Question 1 |

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 _______.

The maximum shear force (in kN, round off to one decimal place) that can be carried by the section, is _______.

393.5 | |

125.4 | |

256.3 | |

485.2 |

Question 1 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}

Question 2 |

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 ________

As per IS: 800-2007, the web bearing strength (in kN, round off to 2 decimal places) of the beam is ________

272.73 | |

185.64 | |

286.92 | |

312.44 |

Question 2 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}

Question 3 |

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

Zero | |

25% | |

50% | |

governed by the area of the flange |

Question 3 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.

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.

Question 4 |

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 0.45f_{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

39 cm | |

42 cm | |

45 cm | |

48 cm |

Question 4 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.

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.

Question 5 |

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 ____.

1000 | |

100 | |

120 | |

1200 |

Question 5 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}

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}

Question 6 |

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?

(i) RS should be provided under the concentrated load only. (ii) PQ should be placed in the tension side of the flange. | |

(i) RS helps to prevent local buckling of the web. (ii) PQ should be placed in the compression side of the flange. | |

(i) RS should be provided at supports. (ii) PQ should be placed along the neutral axis. | |

(i) RS should be provided away from points of action of concentrated loads. (ii) PQ should be provided on the compression side of the flange. |

Question 6 Explanation:

PQ is a horizontal stiffener in the given plate girder. Horizontal stiffeners are also called longitudinal stiffeners. The horizontal stiffener are provided in the compression zone of the web. The first horizontal stiffener is provided at one-fifth of the distance from the compression flange to the tension flange. If required another stiffener is provided at the neutral axis. Horizontal stiffeners are not continuous and are provided between vertical stiffeners.

RS is a vertical stiffener in the given plate girder. Vertical stiffeners are also called transverse stiffeners. It is assumed that the vertical stiffener is not subjected to any load and is selected to provide necessary lateral stiffness only and can therefore, br crimped or joggled for tight fittings. Such stiffeners increase the buckling resistance of the web caused by shear.

RS is a vertical stiffener in the given plate girder. Vertical stiffeners are also called transverse stiffeners. It is assumed that the vertical stiffener is not subjected to any load and is selected to provide necessary lateral stiffness only and can therefore, br crimped or joggled for tight fittings. Such stiffeners increase the buckling resistance of the web caused by shear.

Question 7 |

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

750kN | |

350kN | |

337.5kN | |

300kN |

Question 7 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}

\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}

Question 8 |

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

110mm | |

89mm | |

63mm | |

55mm |

Question 8 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 55mm

Question 9 |

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

P-2 Q-3 R-1 S-5 | |

P-4 Q-2 R-1 S-3 | |

P-3 Q-4 R-2 S-1 | |

P-1 Q-5 R-2 S-3 |

Question 10 |

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,

The tension flange is 'laterally restrained' | |

the compression flange is 'laterally restrained' | |

the web is adequately stiffened | |

the conditions in (A) and (C) are met |

There are 10 questions to complete.