A steel plate, connected to a fixed channel using three identical bolts A, B, and C, carries a load of 6 kN as shown in the figure. Considering the effect of direct load and moment, the magnitude of resultant shear force (in kN) on bolt C is

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PY 8: GATE ME 2017 Official Paper: Shift 2

Option 3 : 17

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

**Concept:**

Eccentric Loading of Riveted Joints:

- An eccentrically loaded joint is one in which line of application of load does not pass through centre of gravity (c.g) of rivets.
- It passes away from c.g axis.
- This has two effects, primary/direct load and secondary load.

Direct Load:

- This load acts parallel to the load acting and vertically downwards.
- The magnitude for the direct load is \(P'=\frac{Load\;acting}{No.\;of\;rivets}\)
- It is represented by P'.

Secondary load:

- It acts perpendicular to the line joining the centre of gravity of rivets assembly to individual rivets.
- The direction of the secondary load is the same as given by external load i.e. the moment produced due to external load is clockwise so the secondary load will also be clockwise.
- The magnitude for the secondary load is \(P"=\frac{P\;\times \;e\;\times\;r_1}{\sqrt{r_1^2\;+\;r_2^2\;+\;r_3^2\;+\;r_4^2}}\) for rivet 1 and similarly for other rivets by changing r2, r3 and r4 in the numerator.

Now, both the load are added vectorially.

**Calculation:**

**Given:**

\({P_{{S_1}}} = Primary\;shear = \frac{6}{3} = 2kN\)

\({P_{{S_2}}} = \frac{{{P_e}{r_c}}}{{r_A^2 + r_B^2 + r_C^2}} = \frac{{6 \times 250 \times 50}}{{{{50}^2} + {0^2} + {{50}^2}}}\)

\({P_{{S_2}}} = 15kN\)

∴ Total resultant shear force

\({P_s} = {P_{{S_1}}} + {P_{{S_2}}} = 15 + 2 = 17kN\)