Question 1 |

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 1 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 2 |

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 2 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 3 |

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 3 Explanation:

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

Question 4 |

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 4 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}

Question 5 |

Four bolts P, Q, R and S of equal diameter are used for a bracket subjected to a load of 130 kN as shown in the figure.

The force in bolt P is

The force in bolt P is

32.50 kN | |

69.33 kN | |

82.50 kN | |

119.32 kN |

Question 5 Explanation:

\begin{aligned} F_{1} &= \frac{P}{n}=\frac{130}{4}=32.5 kN \\ F_{2} &= \frac{Pe}{\sum r^{2}_{i}}\times r_{p} \\ r_{P} &= r_{Q} = r_{R} = r_{S} \\ &= \sqrt{50^{2}+120^{2}} \\&=130 mm \\ F_{2} &=\frac{\left ( 130\times 200 \right )\times 130}{4\times 130^{2}} \\ &= \frac{200}{4} =50 kN \\ \cos \theta &= \frac{50}{130} \\ \Rightarrow \; \; F_{R} &= \sqrt{\left [ \left ( 32.5^{2} \right )+\left ( 50^{2} \right )+\left ( 2\times 32.5\times 50 \right )\times \left ( \frac{50}{130} \right ) \right ]} \\ &= 69.33 kN \end{aligned}

Question 6 |

A fillet weld is simultaneously subjected to factored normal and shear stresses of 120 MPa and 50 MPa, respectively. As per IS 800 : 2007, the equivalent stress (in MPa, up to two decimal places) is ______

178.78 | |

122.45 | |

156.25 | |

147.99 |

Question 6 Explanation:

Factrored normal stress, f_{a}= 120 Mpa

factrored shear stress, q= 50 Mpa

According to IS-800:2007, clause 10.5.10.1.1

The equivalent stress,

\begin{aligned} f_{e}&=\sqrt{f_{a}^{2}+3q^{2}}\leq \frac{F_{u}}{\sqrt{3}y_{mW}} \\ f_{e}&=\sqrt{120^{2}+3\times 50^{2}} \\ &=147.99 \\ \Rightarrow \;\; f_{e}& \leq \frac{f_{u}}{\sqrt{3}y_{mW}}=\frac{400}{\sqrt{3}\times 1.25} \\ &=\left ( 184.75 MPa \right )\end{aligned}

Note: The above check for combination of stresses need not be done for fillet welds where the sum of normal and shear stresses does not exceed f_{wd} [clause 10.5.10.1.2(b)]. Hence, sum of normal and shear stresses =120+50=170 MPa \leq f_{wd} (= 184.75 MPa)

So, the above check need not be done.

So, weld is designed for a resultant shear stress of \left ( f_{r} \right )

\begin{aligned} f_{r}&=\sqrt{f_{a}^{2}+q^{2}} \\ &=\sqrt{120^{2}=50^{2}} \\ &=130 N/mm^{2} \end{aligned}

So, the design stress is 130 N/mm^{2}; however equivalent stress 147.99 MPa.

factrored shear stress, q= 50 Mpa

According to IS-800:2007, clause 10.5.10.1.1

The equivalent stress,

\begin{aligned} f_{e}&=\sqrt{f_{a}^{2}+3q^{2}}\leq \frac{F_{u}}{\sqrt{3}y_{mW}} \\ f_{e}&=\sqrt{120^{2}+3\times 50^{2}} \\ &=147.99 \\ \Rightarrow \;\; f_{e}& \leq \frac{f_{u}}{\sqrt{3}y_{mW}}=\frac{400}{\sqrt{3}\times 1.25} \\ &=\left ( 184.75 MPa \right )\end{aligned}

Note: The above check for combination of stresses need not be done for fillet welds where the sum of normal and shear stresses does not exceed f_{wd} [clause 10.5.10.1.2(b)]. Hence, sum of normal and shear stresses =120+50=170 MPa \leq f_{wd} (= 184.75 MPa)

So, the above check need not be done.

So, weld is designed for a resultant shear stress of \left ( f_{r} \right )

\begin{aligned} f_{r}&=\sqrt{f_{a}^{2}+q^{2}} \\ &=\sqrt{120^{2}=50^{2}} \\ &=130 N/mm^{2} \end{aligned}

So, the design stress is 130 N/mm^{2}; however equivalent stress 147.99 MPa.

Question 7 |

In a fillet weld, the direct shear stress and bending tensile stress are 50 MPa and 150 MPa, respectively. As per IS 800: 2007, the equivalent stress (in MPa, up to two decimal places) will be ______

122.23 | |

145.56 | |

173.3 | |

189.54 |

Question 7 Explanation:

Direct bending tensile stress,

f_{a}=150 Mpa

Direct shear stress,

q=50 Mpa

According to IS 800:2007, clause 10.5.10.1.1

The equivalent stress,

\begin{aligned} f_{e}&=\sqrt{f^{2}_{a}+3q^{2}}\leq \frac{f_{u}}{\sqrt{3}\gamma _{mW}} \\ f_{e}&=\sqrt{150^{2}+3\times 50^{2}} \\&=173.21 MPa\\ \Rightarrow \;\; f_{e} & \leq \frac{f_{u}}{\sqrt{3}\gamma _{mW}} \\&=\frac{400}{\sqrt{3}\times 1.25} \\ &=184.75 MPa \end{aligned}

(Hence OK)

Note: The above check for combination of stressed need not be done for fillet welds where the sum of normal and shear stresses does not exceed f_{wd} [Clause 10.5.10.1.2(b)].

Thus, f_{a}+q=150+50=200 MPa

f_{a}+q \gt f_{wd} (= 184.75 MPa)

\Rightarrow \;\;f_{e}= 173.21 MPa

Thus, the weld is designed for an equivelent stress of 173.21 MPa

f_{a}=150 Mpa

Direct shear stress,

q=50 Mpa

According to IS 800:2007, clause 10.5.10.1.1

The equivalent stress,

\begin{aligned} f_{e}&=\sqrt{f^{2}_{a}+3q^{2}}\leq \frac{f_{u}}{\sqrt{3}\gamma _{mW}} \\ f_{e}&=\sqrt{150^{2}+3\times 50^{2}} \\&=173.21 MPa\\ \Rightarrow \;\; f_{e} & \leq \frac{f_{u}}{\sqrt{3}\gamma _{mW}} \\&=\frac{400}{\sqrt{3}\times 1.25} \\ &=184.75 MPa \end{aligned}

(Hence OK)

Note: The above check for combination of stressed need not be done for fillet welds where the sum of normal and shear stresses does not exceed f_{wd} [Clause 10.5.10.1.2(b)].

Thus, f_{a}+q=150+50=200 MPa

f_{a}+q \gt f_{wd} (= 184.75 MPa)

\Rightarrow \;\;f_{e}= 173.21 MPa

Thus, the weld is designed for an equivelent stress of 173.21 MPa

Question 8 |

Two plates of 8 mm thickness each are connected by a fillet weld of 6 mm thickness as shown in the figure.

The permissible stresses in the plate and the weld are 150 MPa and 110 MPa, respectively. Assuming the length of the weld shown in the figure to be the effective length, the permissible load P (in KN) is

The permissible stresses in the plate and the weld are 150 MPa and 110 MPa, respectively. Assuming the length of the weld shown in the figure to be the effective length, the permissible load P (in KN) is

60 | |

175.51 | |

115.51 | |

55.51 |

Question 8 Explanation:

\begin{aligned}P_{W}&=f_{b}\times l_{eff}\times t_{t} \\ &=110\left ( 100+100+50 \right )\times 0.7\times 6 \\ &=115.5\, kN \\ P_{plate}&=\sigma _{at}\times A_{net} \\ &=150\times \left ( 50\times 8 \right ) \\ &=60\, kN\end{aligned}

So permissible load = 60 kN

So permissible load = 60 kN

Question 9 |

A column is subjected to a load through a bracket as shown in the figure.

The resultant force (in kN, up to one decimal place) in the bolt 1 is_____

The resultant force (in kN, up to one decimal place) in the bolt 1 is_____

3.5 | |

6 | |

2 | |

4 |

Question 9 Explanation:

\begin{aligned} F_{1}&=\frac{P}{n}=\frac{10}{4}=2.5kN \\ F_{2}&=\frac{Pe}{\sum r^{2}_{i}}\times r_{1} \\&=\frac{10\times 15}{4\times 5^{2}}\times 5=7.5 kN \\ F_{R}&=\sqrt{F^{2}_{1}+F_{2}^{2}+2F_{1}F_{2}\cos \theta } \\ &=\sqrt{2.5^{2}+7.5^{2}+2\times 2.5\times 7.5\times \cos 135^{\circ}} \\ &=6kN \end{aligned}

Question 10 |

Two bolted plates under tension with alternative arrangement of bolt holes are shown in figures 1 and 2. The hole diameter, pitch, and gauge length are d, p and g, respectively.

Which one of the following conditions must be ensured to have higher net tensile capacity of configuration shown in Figure 2 than that shown in Figure 1 ?

Which one of the following conditions must be ensured to have higher net tensile capacity of configuration shown in Figure 2 than that shown in Figure 1 ?

p^{2} \gt 2gd | |

p^{2}\lt \sqrt{4gd} | |

p^{2} \gt 4gd | |

p \gt 4gd |

Question 10 Explanation:

Tensile strength of plate in arrangement (2) will be greater than in arrangement (1),

*As per IS code 800:2007 clause 6.3

\begin{aligned} \left ( 0.9A_{net}\frac{f_{up}}{\gamma _{m1}} \right )_{2} & \gt \left ( 0.9A_{net}\frac{f_{up}}{\gamma _{m1}} \right )_{1} \\ \left ( A_{net} \right )_{2} & \gt \left ( A_{net} \right )_{1} \\ \left [ \left ( B-2d+\frac{p^{2}}{4g} \right )t \right ]_{2}& \gt \left [ \left ( B-d \right )t \right ]_{1} \\ B-2d+\frac{p^{2}}{4g} & \gt B-d \\ \frac{p^{2}}{4g} & \gt d \\ p^{2}& \gt 4gd\end{aligned}

There are 10 questions to complete.

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