Question 1 |

A prismatic steel beam is shown in the figure.

The plastic moment, M_{p} calculated for the collapse mechanism using static method and kinematic method is

The plastic moment, M_{p} calculated for the collapse mechanism using static method and kinematic method is

M_{P, \text { static }} \gt \frac{2 P L}{9}=M_{P, \text { kinematic }} | |

M_{P, \text { static }}=\frac{2 P L}{9} \neq M_{P, \text { kinematic }} | |

M_{P, \text { static }}=\frac{2 P L}{9}=M_{P, \text { kinematic }} | |

M_{P, \text { static }} \lt \frac{2 P L}{9}=M_{P, \text { kinematic }} |

Question 1 Explanation:

\begin{aligned} \text{At collapse,} \quad M_{p} \theta+M_{p} \phi&=P \Delta\\ \Rightarrow \quad \quad 3 M_{P} \frac{\Delta}{l}+\frac{3 M_{P} \Delta}{2 l} &=P \Delta \\ M_{P} &=\frac{2 P l}{9}\\ \text{Also,} \qquad\qquad \quad M_{P ,\text { static }}&=M_{P,\text{ kinematic}} \end{aligned}

Question 2 |

A column is subjected to a total load (P) of 60 kN supported through a bracket connection, as shown in the figure (not to scale).

The resultant force in bolt R (in kN,round off to one decimal place) is ________

The resultant force in bolt R (in kN,round off to one decimal place) is ________

28.2 | |

46.8 | |

22.6 | |

38.2 |

Question 2 Explanation:

\begin{aligned} F_{1}&=\frac{P}{n}=\frac{60}{6}=10 \mathrm{kN}\\ F_{2}&=\frac{P \cdot e}{\sum r_{i}^{2}} \times r_{R}=\frac{60 \times 100\mathrm{~mm}}{4 \times 50^{2}+2 \times 40^{2}} \times 40\\ F_{2}&=\frac{60 \times 100 \times 40}{10000+3200}=\frac{60 \times 100 \times 40}{13200}=\frac{200}{11} \mathrm{kN}\\ F_{R}&=F_{1}+F_{2}\\ &=10+\frac{200}{11}=\frac{310}{11}=28.2 \mathrm{kN} \end{aligned}

Question 3 |

Two steel plates are lap jointed in a workshop using 6 mm thick fillet weld as shown
in the figure (not drawn to the scale). The ultimate strength of the weld is 410 MPa.

As per Limit State Design is IS 800 : 2007, the design capacity (in kN, round off to three decimal places) of the welded connection, is _______.

As per Limit State Design is IS 800 : 2007, the design capacity (in kN, round off to three decimal places) of the welded connection, is _______.

125.547 | |

653.247 | |

413.586 | |

212.478 |

Question 3 Explanation:

Design capacity of welded connection

P_s=f_b \times l_{eff}\times t_t

\begin{aligned} P&=\frac{410}{\sqrt{3} \times 1.25} \times 520 \times 0.7 \times 6 \\ &= \frac{716352}{\sqrt{3}}=413586N\\ &= 413.586kN \end{aligned}

P_s=f_b \times l_{eff}\times t_t

\begin{aligned} P&=\frac{410}{\sqrt{3} \times 1.25} \times 520 \times 0.7 \times 6 \\ &= \frac{716352}{\sqrt{3}}=413586N\\ &= 413.586kN \end{aligned}

Question 4 |

The ratio of the plastic moment capacity of a beam section to its yield moment capacity
is termed as

aspect ratio | |

load factor | |

shape factor | |

slenderness ratio |

Question 4 Explanation:

Ratio of \frac{M_p}{M_y}= Shape factor

Question 5 |

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 5 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 6 |

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 6 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 7 |

The critical bending compressive stress in the extreme fibre of a structural steel section is 1000 MPa. It is given that the yield strength of the steel is 250 MPa, width of flange is 250 mm and thickness of flange is 15 mm. As per the provisions of IS:8002007, the non-dimensional slenderness ratio of the steel cross-section is

0.25 | |

0.5 | |

0.75 | |

2 |

Question 7 Explanation:

\lambda =\sqrt{\frac{f_y}{f_{cr}}}=\sqrt{\frac{250}{1000}}=0.5

Question 8 |

For a channel section subjected to a downward vertical shear force at its centroid, which one of the following represents the correct distribution of shear stress in flange and web?

A | |

B | |

C | |

D |

Question 8 Explanation:

Shear flow in horizontal member (flange) is linear with zero at free end and in vertical member (web) it is parabolic.

Question 9 |

A steel column is restrained against both translation and rotation at one end and is restrained only against rotation but free to translate at the other end. Theoretical and design (IS:800-2007) values, respectively, of effective length factor of the column are

1.0 and 1.0 | |

1.2 and 1.0 | |

1.2 and 1.2 | |

1.0 and 1.2 |

Question 9 Explanation:

The given support conditions indicates the following support/ end conditions of column

l_{eff} as per theoretical conditions = 1.0l_{o}

l_{eff} as per IS 800 : 2007 = 1.2 l_{o}

Considering the errors that may occur due to construction of supports on site.

l_{eff} as per theoretical conditions = 1.0l_{o}

l_{eff} as per IS 800 : 2007 = 1.2 l_{o}

Considering the errors that may occur due to construction of supports on site.

Question 10 |

A 16 mm thick gusset plate is connected to the 12 mm thick flange plate of an I-section using fillet welds on both sides as shown in the figure. The gusset plate is subjected to a point load of 350 kN acting at a distance of 100 mm from the flange plate. Size of fillet weld is 10 mm.

The maximum resultant stress (in MPa, round off to 1 decimalplace) on the fillet weld along the vertical plane would be ____

The maximum resultant stress (in MPa, round off to 1 decimalplace) on the fillet weld along the vertical plane would be ____

86.2 | |

105.3 | |

115.8 | |

128.6 |

Question 10 Explanation:

Direct concentrated load (P) = 350 kN

Eccentricity (e) = 100 mm

Depth of weld (d_w) at each face = 500 mm

Size of weld (S) = 10 mm

Throat thickness of weld (t_t) = 0.7S = 7 mm

It is a case of combined shear (P) and Bending moment (M = P.e)

Shear stress acting on weld due to direct shear force P,

\begin{aligned} (q)&=\frac{P}{A_w} \\ &= \frac{P}{2d_wt_t}\\ &=\frac{350 \times 10^3}{2 \times 500 \times 7}=50MPa \end{aligned}

Bending normal stress acting at i^{th} point from Neutral axis [f_i]=\frac{M}{I}y_i

Maximum bending stress at extreme point in weld

\begin{aligned} [f]&=\frac{M}{I}y_{max} \\ &= \frac{Pe}{2t_t \frac{d_w^3}{12}} \times y_{max}\\ &= \frac{350 \times 10^3 \times 100}{2 \times 7 \times \frac{(500)^3}{12}} \times 250\\ &= 60MPa \end{aligned}

As per IS800:2007. Maximum resultant stress an equivalent stress in fllet weld due to combined shear and bending stress

\begin{aligned} (f_{eq})&=\sqrt{f^2+3q^2} \\ &= \sqrt{60^2+3 \times 50^2}\\ &=105.3MPa \end{aligned}

Eccentricity (e) = 100 mm

Depth of weld (d_w) at each face = 500 mm

Size of weld (S) = 10 mm

Throat thickness of weld (t_t) = 0.7S = 7 mm

It is a case of combined shear (P) and Bending moment (M = P.e)

Shear stress acting on weld due to direct shear force P,

\begin{aligned} (q)&=\frac{P}{A_w} \\ &= \frac{P}{2d_wt_t}\\ &=\frac{350 \times 10^3}{2 \times 500 \times 7}=50MPa \end{aligned}

Bending normal stress acting at i^{th} point from Neutral axis [f_i]=\frac{M}{I}y_i

Maximum bending stress at extreme point in weld

\begin{aligned} [f]&=\frac{M}{I}y_{max} \\ &= \frac{Pe}{2t_t \frac{d_w^3}{12}} \times y_{max}\\ &= \frac{350 \times 10^3 \times 100}{2 \times 7 \times \frac{(500)^3}{12}} \times 250\\ &= 60MPa \end{aligned}

As per IS800:2007. Maximum resultant stress an equivalent stress in fllet weld due to combined shear and bending stress

\begin{aligned} (f_{eq})&=\sqrt{f^2+3q^2} \\ &= \sqrt{60^2+3 \times 50^2}\\ &=105.3MPa \end{aligned}

There are 10 questions to complete.