Concept:

For economical spacing of truss,

The cost of truss should be equal to twice the cost of purlins plus the cost of roof covering.

t = 2p + r

where t= cost of truss, p= cost of purlin, r = cost of the roof.

However, the above expression is used to check the spacing of the roof truss. It can not be used to design the spacing as spacing does not occur in the equation. As a guide, the spacing of roof truss can be kept 1/4 of the span up to 15m and 1/5 of span for 15-30 m span of roof trusses.

Concept:

Economical depth of a plate girder is given by,

 \(d = {\left( {\frac{{Mk}}{{{f_y}}}} \right)^{1/3}}\)

\(k = \frac{d}{{{t_w}}} = \frac{{depth}}{{thickness\;of\;web}}\)

Where, M = Moment and f_y = yield stress of steel.

Calculation:

Given,

Load = 80 kN/m

Span (ℓ) = 25 m

k = 200, F_y = 250 MPa

\(B.{M_{max}} = \frac{{w{\ell ^2}}}{8} = \frac{{80 \times {{\left( {25} \right)}^2}}}{8} = 6250\;kNm\)

\(d = {\left( {\frac{{6250 \times {{10}^6} \times 200}}{{250}}} \right)^{1/3}} = 1710\;mm\)

\({t_w} = \frac{d}{k} = \frac{{1710}}{{200}} = 8.55\;mm\)

The effective length of groove weld:

As per IS: 800-2007, Clause 10.5.4.2 

The effective length of butt weld or groove weld shall be taken as the length of the continuous full-size butt weld, but not less than four times the size of the weld or minimum of 40 mm.

Maximum clear spacing between the effective length of the weld:

• For the welds in compression zone = Minimum of 12 × Thickness of the weld or 200 mm
• For the welds in tension zone = Minimum of 16 × Thickness of the weld or 200 mm

Ends of the longitudinal fillet welds shall not be less than the width of the member.

\({\rm{Pitch\;of\;truss\;}} = \frac{{{\rm{Rise}}}}{{{\rm{Span}}}} = \frac{{\rm{h}}}{{\rm{l}}}\)

h = 9 m

L = 1.5 m

\({\rm{Pitch\;}} = \frac{9}{{1.5}} = 6\)

Concept

As per IS 800:2007, the compact section has the capacity to form a plastic hinge but does not have capacity to develop collapse mechanism because of local buckling. However, the plastic moment carrying capacity M_p) of compact section is given as:

M_p = f­_y Z_p

Where,

f_y is the yield strength of material

Z_p is the plastic section modulus and it is given as,

\({Z_p} = \frac{A}{2}\left( {\;{{\bar y}_{1\;}} + \;{{\bar y}_{2\;}}} \right)\)

Where,

A is the total cross-sectional area,

y̅₁ and y̅₂ are centroid distance of compression area and tension area from Plastic Neutral axis

P.N.A is the axis which divides the cross-section into two equal areas i.e.

Total area above P.N.A = Total area below P.N.A

Calculation:

Let P.N.A lies within the flange portion of T-section at a distance ‘y’ from the top fiber.

Now, by equating areas above and below P.N.A

100 × y = (30 × 100) + 100 × (30 - y)

⇒ y = 30 mm ⇒ PNA lies at Junction of web and flange.

\({\bar y_1} = \frac{{30}}{2} = 15\;mm\)   [As both are rectangles]

\({\bar y_2} = \frac{{100}}{2} = 500\;mm\) 

\(\therefore {Z_p} = \frac{A}{2}\left( {{{\bar y}_1} + {{\bar y}_2}} \right) = \left( {\frac{{30 \times 100 + 30 \times 100}}{2}} \right)\left( {15 + 50} \right)\) = 195 × 10³ mm³

∵ M_p = f_y Z_p ⇒ (40 × 10⁶) = (195 × 10³) × f_y⇒ f_y = 205.13 Mpa

Concept:

The Moment, M is given as:

M = C₁Z₁ + C₂Z₂

Where

C₁ is the compression force for the portion where stress distribution is rectangular

C₂ is the compression force for the portion where stress distribution is linear.

Z₁ and Z­₂ are corresponding level arms.

The Plastic Neutral axis for the symmetrical rectangular section will lie at its mid axis.

Calculation:

The stress distribution across the section:

B = 900 mm,h = 400 mm

C₁ = Stress × Area \( = {f_y} \times \frac{h}{3} \times B = \frac{{Bh}}{3}{f_y}\) 

\(\therefore {C_1} = \left( {\frac{{200\; \times \;400}}{3}} \right) \times 250\) = 6666.67 kN

C₁ will act at  distance from the top fiber.

\(\therefore {z_1} = h - \left( {\frac{h}{6} + \frac{h}{6}} \right) = \frac{{2h}}{3} = \frac{{2\; \times \;400}}{3} = \frac{{800}}{3}\;mm\) 

\({C_2} = \frac{1}{2} \times B \times \frac{h}{6} \times {f_y} = \frac{{Bh}}{{12}}{f_y} = \left( {\frac{{200\; \times \;400}}{{12}}} \right) \times 250\) = 1666.67 kN

This will act at a distance of \(\left( {\frac{h}{3} + \frac{h}{{18}}} \right)\) from top fiber

\(\therefore {z_2} = h - 2 \times \left( {\frac{h}{3} + \frac{h}{{18}}} \right) = \frac{{2h}}{9} = \frac{{2\; \times \;400}}{9} = \frac{{800}}{9}\;mm\) 

Now,

M = C₁z₁ + C₂z₂ \(= \left( {6666.67 \times \frac{{800}}{3}} \right) + \left( {1666.67 \times \frac{{800}}{9}} \right)\) = 1777778.67 + 148148.15

∴ M = 1925.93 kN-m

Concept:

Bearing strength of concrete = 0.45 f_ck

\({\rm{Required\;area\;of\;plate}} = \frac{{{\rm{Factored\;load}}}}{{{\rm{Bearing\;strength\;of\;concrete}}}}\)

Maximum moment at critical faction = \(\frac{{{\rm{w}}{{\rm{C}}^2}}}{2}\)

Moment capacity of base plate and angle leg combined: \(\left( {{{\rm{M}}_{\rm{d}}}} \right) = \frac{{1.2 \times {{\rm{f}}_{\rm{y}}} \times {{\rm{z}}_{\rm{e}}}}}{{{{\rm{\gamma }}_{{\rm{mo}}}}}}\) 

Calculation:

Bearing strength of concrete = 0.45 f_ck = 0.45 × 20 = 9 MPa

\({\rm{Required\;area\;of\;plate}} = \frac{{1200 \times {{10}^3}}}{9} = 133.3 \times {10^3}{\rm{\;m}}{{\rm{m}}^2}\)

Maximum moment at critical faction = \(\frac{{{\rm{w}}{{\rm{C}}^2}}}{2} = \frac{{9 \times {{104}^2}}}{2} = 48672{\rm{\;Nmm}}\) 

Moment capacity of base plate and angle leg combined:

\(\left( {{{\rm{M}}_{\rm{d}}}} \right) = \frac{{1.2 \times {{\rm{f}}_{\rm{y}}} \times {{\rm{Z}}_{\rm{e}}}}}{{{{\rm{\gamma }}_{{\rm{mo}}}}}} = \frac{{1.2 \times 250}}{{1.1}} \times \left( {1 \times \frac{{{{\rm{t}}^2}}}{6}} \right) = 45.45{\rm{\;}}{{\rm{t}}^2}\)

Maximum moment at critical section ≥ Moment capacity of gusset base plate

48672 ≤ 45.45 t²

t² > 1070.7

t > 32.7 mm

t ≃ 35 mm

Hence the required area of plate (in mm2) and minimum thickness of base plate (in mm) respectively are 133.33 × 103, and 35

Concept:

In the Plastic method, every fiber of cross-section gets yielded due to the application of load. Therefore, stress distribution across a section is rectangular about the plastic Neutral axis (P.N.A).

P.N.Ais the axis which divided the cross-section into two equal areas i.e.

The total area above P.N.A = Total area below P.N.A

Resultant compressive force (C) passes through the Centroid of area (not through the centroid of stress distribution) because stress is constant throughout the depth but the area is varying because given section triangle.  

Calculation:

Consider a small strip of ‘dy’ thickness at a distance ‘y’ from the apex of the triangle. Let ‘dc’ be the compressive force acting on this strip.

Location of ‘y_p’:

Area of ΔADC = Area of the portion (DECD)

\(= \frac{1}{2} \times {y_p} \times {B_p} = \left( {\frac{{{B_p} + B}}{2}} \right) \times \left( {h - {y_p}} \right)\)         ---(1)

Now,

\(\frac{h}{B} = \frac{{{y_p}}}{{{B_p}}} = \frac{y}{{b'}}\) 

\(\Rightarrow {B_p} = \frac{B}{h}{y_p}\)  and \(b' = \frac{B}{h}y\)

Substitute B_p in equation (1)

\(\frac{1}{2} \times {y_p} \times \frac{B}{h}{y_p} = \left( {\frac{{{y_p}\frac{B}{h} + B}}{2}} \right) \times \left( {h - {y_p}} \right)\) 

\(\frac{{y_p^2}}{h} = \frac{{\left( {{y_p} + h} \right)\left( {h - {y_p}} \right)}}{h}\) 

\( \Rightarrow 2y_p^2 = {h^2} \Rightarrow {y_p} = \frac{h}{{\sqrt 2 }}\) 

Also,

\(dc = \left( {dA} \right)\left( {{f_y}} \right)\) = (b’dy) f_y

Integratig on both the sides,

\(\smallint dc = \mathop \smallint \nolimits_0 \frac{B}{h}y\;dy,\;{f_y} \Rightarrow C = \frac{B}{h}{f_y}\;\mathop \smallint \limits_0^{{y_p}} ydy = \left( {\frac{B}{h}{f_y}} \right)\left( {\frac{{y_p^2}}{2}} \right) = \frac{{Bh{f_y}}}{4}\) 

Since centroid of ADE is at a distance of \(\frac{{{y_p}}}{3}\) from P.N.A. So, it will act at a distance of \(y_c' = \frac{{{y_p}}}{3} = \frac{h}{{3\sqrt 2 }}\) from P.N.A.

Concept:

As per IS 800:2007, Table 6, the permissible deflection in Gantry girder is tabulated below:

Calculation

Case 1: When 100 KN electrically operated crane (EOT) is provided on the gantry girder.

From the above table, the maximum permissible deflection, δ_max­ = L/750

Where, L is the effective span = 8 m given

∴ δ_max = 8000/750 = 10.67 mm

Case 2: When 50 kN hand-operated is provided on the same gantry girder

From the above table, the maximum permissible deflection, δ_max­ =  L/500.

This deflection must be equal to 10.67 mm as  per the condition  given in the question

∴ 10.67 = L/500

⇒ Effective span, L = 5.335 m

Concept

To make a collapse mechanism, no of plastic hinges required = D_s +1.

Where

D_s is static indeterminacy

In this Case, D_s = 1

⇒ Two plastic hinges will be required to form a collapse mechanism. The possible location of the plastic hinge are A and B. There will no plastic hinge at C because at C bending moment is zero due to hinge connection.

Initially, the plastic hinge will be formed at fixed support as given in the question and let load required to form that hinge is P₁.

After that, load if further increase such that second plastic hinge is formed B. By doing so, the collapse mechanism if formed as two plastic hinges have already been developed and there is no need to increase the load further and let load required to form this collapse mechanism is P₂.

The additional increment in load is required at its midpoint so that a complete collapse mechanism is formed in the propped cantilever beam will be P₁ – P₂.

Further,

The load P₁ where only one plastic hinge is formed at fixed support can be evaluated easily by elastic analysis and drawing FBD of the beam as shown in the calculation part.

The collapse load P₂, where plastic hinges at A and B are formed and the collapse mechanism is also formed is evaluated using the principle of virtual work as stated below:

External work done + Internal work done = 0

Calculation:

Calculation of P₁:

Total displacement at joint, C is zero i.e.

\({{\rm{\Delta }}_{1P}} + \frac{L}{2}\theta = {{\rm{\Delta }}_2}\) 

\({{\rm{\Delta }}_{1P}} = \frac{{{P_1}{{\left( {\frac{L}{2}} \right)}^3}}}{{3\;EI}} = \frac{{{P_1}{L^3}}}{{24\;EI}}\) 

\(\theta = \frac{{{P_1}{{\left( {\frac{1}{2}} \right)}^2}}}{{2EI}} = \frac{{{P_1}{L^3}}}{{24EI}}\) 

\({{\rm{\Delta }}_2} = \frac{{R{L^3}}}{{3EI}}\) 

Compaibility Equation:

\(\frac{{{P_1}{L^3}}}{{24EI}} + \frac{L}{2} \times \left( {\frac{{{P_1}{L^2}}}{{8\;EI}}} \right) = \frac{{R{L^3}}}{{3EI}} \Rightarrow R = \frac{{5{P_1}}}{{16}}\) 

Let B.M. at support A is M_A,

∑ M = 0 (about A)

\({M_A} + R \times L = {P_1} \times \frac{L}{2}\) 

\({M_A} + \frac{{5{P_1}}}{{16}} \times L = \frac{{{P_1}L}}{2} \Rightarrow {M_A} = \frac{{3{P_1}L}}{{16}}\) 

Further, since at ‘A’ plastic hinge is formed.

So, M_A = M_p  \(\Rightarrow {M_P} = \frac{{3{P_1}L}}{{16}} \Rightarrow {P_1} = \frac{{16\;{M_P}}}{{3L}}\)

External work done \(= {P_2} \times \frac{L}{2}\theta\) 

\(\begin{array}{*{20}{c}}{Internal\;work\;Done = }&{ - {M_p}\theta }& + &{\left( { - {M_p}} \right)\left( {\theta + \theta } \right)}\\{}& \downarrow &{}& \downarrow \\{}&{at\;A}&{}&{at\;B}\end{array}\) 

By the principle of virtual work,

\({P_2} \times \frac{L}{2}\theta = 3\;{M_p}\theta \Rightarrow {P_2} = \frac{{6{M_p}}}{L}\) 

\(\therefore {P_2} - {P_1} = \frac{{6{M_p}}}{L} - \frac{{16}}{3}\frac{{{M_P}}}{L} = \frac{2}{3}\frac{{{M_P}}}{L}\)