A 6 {m} long simply supported beam is prestressed as shown in the figure. The beam carries a uniformly distributed load 6 {kN} {/} {m} over its entire span. If the effective flexural rigidity E I=2 \times 10^{4} {kNm}^{2} and the effective prestressing force is 200 {kN}, the net increase in length of the prestressing cable (in {mm}, up to two decimal places) is

GATE Civil Engineering (CE) 2018 Shift 2 Solved Paper

Question : 59 of 65

Marks: +1, -0

6 {m}

long simply supported beam is prestressed as shown in the figure.

The beam carries a uniformly distributed load

6 {kN} {/} {m}

over its entire span. If the effective flexural rigidity

E I=2 × 10^{4} {kNm}^{2}

and the effective prestressing force is

200 {kN}

, the net increase in length of the prestressing cable (in

{mm}

, up to two decimal places) is

Solution:

Span of PSC beam

=6 {m}

E I=2 × 10^{4} {kNm}^{2}=2 × 10^{4} × 1000^{3} {N}- {mm}^{2}

=2 × 10^{4} {kNm}^{2}

P=200 {kN}

Total UDL

=6 {kN} {/} {m}

eccentricity

=e=50 {mm}

(a) Slope of beam due to P-force

θ_{1}=\;{P . e ⋅ L}/{2 E I}

\;\; =\;{200 × 10^{3} × 50 × 6000}/{2 × 2 × 10^{13}}

\;\; =1.5^{-3}

(upward)
(b) Slope of beam due to UDL

θ_{2}=\;{w L^{3}}/{24 E I}

\;\; =\;{6 ×(6000)^{3}}/{24 × 2 × 10^{13}}

\;\; =(+) 2.7 × 10^{-3}

(downward)
(c) Net slope of beam

θ=θ_{1}+θ_{2}

\;\; =(-) 1.5 × 10^{-3}+(+) 2.7 × 10^{-3}

\;\; =1.2 × 10^{-3}

(d) Total net increase in length

=2 e θ

=2 × 50 × 1.2 × 10^{-3}=0.12 {mm}

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