Question 1 |

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 1 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 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 |

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 4 Explanation:

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

Question 5 |

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 5 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 5 questions to complete.

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