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GATE CE 2015 Official Paper: Shift 1

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

**Concept:**

Carry-over factor (C.O.F) is given by,

\({\rm{C}}.{\rm{O}}.{\rm{F}} = \frac{{{\rm{Carry\;over\;moment}}}}{{{\rm{Applied\;moment}}}}\)

The applied moment is the moment due to **external load **at the **far end of the beam.**

Carry over moment will be moment acting at the **fixed support** of the beam.

**Calculation:**

By **symmetry** we can consider two parts of beam separately with half the load.

Moment at joint B = 10 × 1 = 10 kNm

∴ Applied moment = 10 kNm

Carry over factor (C.O.F) = 1/2

**∴ Carry over moment = C.O.F × Applied moment**

Carry over moment = 1/2 × 10 = 5 kNm

**Hence carry-over moment at joint A due to 10 kNm moment at the propped end will be 5 kNm**.