Structural Analysis Test 3

Result

Structural Analysis Test 3

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Weekly Quiz Competition

Question 1
1 / -0

A two-hinged semi-circular arch of radius 12 m is subjected to rise of temperature of 50°C due to which horizontal thrust developed will be “H”. How much rise of temperature is allowed to the above arch such that the horizontal thrust developed will get reduced to 50% if radius of arch is reduced to 9 m now.

A
44.45°C

B
25°C

C
18.75°C

D
14.06°C

Solution

Concept:
Horizontal thrust due to rise of temperature is given by:
\({\rm{H}} = \frac{{4{\rm{EI\alpha T}}}}{{{\rm{\pi \;}}{{\rm{R}}^2}}}\)
E = Modulus of Elasticity of the material
I = Moment of Inertia of the arch section
α = Coefficient of thermal expansion
T = Rise in temperature
R = Radius of the arch
Calculation:
As, E, I, α, π remains same
\({\rm{H}} \propto \frac{{\rm{T}}}{{{{\rm{R}}^2}}}\)
\({{\rm{H}}_1} = \frac{{{{\rm{T}}_1}}}{{{\rm{R}}_1^2}}\) …(i)
\({{\rm{H}}_2} = \frac{{{{\rm{T}}_2}}}{{{\rm{R}}_2^2}}\) …(ii)
R₁ = 12 m
H₂ = 0.50 H₁
R₂ = 9 m
From (i) and (ii)
\(\frac{{{{\rm{H}}_1}}}{{{{\rm{H}}_2}}} = \frac{{{{\rm{T}}_1}{\rm{R}}_2^2}}{{{{\rm{T}}_2}{\rm{R}}_1^2}}\)
\(\frac{{{{\rm{H}}_1}}}{{0.50{\rm{\;}}{{\rm{H}}_1}}} = \frac{{50 \times {9^2}}}{{{{\rm{T}}_2} \times {{12}^2}}}\)
T₂ = 14.06°
Question 2
1 / -0

The flexibility matrix of a Beam element is given as \(\frac{L}{{4EI}}\left[ {\begin{array}{*{20}{c}} { - 1}&2\\ { - 2}&1 \end{array}} \right]\). Then the stiffness matrix is

A
\(\left[ {\begin{array}{*{20}{c}} {\frac{{4EI}}{{3L}}}&{\frac{{-8EI}}{{3L}}}\\ {\frac{{ 8EI}}{{3L}}}&{\frac{{ - 4EI}}{{3L}}} \end{array}} \right]\)

B
\(\left[ {\begin{array}{*{20}{c}} {\frac{{3EI}}{{4L}}}&{\frac{{6EI}}{{4L}}}\\ {\frac{{2EI}}{{3L}}}&{\frac{{4EI}}{{3L}}} \end{array}} \right]\)

C
\(\left[ {\begin{array}{*{20}{c}} {\frac{{ - 4EI}}{{3L}}}&{\frac{{8EI}}{{3L}}}\\ {\frac{{ - 8EI}}{{3L}}}&{\frac{{4EI}}{{3L}}} \end{array}} \right]\)

D
\(\left[ {\begin{array}{*{20}{c}} {\frac{{8EI}}{{3L}}}&{\frac{{ - 8EI}}{{3L}}}\\ {\frac{{ 4EI}}{{3L}}}&{\frac{{-4EI}}{{3L}}} \end{array}} \right]\)

Solution

k → stiffness matrix
f → flexibility matrix
k = [f]^–1
[f] = \(\frac{L}{{4EI}}\left[ {\begin{array}{*{20}{c}} { - 1}&2\\ { - 2}&1 \end{array}} \right]\)
[f] = \(\left[ {\begin{array}{*{20}{c}} {\frac{{ - L}}{{4EI}}}&{\frac{L}{{2EI}}}\\ {\frac{{ - L}}{{2EI}}}&{\frac{L}{{4EI}}} \end{array}} \right]\)
[f]^–1 = \(\frac{{adj\left[ f \right]}}{{\left| f \right|}}\)
|f| = \(\frac{{ - {L^2}}}{{16{{\left( {EI} \right)}^2}}} + \frac{{{L^2}}}{{4{{\left( {EI} \right)}^2}}}\)
|f| = \(\frac{{3{L^2}}}{{16{{\left( {EI} \right)}^2}}}\)
adj[f] = \(\left[ {\begin{array}{*{20}{c}} {\frac{L}{{4EI}}}&{\frac{{ - L}}{{2EI}}}\\ {\frac{L}{{2EI}}}&{\frac{{ - L}}{{4EI}}} \end{array}} \right]\)
[f]^–1 = \(\frac{{16{E^2}{I^2}}}{{3{L^2}}}\left[ {\begin{array}{*{20}{c}} {\frac{{ L}}{{4EI}}}&{\frac{-L}{{2EI}}}\\ {\frac{{ L}}{{2EI}}}&{\frac{-L}{{4EI}}} \end{array}} \right]\)
[f]^–1 = \(\left[ {\begin{array}{*{20}{c}} {\frac{{ 4EI}}{{3L}}}&{\frac{{-8EI}}{{3L}}}\\ {\frac{{ 8EI}}{{3L}}}&{\frac{{-4EI}}{{3L}}} \end{array}} \right]\)