GATE Civil Engineering (CE) 2019 Shift 1 Solved Paper

Question : 46 of 65

Marks: +1, -0

A portal frame shown in figure (not drawn to scale) has a hinge support at joint P and a roller support at joint R. A point load of

50 {kN}

is acting at joint

{R}

in the horizontal direction. The flexural rigidity. EI, of each member is

10^{6} {kNm}^{2}

. Under the applied load, the horizontal displacement (in

{mm}

, round off to 1 decimal place) of joint

{R}

would be _____________

Solution:

Σ {F}_{ {x}}=0

⇒ {H}_{ {p}}= {F}=0 ⇒ {H}_{ {p}}=- {F}

Σ {M}_{ {p}}=0

⇒ {V}_{ {R}} × 5+ {F} × 10=0 ⇒ {V}_{ {R}}=-2 {F}

Σ {F}_{ {y}}=0

⇒ {V}_{ {p}}+ {V}_{ {R}}=0 ⇒ {V}_{ {p}}=2 {F}

∆ {R}]

horizontal

=\;{∂ {U}}/{∂ {F}}=\;{ {VU}_{ {PQ}}}/{∂ {F}}+\;{∂ {V}_{ {OR}}}/{∂ {F}}(

Castigliano's theorem

)

{U}_{ {PQ}}=∫ \;{ {M}^{2} {PQ}}/{2 {EI}} {d} {x} ; {M}_{ {PQ}}= {V}_{ {p}} {x}=2 {F}_{ {x}} ; {x} ∈[0,5]

⇒ \;{∂ {U}_{ {PQ}}}/{∂ {F}}=\;{1}/{ {EI}} ∫ \;{∂ {M}_{ {PQ}}}/{∂ {F}} × {M}_{ {PQ}} {d} {x}=\;{1}/{ {EI}} ∫↙{0}↖{5} 2 {F} {x} × 2 {x} {dx}

\;\; =\;{4 {F}}/{ {EI}}[\;{ {x}^{3}}/{3}]_{0}^{5}=\;{4 × 50}/{10^{6}} × \;{125}/{3}=\;{25}/{3000}

{U}_{ {QR}}=∫ \;{ {M}_{ {QR}}^{2}}/{2 {EI}} {d} {x} ; {M}_{ {QR}}= {F}_{ {x}} ; {x} ∈[0,10]

⇒ \;{∂ {U}_{ {QR}}}/{∂ {F}}=\;{1}/{ {EI}} ∫ \;{∂ {M}_{ {QR}}}/{∂ {F}} {M}_{ {QR}} {dx}=\;{1}/{ {EI}} ∫↙{0}↖{10} {F} {x} × {xdx}

\;\; =\;{ {F}}/{ {EI}}[\;{ {x}^{3}}/{3}]_{0}^{10}=\;{50}/{10^{6}} × \;{1000}/{3}=\;{50}/{3000}

⇒ \;{∂ {U}}/{∂ {F}}=\;{25}/{3000}+\;{50}/{3000}=\;{25}/{1000}=0.025 {m}=25 {mm}

∴ ∆ {R}]

Horizontal

=\;{∂ {U}}/{∂ {F}}=25 {mm}

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