Explanation:

For a cantilever beam carrying uniformly distributed load throughout span, standard formula for deflection (δ) and slope (θ) at free end are given below:

\({\rm{δ }} = {\rm{\;}}\frac{{{\rm{w}}{{\rm{l}}^4}}}{{8{\rm{EI}}}}\) and \({\rm{\ta }} = {\rm{\;}}\frac{{{\rm{w}}{{\rm{l}}^3}}}{{6{\rm{EI}}}}\)  (θ is in radian)

Where w is uniformly distributed load.

As per given in question, length of beam is doubled n change in deflection,

\(δ ' = \frac{{w{{\left( {2l} \right)}^4}}}{{8EI}} = \frac{{16\; × \;w{l^4}}}{{8EI}}\)

⇒ δ' = 16 × δ 

Additional Information

Explanation:

In case of simply supported beam as shown in figure below:

\({\delta_c} = \frac{{P{L^3}}}{{48EI}}\)

\({\ta _B} = \ta _A=\frac{{w{L^2}}}{{16EI\;}}\)

When a central point load is acting, we have following boundary conditions,

•  Deflections at supports are assumed zero unless orwise stated, while slope will be maximum.
• The slope at center of symmetrically loaded and supported beams is zero.
• The deflection will be maximum at center of loaded beam.

Additional Information

Deflection and slope of various beams are given by:

Where, y = Deflection of beam, θ = Slope of beam

Explanation:

The deflection at free end of cantilever due to uniformly distributed load is given as:

\(\Delta =\frac{wL^4}{8EI}\)

Given that 

E = 2 × 10⁵ N/mm² 

I = 8 × 10⁷ mm⁴

L = 2500 mm

Δ  = 5 mm

Let deflection is 

\(5 = \frac{w\times (2500)^4}{8\times 2\times 10^5\times 8\times 10^7}\)

w = 16.38 N/mm

w =  \( \frac{16.38 \times 10^{-3}kN}{10^{-3} \ m}\) = 16.38 KN/m.

Concept:

The deflection of beam at free end due to load P is \(\Delta = \frac{{P{l^3}}}{{3EI}}\)

Since deflection in beam and spring are equal, refore stiffness of both i.e. beam and spring are in parallel connection.

For parallel combination:

k_eq = k₁ + k₂

where k₁ = stiffness of spring and k₂ = stiffness of beam.

Calculation:

Given:

Deflection of beam due to point load P if spring is not re \(\Delta = \frac{{P{l^3}}}{{3EI}}\)

Stiffness \({k_2} = \frac{P}{\Delta }\)

\({k_2} = \frac{P}{{\frac{{P{l^3}}}{{3EI}}}} = \frac{{3EI}}{{{l^3}}}\)

\(\refore \;{k_{eq}} = k + \frac{{3EI}}{{{L^3}}} = \frac{{k{L^3}\;+\;3EI}}{{{L^3}}}\)

∴ Transverse deflection under load

\(\Rightarrow \Delta_{net}= \frac{P}{{{K_{eq}}}} = \frac{{P{L^3}}}{{K{L^3} + 3EI}}\)

\( \Rightarrow \Delta_{net} = \frac{{P{L^3}}}{{3EI}}\;\left( {\frac{1}{{\frac{{K{L^3}}}{{3EI}}\;+\;1}}} \right)\)

\(\Rightarrow \Delta = \frac{{P{L^3}}}{{3EI}}\;\left( {\frac{{3EI}}{{K{L^3}\;+\;3EI}}} \right)\)

Shortcut Method:

Deflection of beam = Deflection of spring

\(\frac{{\left( {P\;-\;R} \right){L^3}}}{{3EI}} = \frac{R}{K}\)

\( \Rightarrow \left( {\frac{1}{K} + \frac{{{L^3}}}{{3EI}}} \right)R = \frac{{P{L^3}}}{{3EI}}\)

\(R = \frac{{P{L^3}}}{{3EI}} \times \frac{{3EIK}}{{3EI\;+\;{L^3}}}\)

\(\Delta_{net} = \frac{R}{K} = \frac{{P{L^3}}}{{3EI}}\;\left( {\frac{{3EI}}{{3EI\;+\;K{L^3}}}} \right)\)

Concept:

Deflection at centre of a simply supported beam due to point load (W):

\(δ_1=\frac{WL^3}{48EI}\)

Deflection at centre of a simply supported beam due to uniformly distribute load (W = wL):

\(δ_2=\frac{5wL^4}{384EI}=\frac{5WL^3}{384EI}\;\;\;(\because w=\frac{W}{L})\)

Calculation:

Given:

\(\frac{\delta_2}{\delta_1}=\frac{5WL^3}{384EI}\times \frac{48EI}{WL^3}=\frac{5}{8}\)

Hence,

The deflection will be reduced by

\(={\delta _1} - {\delta _2} = \left( {1 - \frac{5}{8}} \right){\delta _1} = \frac{3}{8}{\delta _1}\)

I^st condition:

For a cantilever beam subjected to load W at distance of L/3 from free end, deflection is given by:

\({\rm{y}}{{\rm{c}}_1} = \frac{{\rm{W}}}{{3{\rm{EI}}}} \times {\left( {\frac{{2{\rm{L}}}}{3}} \right)^3} + \frac{{\rm{W}}}{{2{\rm{EI}}}}{\left( {\frac{{2{\rm{L}}}}{3}} \right)^2} \times \frac{{\rm{L}}}{3} = {\frac{{28{\rm{WL}}}}{{162{\rm{EI}}}}^3}\)

II^nd condition:

For a cantilever beam subjected to load W at distance of 2L/3 from free end:

\({\rm{y}}{{\rm{c}}_2} = \frac{{\rm{W}}}{{3{\rm{EI}}}} \times {\left( {\frac{{\rm{L}}}{3}} \right)^3} + \frac{{\rm{W}}}{{2{\rm{EI}}}}{\left( {\frac{{\rm{L}}}{3}} \right)^2} \times \frac{{2{\rm{L}}}}{3} = \frac{{8{\rm{W}}{{\rm{L}}^3}}}{{162{\rm{EI}}}}\)

\({\rm{Ratio\;}}\left( {\rm{r}} \right) = \frac{{{\rm{y}}{{\rm{c}}_1}}}{{{\rm{y}}{{\rm{c}}_2}}} = \frac{{{{\frac{{28{\rm{WL}}}}{{162{\rm{EI}}}}}^3}}}{{\frac{{8{\rm{W}}{{\rm{L}}^3}}}{{162{\rm{EI}}}}}} = \frac{{28}}{8} = \frac{7}{2}\)

\(\refore \frac{{{\rm{y}}{{\rm{c}}_2}}}{{{\rm{y}}{{\rm{c}}_1}}} = \frac{2}{7}\)

Explanation:

Deflection of end B = 0

(Downward deflection due to uniformly distributed load) – (upward deflection due to RB) = 0

\(\frac{{w{l^4}}}{{8EI}} - \frac{{{R_B}{l^3}}}{{3EI}} = 0\)

\(\begin{array}{l} {R_B} = \frac{3}{8}wl\\ {R_A} = wl - \frac{3}{8}wl = \frac{5}{8}wl \end{array}\)

As asked at propped end i.e. at B

 ∴ RB = 3wl/8

Important Points Deflection of various beams is given by: 

Concept:

Area moment method

Theorem 1: The difference of slope of any two points of a beam is equal to area of\(\frac{M}{{EI}}\) diagram between those points.

θ_B – θ_A = Area of \(\frac{M}{{EI}}\) diagram between B and A.

Theorem 2: The difference of deflection of two points of a beam is equal to moment of area of\(\frac{M}{{EI}}\)diagram between those points.

Y_B – Y_A = Moment of area of\(\frac{M}{{EI}}\)diagram between B and A.

Y_B – Y_A = (Ax?)

Calculation:

δ_max = Ax? 

\({δ _{max}} = \frac{1}{3} \times L \times \frac{{W{L^2}}}{{2EI}} \times \frac{3}{4}L\)

\(\Rightarrow {δ _{max}} = \frac{{W{L^4}}}{{8EI}}\)

Explanation:

Concept:

The maximum deflection (at centre) of a simply supported beam with uniformly distributed load (UDL) is given by

\(\delta = {5\over 384}\frac{{w{L^4}}}{{EI}}\)

where w is weight per unit length,

Additional Information 

Deflection and slope of various beams is given by:

Where, y = Deflection of beam, θ = Slope of beam