Q1GATE 2025NAT

A steel beam supported by three parallel pin-jointed steel rods is shown in the figure. The moment of inertia of the beam is $8 \times 10^7 , \text{mm}^4$ . Take modulus of elasticity of steel as $210 , \text{GPa}$ . The beam is subjected to a uniformly distributed load of $6.25 , \text{kN/m}$ , including its self-weight. The axial force (in kN) in the centre rod CD is ______ (round off to one decimal place).

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📖 Explanation

\begin{aligned} w &= 6.25 , \text{kN/m} \\ L &= 4 , \text{m} \quad (\text{Span of the beam}) \\ E &= 210 , \text{GPa} = 210 \times 10^9 , \text{Pa} \\ I &= 8 \times 10^7 , \text{mm}^4 = 8 \times 10^{-5} , \text{m}^4 \\ d_{CD} &= 30 , \text{mm}, \quad d_{AB} = d_{EF} = 12 , \text{mm} \end{aligned}

Since the beam is supported by three rods A, C, and E, and the middle rod CD is stiffer due to a larger diameter, it attracts more force. We assume: Axial force in rod CD = $F$ Let vertical deflection in beam at point C due to UDL and axial shortening of CD be equal Step 1: Area of rods

\begin{aligned} A_{CD} &= \frac{\pi}{4} (0.03)^2 = 7.07 \times 10^{-4} , \text{m}^2 \\ L_{rod} &= 1 , \text{m} \end{aligned}

Step 2: Deflection of beam at midspan (C) due to UDL alone For a simply supported beam with UDL: $\delta_B = \frac{5wL^4}{384EI}$ Substitute values: $\delta_B = \frac{5 \times 6.25 \times 4^4}{384 \times 210 \times 10^9 \times 8 \times 10^{-5}} \ \delta_B = \frac{5 \times 6.25 \times 256}{384 \times 210 \times 8 \times 10^4} \ \delta_B \approx 0.0003095 , \text{m}$ Step 3: Axial shortening of rod CD = $\delta_{rod} = \frac{F L}{AE}$ Equating both deflections: $\frac{F \cdot 1}{7.07 \times 10^{-4} \cdot 210 \times 10^9} = 0.0003095$ Solve for $F$ : $F = 0.0003095 \times 7.07 \times 10^{-4} \times 210 \times 10^9 \ F \approx 16.3 , \text{kN}$

Q2GATE 2023MCQ

Consider the horizontal axis passing through the centroid of the steel beam cross-section shown in the figure. What is the shape factor (rounded off to one decimal place) for the cross-section ?

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📖 Explanation

\begin{aligned} Z_{P} & =\frac{A}{2}\left(\bar{y}_{1}+\bar{y}_{2}\right) \\ & =\left(3 b \times \frac{b}{2}+b \times b\right)\left(2 \times \frac{11}{20} b\right) \\ & =\frac{11}{4} b^{3} \end{aligned}

$z_{e}=\frac{1}{y_{max }}=\frac{\frac{b \times (3 b)^{3}}{12}+\frac{2 b(b)^{3}}{12}}{\frac{3 b}{2}}$ $z_{e}=\frac{29}{18} b^{3}$ Shape factor $(S)=\frac{Z_{P}}{Z_{e}}=\frac{\frac{11}{4} b^{3}}{\frac{29}{18} b^{3}}=1.7$

Q3GATE 2020NAT

The flange and web plates of the doubly symmetric built-up section are connected by continuous 10 mm thick fillet welds as shown in the figure (not drawn to the scale). The moment of inertia of the section about its principal axis X-X is $7.73 \times 10^{6} mm^4$ . The permissible shear stress in the fillet welds is 100 $N/ mm^2$ . The design shear strength of the section is governed by the capacity of the fillet welds. The maximum shear force (in kN, round off to one decimal place) that can be carried by the section, is _______.

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📖 Explanation

q = Shear stress at the level mn in the weld = 100 MPa $=\frac{SA\bar{y}}{Ib}$ Shear force at the given section A = Area of the cross-section above the level $mn=100 \times 10 mm^2$ $\bar{y}=$ C.G. of shaded area above the level mn = 60-5 = 55 m $I=7.73 \times 10^6 mm^4$ b = Width of weld at mn (4 welds) = 4 x t = 4 x 7 = 28 mm t = Throat thickness

\begin{aligned} &=0.7 \times s=0.7 \times 10 \times 4=28 mm \\ 100&= \frac{F \times (100 \times 10) \times 55}{7.73 \times 10^6 \times 28}\\ F&=\frac{100 \times 7.73 \times 10^6 \times 28}{1000\times 55} \\ &=393.527kN \end{aligned}

Q4GATE 2019NAT

A rolled I-section beam is supported on a 75 mm wide bearing plate as shown in the figure. Thicknesses of flange and web of the I-section are 20 mm and 8 mm, respectively. Root radius of the I-section is 10 mm. Assume: material yield stress, $f_y=250 MPa$ and partial safety factor for material, $\gamma _{mo}=1.10$ . As per IS: 800-2007, the web bearing strength (in kN, round off to 2 decimal places) of the beam is ________

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📖 Explanation

Web bearing strength

\begin{aligned} &=[b+2.5(t_f+R)]\times t_w \times \frac{f_y}{\gamma _{mo}} \\ &= [75+2.5(20+10)] \times 8 \times \frac{250}{1.1}\\ &= 272.73kN \end{aligned}

Q5GATE 2019MCQ

Assuming that there is no possibility of shear buckling in the web, the maximum reduction permitted by IS 800-2007 in the (low-shear) design bending strength of a semi-compact steel section due to high shear is

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📖 Explanation

As per IS 800 : 2007 For semi compact section (i) In low shear case ( $V \leq 0.6 V_d$ ) $M_d = Z_ef_y/\gamma _{mo}$ (ii) In high shear case ( $V \gt 0.6 V_d$ ) $M_d = Z_ef_y/\gamma _{mo}$ So reduction is zero.

Q6GATE 2018MCQ

A steel column of ISHB 350 @72.4 kg/m is subjected to a factored axial compressive load of 2000 kN. The load is transferred to a concrete pedestal of grade M20 through a square base plate. Consider bearing strength of concrete as 0.45 $f_{ck}$ , where $f_{ck}$ is the characteristic strength of concrete. Using limit state method and neglecting the self weight of base plate and steel column, the length of a side of the base plate to be provided is

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📖 Explanation

\begin{aligned} &\text{Area required for base plate}\\ &=\frac{\text{Factored load}}{\text{Bearing capacity of concrete}}\\ &=\frac{2000\times 10^{3}}{0.45\times 20}=222222.222mm^{2} \end{aligned}

So, side of base plate $=\sqrt{\text{Area}}=471.4 mm= 47.14 cm$ Since, provided area must be more than required So, answer should be 48 cm.

Q7GATE 2016NAT

The semi-compact section of a laterally unsupported steel beam has an elastic section modulus, plastic section modulus and design bending compressive stress of $500 \; cm^3, 650 \; cm^3$ and 200MPa, respectively. The design flexural capacity (expressed in kNm) of the section is ____.

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📖 Explanation

Design flexural capacity, $M_{d}= \beta _{b}Z_{p}\frac{f_{y}}{\gamma _{m0}}$ $\beta _{b}= 1.0$ (for plastic and compact sections) $\; \; \; \; = \frac{Z_{e}}{Z_{p}}$ (for semi-compact sections) $Z_{p}=\,$ Plastic section modulus $Z_{e}=\,$ elastic section modulus $f_{y}=\,$ yeild stress of steel provided $\gamma_{m0}=\,$ material factor of safety of steel against yeilding As, design bending stress is directly given,

\begin{aligned} M_{d}&= \beta _{b}Z_{p}f_{d} \\ &= \frac{Z_{e}}{Z_{p}}\times Z_{p}\times f_{d} \\ &= Z_{e}\times f_{d} \\ &= 500\times 10^{3}\times 200\times 10^{-6} \\ &= 100 kN-m \end{aligned}

Q8GATE 2011MCQ

The figure shows a schematic representation of a steel plate girder to be used as a simply supported beam with a concentrated load. For stiffeners, PQ (running along the beam axis) and RS (running between the top and bottom flanges), which of the following pairs of statements will be TRUE?

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Q9GATE 2005MCQ

An unstiffened web I-section is fabricated from a 10 mm thick plate by fillet welding as shown in the figure. If yield stress of steel is 250 MPa, the maximum shear load that section can take is

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📖 Explanation

The web depth to thickness ratio

\begin{aligned} &=\frac{300-20}{10}=28 \lt 85 \\ \therefore \;\; V&=f_{s}t_{w}d \\ d&=300 mm, \;f_{s}=0.4f_{y} \\ f_{y}&=0.4 \times 250=100 N/mm^{2} \\ t_{W}&=10mm\\ \Rightarrow \;\; V&=100\times 10\times 300=300 kN \end{aligned}

Q10GATE 2004MCQ

A square steel slab base of area 1 $m^{2}$ is provided for a column made of two rolled channel sections. The $300mm\times 300mm$ column carries an axial compressive load of 2000 kN. The line of action of the load passes through the centroid of the column section as well as of the slab base. The permissible bending stress in slab base is 185 MPa. The required minimum thickness of the slab base is

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📖 Explanation

$300mm\times 300mm$ column leaves projections as, $\; a\, =\, b\, =\frac{1000-300}{2}\, =350mm$ Pressure on the underside of base slab $\; W=\frac{2000}{1}=2000\, kN/m^{2}=2\, N/mm^{2}$ Thickness of slab base, $\; t=\sqrt{\frac{3W}{F_{b}}\left ( a^{2}-\frac{b^{2}}{4} \right )}=\sqrt{\frac{3\times 2}{185}\left ( 350^{2}-\frac{350^{2}}{4} \right )}$ $\; t\simeq 55$ mm

Q11GATE 2003MCQ

Group-I contains some elements in design of a simply supported plate girder and Group-II give some qualitative locations on the girder. Match the items of two lists as per good design practice and relevant codal provisions

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Q12GATE 2002MCQ

An ISMB 500 is used as a beam in a multi-storey construction. From the viewpoint of structural design, it can be considered to be 'laterally restrained' when,

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Q13GATE 2002MCQ

When designing steel structures, one must ensure that local buckling in webs does not take place. This check may not be very critical when using rolled steel sections because.

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