Question 1 |

For the square steel beam cross-section shown in the figure, the shape factor about
z- z axis is S and the plastic moment capacity is M_P. Consider yield stress
f_y = 250 MPa and a = 100 mm.

The values of S and M_P (rounded-off to one decimal place) are

The values of S and M_P (rounded-off to one decimal place) are

S = 2.0, M_P= 58.9 kN-m | |

S = 2.0, M_P=100.2 kN-m | |

S = 1.5, M_P= 58.9 kN-m | |

S = 1.5, M_P=100.2 kN-m |

Question 1 Explanation:

Shape factor for diamond shaped section = 2

\begin{aligned} S&=\frac{M_P}{M_y}=\frac{f_yZ_P}{f_yZ_e}\\ M_P&=S.M_y=S.f_y.Z_e \end{aligned}

\begin{aligned} I_{ZZ}&=\frac{a^4}{12}\\ Z_{ZZ}&=\frac{a^4 \times \sqrt{2}}{12 \times a }=\frac{\sqrt{2}a^3}{12}mm^3\\ M_P&=\left [ 2 \times 250 \times \frac{\sqrt{2} \times (100)^3}{12} \right ] \times 10^{-6} kNm\\ &=58.93kNm \end{aligned}

\begin{aligned} S&=\frac{M_P}{M_y}=\frac{f_yZ_P}{f_yZ_e}\\ M_P&=S.M_y=S.f_y.Z_e \end{aligned}

\begin{aligned} I_{ZZ}&=\frac{a^4}{12}\\ Z_{ZZ}&=\frac{a^4 \times \sqrt{2}}{12 \times a }=\frac{\sqrt{2}a^3}{12}mm^3\\ M_P&=\left [ 2 \times 250 \times \frac{\sqrt{2} \times (100)^3}{12} \right ] \times 10^{-6} kNm\\ &=58.93kNm \end{aligned}

Question 2 |

A weld is used for joining an angle section ISA 100 mm x 100 mm x 10 mm to
a gusset plate of thickness 15 mm to transmit a tensile load. The permissible
stress in the angle is 150 MPa and the permissible shear stress on the section
through the throat of the fillet weld is 108 MPa. The location of the centroid of
the angle is represented by C_{yy} in the figure, where C_{yy}=28.4mm. The area of
cross-section of the angle is 1903 mm^2. Assuming the effective throat thickness of the weld to be 0.7 times the given weld size, the lengths L_1 and L_2 (rounded-
off to the nearest integer) of the weld required to transmit a load equal to the full strength of the tension member are, respectively

541 mm and 214 mm | |

214 mm and 541 mm | |

380 mm and 151 mm | |

151 mm and 380 mm |

Question 2 Explanation:

As per IS:800-2007 Clause no. 11.2.1

Full strength of tension member,

\begin{aligned} P&=\sigma _{at} \times A_g\\ P&=150 \times 1903\\ P&=285450N \end{aligned}

Weld strength of L_1,

\begin{aligned} P_1&=L_1\times (0.7 \times s) \times 108\\ P_1&=L_1\times (0.7 \times 5) \times 108\\ P_1&=378L_1 \end{aligned}

Weld strength of L_2,

\begin{aligned} P_2&=L_2\times (0.7 \times s) \times 108\\ P_2&=L_2\times (0.7 \times 5) \times 108\\ P_2&=378L_2 \end{aligned}

Applying moment CG equal to zero,

\begin{aligned} P_1 \times 28.4 &=P_2 \times (100-28.4)\\ \frac{P_1}{P_2}&=\frac{71.6}{28.4}\\ \frac{L_1 \times 378}{L_2 \times 378}&=\frac{71.6}{28.4}=2.521\\ L_1&=2.521L_2\\ \text{Total weld strength}&=\text{Tensile strength}\\ (L_1+L_2) \times 3.5 \times 108&=285450\\ L_1+L_2&=755.158\\ 2.521L_2+L_2&=755.152\\ L_2&=214.472mm\\ L_1&=540.685mm\\ \end{aligned}

Full strength of tension member,

\begin{aligned} P&=\sigma _{at} \times A_g\\ P&=150 \times 1903\\ P&=285450N \end{aligned}

Weld strength of L_1,

\begin{aligned} P_1&=L_1\times (0.7 \times s) \times 108\\ P_1&=L_1\times (0.7 \times 5) \times 108\\ P_1&=378L_1 \end{aligned}

Weld strength of L_2,

\begin{aligned} P_2&=L_2\times (0.7 \times s) \times 108\\ P_2&=L_2\times (0.7 \times 5) \times 108\\ P_2&=378L_2 \end{aligned}

Applying moment CG equal to zero,

\begin{aligned} P_1 \times 28.4 &=P_2 \times (100-28.4)\\ \frac{P_1}{P_2}&=\frac{71.6}{28.4}\\ \frac{L_1 \times 378}{L_2 \times 378}&=\frac{71.6}{28.4}=2.521\\ L_1&=2.521L_2\\ \text{Total weld strength}&=\text{Tensile strength}\\ (L_1+L_2) \times 3.5 \times 108&=285450\\ L_1+L_2&=755.158\\ 2.521L_2+L_2&=755.152\\ L_2&=214.472mm\\ L_1&=540.685mm\\ \end{aligned}

Question 3 |

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 3 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 4 |

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 4 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 5 |

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 5 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 6 |

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 6 Explanation:

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

Question 7 |

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 7 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 8 |

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 8 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 9 |

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 9 Explanation:

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

Question 10 |

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 10 Explanation:

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

There are 10 questions to complete.