Question 1 |

A steel member 'M' has reversal of stress due to live loads, whereas another member 'N' has reversal of stress due to wind load. As per IS 800:2007, the maximum slenderness ratio permitted is:

less for member 'M' than that of member 'N' | |

more for member 'M' than for member 'N' | |

same for both the members | |

not specified in the Code |

Question 1 Explanation:

As per IS 800:2007

Question 2 |

A steel flat of rectangular section of size 70\times 6 mm is connected to a gusset plate
by three bolt each having a shear capacity of 15 kN in holes having diameter 11.5
mm. If the allowable tensile stress in the flat is 150 MPa, the maximum tension
that can be applied to the flat is

42.3kN | |

52.65kN | |

59.5kN | |

63kN |

Question 2 Explanation:

Along (1)-(1)

\begin{aligned} A_{net}&=\left ( 70-11.5 \right )\times 6 \\ &=351\, mm^{2} \\ P&=\left ( A_{net} \right )\times \left ( f_{y} \right ) \\ &=351\times 150 =52.65\, kN \end{aligned}

Along (2)-(2)

\begin{aligned} A_{net}&=\left ( 70-2\times 11.5 \right )\times 6 \\ &=282\, mm^{2} \\ P&=\left ( A_{net} \right )\times f_{y} \\ &=282\times 150 N=42.3 kN\end{aligned}

Question 3 |

In the design of welded tension members, consider the following statements :

I. The entire cross-sectional are of the connected leg is assumed to contribute to the effective area in case of angles.

II. Two angles back-to-back and tack-welded as per the codal requirements may be assumed to behave as a tee section.

III. A check on slenderness ratio may be necessary in some cases.

The TRUE statements are

I. The entire cross-sectional are of the connected leg is assumed to contribute to the effective area in case of angles.

II. Two angles back-to-back and tack-welded as per the codal requirements may be assumed to behave as a tee section.

III. A check on slenderness ratio may be necessary in some cases.

The TRUE statements are

Only I and II | |

Only II and III | |

Only I and III | |

I, II and III |

Question 4 |

The permissible stress in axial tension \sigma _{st}
in steel member on the net effective
area of the section shall not exceed the following value (f_{y} is the yield stress)

0.80 f_{y} | |

0.75 f_{y} | |

0.60 f_{y} | |

0.50 f_{y} |

Question 5 |

Two equal angles ISA 100mm\times 100mm
of thickness 10 mm are placed back to back and connected to the either side of a gusset plate through a single row of 16 mm diameter rivets in double shear. The effective areas of the connected and unconnected legs of each of these angles are 775 mm^{2} and 950 mm^{2}, respectively. If the angles are NOT tack riveted, the net effective area of this pair of angles is

3650mm^{2} | |

3450mm^{2} | |

3076mm^{2} | |

2899mm^{2} |

Question 5 Explanation:

When angles are not tack riveted, they will be considered as single angles connected on one side of gusset plate,

\begin{aligned} A_{e}&=\left ( A_{1}+kA_{2} \right )\times 2 \\ k&=\frac{3A_{2}}{3A_{1}+A_{2}} \\ &=\frac{3\times 775}{3\times 775+950} \\ &=0.71 \\ A_{e}&=\left ( 775+0.71\times 950 \right )\times 2 \\ &=2899\, mm^{2}\end{aligned}

\begin{aligned} A_{e}&=\left ( A_{1}+kA_{2} \right )\times 2 \\ k&=\frac{3A_{2}}{3A_{1}+A_{2}} \\ &=\frac{3\times 775}{3\times 775+950} \\ &=0.71 \\ A_{e}&=\left ( 775+0.71\times 950 \right )\times 2 \\ &=2899\, mm^{2}\end{aligned}

Question 6 |

A truss tie consisting of 2 ISA 75x75x8 mm carries a pull of 150 kN. At ends
the two angels are connected, one each on either side of a 10 mm thick gusset
plate, by 18 mm diameter rivets arranged in one row. The allowable stresses in
rivet are f_s=90.0N/mm^2 and f_{br}=250N/mm^2

Minimum number of rivets required at each end is

Minimum number of rivets required at each end is

2 | |

3 | |

4 | |

5 |

Question 6 Explanation:

Strength of rivets in shearing,

=\frac{2\times \pi \left ( 18+1.5 \right )^{2}\times 90}{4}=53.76 kN

Strength of rivets in bearing,

=\left ( 18+1.5 \right )\times 10\times 250 =48.75 kN

\therefore Rivet value =48.75 kN

Number of rivets =\frac{150}{48.75}=3.08 \approx 4

=\frac{2\times \pi \left ( 18+1.5 \right )^{2}\times 90}{4}=53.76 kN

Strength of rivets in bearing,

=\left ( 18+1.5 \right )\times 10\times 250 =48.75 kN

\therefore Rivet value =48.75 kN

Number of rivets =\frac{150}{48.75}=3.08 \approx 4

Question 7 |

A truss tie consisting of 2 ISA 75x75x8 mm carries a pull of 150 kN. At ends
the two angels are connected, one each on either side of a 10 mm thick gusset
plate, by 18 mm diameter rivets arranged in one row. The allowable stresses in
rivet are f_s=90.0N/mm^2 and f_{br}=250N/mm^2

Maximum tensile stress in the tie in N/mm^{2} is

Maximum tensile stress in the tie in N/mm^{2} is

93.6 | |

87.5 | |

77.2 | |

66 |

Question 7 Explanation:

(i) Assuming both thr angles as tack riveted, Gross area,

A_{g}=\left ( 75-4 \right )\times 8\times 2=1136 mm^{2}

For angles, connected on both sides of gussets plate, Effective area,

\begin{aligned} A_{e}&=A_{1}+kA_{2} \\ A_{1}&= \text{ Net area of connected leg} \\ &=\frac{1136}{2}-\left ( 18+1.5 \right )\times 8 \\ &=412 mm^{2} \\ A_{2}&=\text{ Area of outstanding leg} \\ &=\frac{1136}{2}=568 mm^{2} \\ k&=1.0 \\ A_{e}&=412+\left ( 1.0\times 568 \right ) \\ &=980 mm^{2} \end{aligned}

\therefore\;\; Total area for both angles =2\times 980=1960\, mm^{2}

\therefore Maximum tensile stress =\frac{150\times 10^{2}}{1960}=76.5 N/mm^{2}

(ii) Also assuming both the angle sections are not tack riveted,

\begin{aligned} \therefore \;\; k_{1}&=\frac{3A_{1}}{3A_{1}+A_{2}} \\ &=\frac{3\times 412}{3\times 412+568} \\ &=0.685 \\ \therefore \;\; A_{e}&=\left ( A_{1}+k_{1}A_{2} \right )\times 2 \\ &=\left ( 412+0.685\times 568 \right )\times 2 \\ &=1602.2 mm^{2} \end{aligned}

\therefore Maximum tensile stress =\frac{150\times 10^{3}}{1602.3}=93.6 N/mm^{2}

Question 8 |

ISA 100x100x10 mm (Cross sectional area = 1908 mm^{2}) serves as tensile
member. This angle is welded to a gusset plate along A and B appropriately
as shown. Assuming the yield strength of the steel to be 260 N/mm^{2}. The tensile strength of this member can be taken to be approximately,

500kN | |

300kN | |

225kN | |

375kN |

Question 8 Explanation:

\begin{aligned} A_{1} &= 1000 \\ A_{2} &= 1000 \\ \therefore \; \; k&=\frac{3A_{1}}{3A_{1}+A_{2}}=0.75 \\ A&=A_{1}+kA_{2} \\ &=1750 mm^{2} \end{aligned}

Tensile strength of member

=0.6\times f_{y}\times A =0.6\times 280\times 1750=273 kN

So, tensile strength can be taken as 225 kN.

There are 8 questions to complete.