.8.2

For structural member exposed to weather directly and non-accessible for painting, the minimum thickness of structural member its 8 mm.

Using Unwin’s equation:

$\varphi \ge 6.04\sqrt{{\mathrm{t}}_{\mathrm{m}\mathrm{m}}}$ 

$\therefore {\varphi }_{\mathrm{m}\mathrm{i}\mathrm{n}}\ge 6.04\times \sqrt{8}\ge 17.08\approx 18\mathrm{m}\mathrm{m}$

Concept:

The throat thickness of the welded connection can be written as:

t = KS

Where,

K is the factor that depends on the angle between fusion faces.

S is the size of the weld

For t to be minimum, the size of the weld is to be minimum and the minimum size of the weld, as per IS 800:2007 codal provisions, depends  on the  thickness of the thicker plate being connected as shown in below tabulated form:

For ‘t’ to be maximum, the size of the weld is to be maximum and as per IS 800: 2007, the maximum Size of the weld for square edge is given as:

S_max = Thickness of plate - 1.5

Calculation:

As per IS 800:2007, if the angle between fusion faces lies between 91-100^o then K = 0.65.

Thickness of thicker plate = 12 mm ⇒ S­_min = 5 mm

Minimum throat thickness, t_min = 0.65 × 5 = 3.25 mm

Maximum size of weld, S_max = max (12, 8) – 1.5 = 10.5 mm

Maximum throat thickness, t_max = 0.65 × 10.5 = 6.825 mm

For the calculation of Design wind load on structure IS 875-1987 part-III the relates the intensity of

wind pressure to the basic maximum basic wind speed Vb in m/sec. This wind speed is modified to

include risk level, terrain roughness, height and size of structure and local topography,

The design wind velocity Vz at any height for the structure is given below

Vz = k1. k2. k3.Vb

k1 = Risk coefficient or probability coefficient

k2 = Terrain, height and structure size factor.

Using standard empirical formula for this:

$\frac{M}{M_y}=\mathrm{S}\mathrm{h}\mathrm{a}\mathrm{p}\mathrm{e}\mathrm{F}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}-\frac{1}{2{(\frac{\mathrm{S}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{t}\mathrm{y}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}\mathrm{i}\mathrm{n}\mathrm{g}}{\mathrm{S}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{i}\mathrm{n}\mathrm{b}\mathrm{e}\mathrm{y}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{y}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}\mathrm{i}\mathrm{n}\mathrm{g}})}^2}$

For rectangular section, Shape factor = 1.5

Strain at yielding = ε_y

Strain beyond yielding = 2ε_y

$\frac{\mathrm{M}}{{\mathrm{M}}_{\mathrm{y}}}=1.5-\frac{1}{2{(\frac{{\in }_{\mathrm{y}}}{2{\in }_{\mathrm{y}}})}^2}$

$\frac{\mathrm{M}}{{\mathrm{M}}_{\mathrm{y}}}=1.5-\frac{1}{8}=1.375$

Or

M = 1.375M_y

Concept

IS 800:2007 classified the cross-sections into four classes depending upon the material yield strength and width to thickness ratio of the individual components within the cross-section and loading arrangement. The characteristics of these four classes are given below:

Plastic or Class I section:   

1. It can develop a plastic hinge and collapse mechanism.

2. They are fully effective in pure compression and capable of reaching the full plastic moment in bending and hence, used in plastic design.

Compact or Class 2 Section:

1. It can form a plastic hinge but does not have the capacity to develop a collapse mechanism because of local buckling.

2. They have lower deformation capacity but also fully effective in pure compression and are capable of reaching their full plastic moment in bending.

Semi- Compact or Class 3 Section:

1. It has the capacity to develop yield moment only and local buckling is liable to prevent the development of the plastic moment resistance.

2. They are also fully effective in pure compression but local buckling prevents the attainment of the full plastic moment in bending.  Bending moment resistance in these cross-sections is limited to yield moment only.

Slender or Class 4 Section:

1. This section fails before reaching the yield stress i.e. local buckling will occur even before the attainment of the yield stress in extreme fiber.

ISA 90 × 90 × 10 mm

Diameter of rivet, (d) = 18 mm

Diameter sectional area, A_g = (90 + 90 - 10) × = 1700 mm

For Fe 410 grade steel,

f_u. = 410 Mpa, γ_m1 = 1.25, f_y = 250 Mpa, γ_mo = 1.1

Tensile strength due to gross section yielding,

$T_{dg}=\frac{A_g\times f_y}{{\gamma }_{mo}}=\frac{1700\times 250}{1.1}=386.36kN$ 

Concept:

The moment capacity of the base plate is given by,

$M_d=1.2\times \frac{f_y}{{\gamma }_{m0}}\times Z_e$ 

Calculation:

Given: f_y = 250 mpa, γ_m0 = 1.1

$M_d=1.2\times \frac{250}{1.1}\times \left(\frac{1\times b}{6}\right)$ 

$\therefore M_d=1.2\times \frac{250}{1.1}\times \frac{1\times {30}^2}{6}=40.91kNm$ 

The design of battened column is prescribed under section 7.7 of IS 800 : 2007. As per

1) 7.7.1.3 – The number of battens shall be of a minimum of three bays members.

2) 7.7.1.4 – The effective slenderness ratio of should be taken 1.1 times the actual slenderness ratio.

3) 7.7.2.1 – Battens should be designed to carry 2.5% of the axial load of its transverse shear capacity.

4) 7.7.2.3 – The thickness of the batten or the tie plates shall not be less than one-fiftieth of the distance between innermost connecting lines of connection.

5) 7.7.3 – The spacing shall be such that the slenderness ratio shall neither be greater than 50 nor greater than 0.7 × slenderness ratio of member as a whole.

Hence ii, iii and iv are incorrect statements.

Transverse Stiffeners are provided to increase buckling resistance of the web due to shear, It is provided vertically along the span.

Horizontal Stiffener / Longitudinal Stiffener is designed to prevent web buckling due to bending compression.

End Bearing Stiffeners are provided at the supports & Load Bearing Stiffeners are provided at the points of concentrated loads.

A channel is used on the compression face which improves the lateral buckling strength of the beam and provides larger moment of inertia about the vertical axis against the lateral loads.

To increase the torsional stiffness of the girder, the channel is used just below the compression flange of the wide flange beam and is supported by brackets as shown in figure below.

Concept:

Purlins: These are the members which are spanning on the roof frames to support the roof coverings and runs parallel to ridge to connect different trusses situated in the longitudinal direction.

Rafters: These are a series of sloped structural members that extend from the ridge to the downslope perimeter or to the bottom chords and are designed to support the roof deck and its associated loads.

Now, the principal rafter of a roof truss with purlins placed at intermediate parts on panel length is essentially becomes a continuous beam with several intermediate supports and with a fixed end support. These kind of beams can be analysed easily to find bending moment and shear force with the help of moment distribution method. The whole concept in a pictorial view has shown in the following diagram:

As per IS 800:2007, CL 8.6.1.2, in order to avoid buckling of the compression flange into the web, web thickness shall satisfy the following:

1) When transverse stiffeners are not provided:

$\frac{d}{t_w}\le 345{ϵ}_f^2$

2) When transverse stiffeners are provided: (C = spacing between stiffners)

a. When C ≥ 1.5d

$\frac{d}{t_w}\le 345{ϵ}_f^2$

b. When C < 1.5 d

For 410 grade of steel, f_u = 410 Mpa

A_nb = 250 mm², f_ub = 500 Mpa

A_n = (120 ×10) - (2 × 22) = 1156 mm²

Strength of both in single shear:

$V_{sb}=A_{nb}\times \frac{fu}{\sqrt{3}{\gamma }_{mb}}=250\times \frac{500}{\sqrt{3}\times 1.25}$ 

⇒ V_sb = 57.75 kN

Strength of bolt in bearing:

$V_{pb}=2.5k_b{d}_t\times \frac{fu}{{\gamma }_{mb}}$ $=2.5\times 1\times 20\times 10\times \frac{410}{1.25}$ 

⇒ V_pb = 164 kN

Strength of plate $=\frac{0.9f_u{A}_n}{{\gamma }_{m1}}$$=\frac{0.9\times 410\times 1156}{1.25}$ = 341.25 kN

∴ The strength of the connection is the minimum of the three values i.e 57.75 kN.

Concept

As per IS 800: 2007, Cl- 3.8, the Slenderness ratio of steel sections is limited to avoid buckling in compression members and its value is 180 for pure compression members.

Now, to predict in which direction buckling will occur, it is desired to determine the actual slenderness ratio about both major and minor axis.

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

$\lambda =\frac{l_{eff}}{r}$

Further, the effective length under various conditions as per IS 800:2007 is given below:

Calculation

Case 1: Buckling about major axis

Column is restrained in position but no in direction at both ends i.e. its effective length

L_e1 = 1.00 L = 1 × 15 = 15 m

Radius of gyration about major axis, r₁ = 50 mm

Slenderness ratio about major axis, λ₁ = 15000/50 =  300

λ_max = 180 for pure compression member

Since the actual value of slenderness exceeds the maximum permitted value of SR ⇒ Buckling may occur about the major axis.

Case 1: Buckling about the minor axis

the column has fixed against rotation and translation at both ends i.e. its effective length

L_e1 = 0.50 L = 0.5 × 15 = 7.5 m

The radius of gyration about the minor axis, r₂ = 20 mm

Slenderness ratio about major axis, λ₂ = 7500/20 =  375

λ_max = 180 for pure compression member

Since the actual value of slenderness exceeds the maximum permitted value of Slenderness Ratio ⇒ Buckling may also occur about the minor axis.

∴ Buckling may occur either about the major axis or minor axis.

Concept:

As per clause number 8.2 of IS 800: 2007, the non-dimensional lateral torsional slenderness ratio is given by,

${\lambda }_{\mathrm{L}\mathrm{T}}=\sqrt{\frac{{\beta }_{\mathrm{b}}{\mathrm{Z}}_{\mathrm{p}}{\mathrm{f}}_{\mathrm{y}}}{{\mathrm{M}}_{\mathrm{c}\mathrm{r}}}}$ 

Where, ${\lambda }_{\mathrm{L}\mathrm{T}}$is the non-dimensional lateral torsional slenderness ratio

${\beta }_{\mathrm{b}}=1$ for plastic section

${\mathrm{Z}}_{\mathrm{p}}$ is plastic sectional modulus

${\mathrm{f}}_{\mathrm{y}}$ is yield stress of the material   

${\mathrm{M}}_{\mathrm{c}\mathrm{r}}$ is elastic critical moment corresponding to lateral torsional buckling of the beam 

Calculation:

Given, ${\beta }_{\mathrm{b}}=1,{\mathrm{Z}}_{\mathrm{p}}=3948812\mathrm{m}{\mathrm{m}}^3,{\mathrm{M}}_{\mathrm{c}\mathrm{r}}=16866\times {10}^6\mathrm{N}-\mathrm{m}\mathrm{m},{\mathrm{f}}_{\mathrm{y}}=250\mathrm{N}/\mathrm{m}{\mathrm{m}}^2$

$\therefore {\lambda }_{\mathrm{L}\mathrm{T}}=\sqrt{\frac{1\times 3948812\times 250}{16866\times {10}^6}}=0.242$ 

∴ The non-dimensional lateral torsional slenderness ratio will be nearly 0.242.

Concept:

The block shear strength as per IS 800:2007 is given as:

$T_{db1}=\frac{A_{vg}{f}_y}{\surd 3{\gamma }_{mo}}+\frac{0.9f_u{A}_{tn}}{{\gamma }_{m1}}$

$T_{db2}=\frac{0.9f_u{A}_{vn}}{\surd 3{\gamma }_{m1}}+\frac{f_y{A}_{tg}}{{\gamma }_{m0}}$

Where,

A_vg = Minimum  gross area in shear along the line of transmitted force

A_vn = Minimum net area in shear along the line of transmitted force

A_tg = Minimum gross area in tension perpendicular to the line of force from hole to the toe of angle or next last row of bolts in plates.

A_tn = Minimum net area in tension perpendicular to the line of force from hole to the toe of angle or next last row of bolts in plates.

Since the member is welded to the gusset plate, no net areas are involved and hence, A_vn and A_tn in above equation should be taken as corresponding gross areas.

Since length of weld is different on both sides, average length of weld on each side is considered.

Calculation:

Given:

f_u = 410 MPa, f_y = 250 Mpa, ISA – 100 × 25 × 6

Area of connected leg, $A_{nc}=\left(100-\frac{6}{2}\right)\times 6=582m{m}^2$

Area of outstanding leg, $A_0=\left(75-\frac{6}{2}\right)\times 6=432m{m}^2$

Average length of weld on each side, $L_{avg}=\left(\frac{150+300}{2}\right)$ = 225 mm

A_vg = (225 × 2) × 6 = 2700 mm²

A_vn = A_vg = 2700 mm²

A_tg = 100 × 6 = 600 mm²

A_tn = A_tg = 600 mm²

$T_{db1}=\frac{A_{vg}{f}_y}{\sqrt{3}{r}_{m0}}+\frac{0.9f_u{A}_{fn}}{r_{m1}}$  $=\left(\frac{2700\times 250}{\sqrt{3}\times 1.1}\right)+\frac{0.9\times 410\times 600}{1.25}$= 354.28 + 177.12 = 531.4 kN

$T_{db2}=\frac{0.9f_u{A}_{0n}}{\sqrt{3}{r}_{m1}}+\frac{A_{fg}{f}_y}{r_{m0}}$ $=\left(\frac{0.9\times 410\times 2700}{\sqrt{3}\times 1.25}\right)+\frac{600\times 250}{\left(1.1\right)}$ = 460.17 + 136.36 = 596.53 kN

T_d = Min (531.4, 596.53) = 531.4 kN

The maximum resultant shear stress (q) is the result of the resultant of the direct stress and shear stress due to the bending effect of the eccentric load.

Direct stress due to vertical load (q­_v) = $\frac{P}{WeldedArea}$

P = 250 kN

Welded area = 2 × (200 + 500) × 0.7 × 10 

$q_v=\frac{250\times 1000}{2\left(200+500\right)\times 0.7\times 10}=25.51N/m{m}^2$ 

Shear stress due to bending moment produced due to eccentric loading:

$q_b=\frac{P\times e\times r}{I_p}$ 

Where, I_p is polar moment of inertia.

I_p = I_xx + I_yy = 1.8 × 10⁸ + 7 × 10⁷ = 2.5 × 10⁸ mm⁴

Eccentricity is calculated from the C.G axis and it is given as e = 350 + 200/2 = 450 mm

r = √ (100² + 257²) = 275.77 mm

Now,

$q_b=\frac{250\times {10}^3\times 450\times 275.77}{2.5a\times {10}^3}=124.09N/m{m}^2$

Now,

$q=\surd q_v^2+q_b^2+2q_v{q}_{b}cos\theta$

$q=\surd {25.51}^2+{124.09}^2+2\times 25.51\times 124.09\times \frac{200}{275.77}=143.67\frac{N}{m{m}^2}$

Concept:

The force resisted by any weld having throat thickness t and length L is given as:

P = Effective area of weld × permissible stress in the weld 

Effective area of weld = throat thickness × Effective length of weld  i.e. A_e = t × L

The axial force T in the member is resisted by the forces P₁, P_2, and P₃ developed by the weld lines.

The forces P₁ and P₂ are resisting forces developed in longitudinal weld and P₃ is the force developed in transverse welding.

The forces P₁ and P₂ are act at edges of the angle and force P₃ will acts at the centroid of the given weld length.

For joints to be in equilibrium:

The moment at any point (say A located at bottom edge) must be zero i.e. Σ M_A = 0 and for horizontal Equilibrium, ΣF_H = 0.  Using this, the relation between forces P₁, P_2, and P₃ can be predicted.

Calculation:

∑F_H = 0 ⇒ P₁ + P₂ + P₃ T = 600 kN         ---(1)

P₁ = (7 × L₁) × 400= 2800 L₁ N

P₂ = 7 × L₂ × 400 = 2800 L₂ N

P₃ = (7 × 65) × 400 = 182 kN         ---(2)

∵ ∑M_CG = 0

P₁ × 47 = P₂ × 18

$\Rightarrow \frac{P_1}{P_2}=\frac{18}{47}$         ---(3)

From (1), (2) and (3)

$\left(\frac{18}{47}{P}_2\right)+P_2+18=600$

⇒ P₂ = 302.24 kN

$P_1=\frac{18}{47}\times P_2=\frac{18}{47}\times 302.24$ ⇒ P₁ = 115.75 kN

P₃ = 182 kN

Length of weld for its lower portion is given as

P₂ = 2800 L₂ ⇒ (302.24 × 10³) = 2800 × L₂ ⇒ L₂ = 108 mm

Concept:

Collapse load will be the minimum of the following

1. Collapse load due to Beam mechanism in span BD.

2. Collapse load due to sway mechanism in portal frame.

3. Collapse load due to combined mechanism in portal frame.

No. of plastic hinge required to form combined mechanism, N = degree of static indeterminacy +1.

Since, portal frame sways towards right due to horizontal load at B, plastic hinge will not develop at left corner i.e. at B. Further, No plastic hinge develops at E because it is a pin joint.

The desired collapse load is calculated by using principle of virtual work, which states that for a body which is in equilibrium (at rest), the summation external and internal work done zero. 

Calculation:

Collapse based on beam Mechanism in span BD:

Plastic hinge will be formed at B, C and D.

Refer the following figure:

External work done $={\mathrm{W}}_{\mathrm{u}1}\times \frac{\mathrm{L}}{2}\times \theta$

Internal Work done = - (M_p × θ (for PH at B) + 2 × M_p × θ (for PH at C) + Mp × θ (for PH at D))

∴ Internal Work done = -4M_p × θ (PH meAns: Plastic Hinge)

(Here we use minus sign because Mp­ and  both are in opposite direction)

By Principle of Virtual work

External Work Done + Internal Work Done = 0

${\mathrm{W}}_{\mathrm{u}1}\times \frac{\mathrm{L}}{2}\theta -4\mathrm{M}\mathrm{p}\theta =0$

$\therefore {\mathrm{W}}_{\mathrm{u}1}=\frac{8\mathrm{M}\mathrm{p}}{\mathrm{L}}$

Collapse based on sway Mechanism in portal frame:

Plastic hinge will be formed at A, B and D.

Refer the following figure:

External work done $={\mathrm{W}}_{\mathrm{u}2}\times \frac{\mathrm{L}}{2}\times \theta \left(\mathrm{a}\mathrm{t}\mathrm{B}\right)+{\mathrm{W}}_{\mathrm{u}2}\times 0={\mathrm{W}}_{\mathrm{u}2}\frac{\mathrm{L}}{2}\theta$

Internal Work done = - {Mp × θ (for PH at A) + Mp × θ (for PH at B) + Mp × θ (for PH at D)} = - 3 Mp × θ

(Here we use minus sign because Mp­ and θ both are in opposite direction)

By Principle of Virtual work

External Work Done + Internal Work Done = 0

${\mathrm{W}}_{\mathrm{u}2}\frac{\mathrm{L}}{2}\theta -3\mathrm{M}\mathrm{p}\theta =0$

$\therefore {\mathrm{W}}_{\mathrm{u}2}=\frac{6\mathrm{M}\mathrm{p}}{\mathrm{L}}$

Collapse based on Combined Mechanism in portal frame:

The plastic hinges will be formed at A, C and D.

Refer the following figure:

External work done $={\mathrm{W}}_{\mathrm{u}3}\times \frac{\mathrm{L}}{2}\times \theta +{\mathrm{W}}_{\mathrm{u}3}\times \frac{\mathrm{L}}{2}\times \theta ={\mathrm{W}}_{\mathrm{u}3}\mathrm{L}\theta$

Internal Work done = - {Mp × θ (for PH at A) + 2 × Mp × θ (for PH at C) + Mp × θ (for PH at D)} = - 5 Mp × θ

(Here we use minus sign because Mp­ and θ both are in opposite direction)

By Principle of Virtual work

External Work Done + Internal Work Done = 0

W_u3L θ - 5 Mp θ = 0 

${\mathrm{W}}_{\mathrm{u}3}=\frac{5\mathrm{M}\mathrm{p}}{\mathrm{L}}$

True Collapse Load, W_u = min (W_u1, W_u2, W_u3)

$\mathrm{C}\mathrm{o}\mathrm{l}\mathrm{l}\mathrm{a}\mathrm{p}\mathrm{s}\mathrm{e}\mathrm{L}\mathrm{o}\mathrm{a}\mathrm{d}=min\left(\frac{8\mathrm{M}\mathrm{p}}{\mathrm{L}},\frac{6\mathrm{M}\mathrm{p}}{\mathrm{L}},\frac{5\mathrm{M}\mathrm{p}}{\mathrm{L}}\right)=\frac{5\mathrm{M}\mathrm{p}}{\mathrm{L}}$

$\therefore {\mathrm{W}}_{\mathrm{u}}=\frac{5\mathrm{M}\mathrm{p}}{\mathrm{L}}$

Concept:

The design bending strength of such section is given by:

M_d = β_bZ_pf_y/γ_mo

Where

β_b = 1.0 (for plastic & compact section) or Z_e/Z_p (for semi-compact section)

Z_e = Elastic section modulus of the cross section

Z_p = Plastic section modulus of the cross section

f_y = yield stress of the material

γ_mo = 1.1, the partial safety factor

The design strength obtained from the above equation should be for less than 1.2 Z_ef_y/γ_mo for simply supported beam and 1.5 Z_ef_y/γ_mo for cantilever beams.

Calculation:

For laterally supported beam:

Bending capacity (M_d­) is given by:

${\text{M}}_{\text{d}}={\beta }_{\text{b}}\times {\text{Z}}_{\text{pz}}\times \frac{{\text{f}}_{\text{y}}}{{\gamma }_{\text{mo}}}$

For plastic section, β_b = 1

$\therefore {\text{M}}_{\text{d}}=1\times 851.11\times {10}^3\times \frac{250}{1.1}=193.43\text{kNm}$

Check: ${\text{M}}_{\text{d}}<1.2{\text{Z}}_{\text{e}}\frac{{\text{f}}_{\text{y}}}{{\gamma }_{\text{mo}}}$

$1.2{\text{Z}}_{\text{e}}\frac{{\text{f}}_{\text{y}}}{{\gamma }_{\text{mo}}}=1.2\times 751.9\times {10}^3\times \frac{250}{1.1}=205.06\text{kNm}$

∵ 193.43 < 205.06 ⇒ Safe

Diameter of rivet = 22 mm

Efficiency $\left(\eta \right)=\frac{{\mathrm{A}}_{\mathrm{n}}}{{\mathrm{A}}_{\mathrm{g}}}\times 100$

$\eta =\frac{\left(\mathrm{p}-\mathrm{d}\right)\times \mathrm{t}}{\mathrm{p}\times \mathrm{t}}=\frac{\left(\mathrm{p}-\mathrm{d}\right)}{\mathrm{p}}\times 100$

$\therefore \eta =\frac{\left(2.5\varphi -23.5\right)}{2.5\varphi }\times 100=\frac{\left(2.5\times 22-23.5\right)}{2.5\times 22}\times 100$

Concept:

The gross-sectional area of the beam (A) is given, by the relationship:

$\mathrm{A}=\frac{2{\mathrm{M}}_{\mathrm{z}}}{\mathrm{d}{\mathrm{f}}_{\mathrm{y}}}+\frac{{\mathrm{d}}^2}{\mathrm{k}}$

The optimum value of d may be obtained by differentiation the above equation with respect to d and equating it to zero.

$\frac{\partial \left(\mathrm{A}\right)}{\partial \mathrm{d}}=\frac{\partial }{\partial \mathrm{d}}\left(\frac{2{\mathrm{M}}_{\mathrm{z}}}{\mathrm{d}{\mathrm{f}}_{\mathrm{y}}}\right)+\frac{\partial }{\partial \mathrm{d}}\left(\frac{{\mathrm{d}}^2}{\mathrm{k}}\right)$

$0=\frac{-2{\mathrm{M}}_{\mathrm{z}}}{{\mathrm{d}}^2{\mathrm{f}}_{\mathrm{y}}}+\frac{2\mathrm{d}}{\mathrm{k}}$

$\frac{2{\mathrm{M}}_{\mathrm{z}}}{{\mathrm{d}}^2{\mathrm{f}}_{\mathrm{y}}}=\frac{2\mathrm{d}}{\mathrm{k}}$

${\mathrm{d}}^3=\frac{{\mathrm{M}}_{\mathrm{z}}\mathrm{k}}{{\mathrm{f}}_{\mathrm{y}}}$

$\mathrm{d}={\left(\frac{{\mathrm{M}}_{\mathrm{z}}\mathrm{k}}{{\mathrm{f}}_{\mathrm{y}}}\right)}^{0.33}$

Optimum depth of plate girder $\left(\mathrm{d}\right)={\left(\frac{{\mathrm{M}}_{\mathrm{z}}\mathrm{k}}{{\mathrm{f}}_{\mathrm{y}}}\right)}^{0.33}$

$\mathrm{d}={\left(\frac{12000\times {10}^6\times 180}{250}\right)}^{0.33}=2051.97\mathrm{m}\mathrm{m}=2.05\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}$

Concept:

For the given propped cantilever beam:

Let the plastic hinge is formed at distance ‘x’ from the end supports. Taking two sections for the given beam as shown below:

For the first half-section:

$\mathrm{\Sigma }{\text{M}}_{\text{A}}=0\Rightarrow {\text{M}}_{\text{p}}+{\text{M}}_{\text{p}}-{\text{w}}_{\text{c}}\times \frac{{(\text{L}-\text{x})}^2}{2}$

${\text{M}}_{\text{p}}=\frac{{\text{w}}_{\text{c}}{(\text{L}-\text{x})}^2}{4}$                   ………. (i)

For the right half section:

ΣM_B = 0

$\therefore {\text{M}}_{\text{p}}-\frac{{\text{w}}_{\text{c}}\times {\text{x}}^2}{2}=0$

$\Rightarrow {\text{M}}_{\text{p}}=\frac{{\text{w}}_{\text{c}}\times {\text{x}}^2}{2}$                  ………. (ii)

From (i) and ii)

$\frac{{\text{w}}_{\text{c}}\times {\text{x}}^2}{2}=\frac{{\text{w}}_{\text{c}}{(\text{L}-\text{x})}^2}{4}$

∴ 2x² = L² – 2 × x × L + x²

∴ x = 0.414 L

$\therefore {\text{M}}_{\text{p}}=\frac{{\text{w}}_{\text{c}}\times {\left(0.414\text{L}\right)}^2}{2}$

$\therefore {\text{M}}_{\text{p}}=\frac{{\text{w}}_{\text{c}}\times {\text{L}}^2}{11.656}$

Calculation:

$M_p=\frac{12\times {10}^2}{11.656}$

⇒ M_p = 102.95 kNm

Tips and Tricks:

For propped cantilever beam loaded uniformly (W_c)

Direct formulae: ${\text{w}}_{\text{c}}=\frac{11.656{\text{M}}_{\text{p}}}{{\text{L}}^2}$

Moment of resistance (M­₁) = Ze × 0.66 f_y = 410 × 10³ × 0.66 × 250

∴ M₁ = 67.65 × 10⁶ Nmm = 67.65 kNm

With two flanges plates on either side of ISMB 250, Moment of Inertia of the built-up beam depends upon the arrangement of the plates.

Arrangement I is not recommended for built up beam and arrangement II is followed.

∴ Moment of Inertia = 9.12 × 10⁷ mm⁴

∴ Section Module (Ze) = 729609.6 mm³

∴ moment capacity = 0.66× 250 × 729609.6 = 120 kNm

∴ Enhanced moment capacity = 120 – 67.65 = 52.74 kNm

For M 20 bolts of grade 4.6

f_ub = 400 MPa

Shear capacity of bolts (V_dsb)

$V_{dsb}=\frac{f_{ub}}{\sqrt{3}\times 1.25}\left(n_s{A}_{sb}+n_n{A}_{nb}\right)$

n_n = 0 [given]

$A_{sb}=\frac{\pi }{4}\times {\left(20\right)}^2=314.16m{m}^2$

n_s = 1 [single shear]

$V_{dsb}=\frac{400}{\sqrt{3}\times 1.25}\times \left(1\times 314.16\right)$

V_dsb = 58.04 kN

Bearing strength of Bolt:

Design Bearing strength of bolt (V_dbp)

$V_{dbp}=\frac{2.5k_{b}t{f}_u\times d}{{\gamma }_{mb}}$

γ_mb = partial safety factor of material = 1.25

k_b is smaller of $\frac{e}{3d_n},\frac{p}{3d_n}-0.25,\frac{f_{ub}}{f_u},1.0$

d_n → diameter of hole = 20 + 2 = 22 mm

$k_b=min\left[\frac{25}{3\times 22},\frac{75}{3\times 22}-0.25,\frac{400}{410},1.0\right]$

k_b = [0.378, 0.886, 0.975, 1.0]

Take k_b = 0.378

Take t = 7.6 mm (lesser of 12 and 7.6 mm)

$V_{dbp}=\frac{2.5\times 0.378\times 7.6\times 410\times 20}{1.25\times 1000}$

V_dbp = 47.11 kN

Design strength will be minimum of V_dbp and V_dsb = 47.11 kN

Force in extreme bolt $\left(F_1\right)=\frac{P}{5}$ kN

$r=\sqrt{{75}^2+{50}^2}$

r = 90.14 mm

r₁ = r₂ = r₃ = r₄ = 90.14 mm

r₅ = 0 (for centre Bolt)

Force in extreme bolt due to moment (P_e)

$F_2=\frac{Pe\times r}{r^2}$

$F_2=\frac{P\times 200\times 90.14}{4\times {(90.14)}^2}$

F₂ = 0.555 P kN

$F_R=\sqrt{{F_1}^2+{F_2}^2+2F_1{F}_{2}cos\theta }$

$47.11=F_R=\sqrt{\frac{P^2}{25}+{\left(0.555P\right)}^2+2\times \frac{P}{5}\times 0.555P\times cos\theta }$

$cos\theta =\frac{50}{r}=\frac{50}{90.14}$

${\left(47.11\right)}^2=\frac{P^2}{25}+{\left(0.555P\right)}^2+2\times \frac{P}{5}\times 0.555P\times \frac{50}{90.14}$

P = 68.29 × 5 = 341.14 kN