Analysis by the Strain Energy Method

Strain energy due to shear is the energy stored because of the shear deformation of the beam. For a rectangular section the shear stress varies parabolically, so the energy is found by integrating over the section.

Derivation

Take a beam of rectangular section (width $b$, depth $d$) carrying shear force $V$ at a section. Shear stress at a distance $y$ from the neutral axis:

$$\tau = \frac{V A\bar y}{I b} = \frac{V}{2I}\left(\frac{d^2}{4} - y^2\right)$$

Strain energy per unit volume $= \dfrac{\tau^2}{2G}$. For an element of length $dx$, with area element $dA = b\,dy$:

$$dU = \int_{-d/2}^{d/2} \frac{\tau^2}{2G}\, b\,dy\,dx$$

Substituting $\tau$:

$$\begin{aligned} dU &= \frac{V^2 b\,dx}{8GI^2}\int_{-d/2}^{d/2}\left(\frac{d^2}{4}-y^2\right)^2 dy \\ &= \frac{V^2 b\,dx}{8GI^2}\cdot\frac{d^5}{30} \end{aligned}$$

Using $I = \dfrac{bd^3}{12}$, so $I^2 = \dfrac{b^2d^6}{144}$:

$$dU = \frac{V^2 b\, d^5 \times 144}{8G\,b^2 d^6\times 30}\,dx = \frac{3V^2}{5\,G\,bd}\,dx = \frac{6}{5}\cdot\frac{V^2}{2GA}\,dx$$

where $A = bd$. Hence the shear strain energy for the whole beam is

$$U_s = \int_0^L \frac{6}{5}\,\frac{V^2}{2GA}\,dx = \int_0^L \frac{3V^2}{5GA}\,dx$$

The factor $\dfrac{6}{5} = 1.2$ is the form factor (shape factor) for shear $\alpha$ of a rectangle. If shear were uniform, $dU = \dfrac{V^2}{2GA}dx$.

Example (check): simply supported beam with central load $P$: $V = \pm P/2$,

$$U_s = \frac{6}{5}\times\frac{(P/2)^2}{2GA}\times L = \frac{3P^2L}{20\,GA}$$

The deflection under the load (from $\delta = \partial U/\partial P$) is $\delta_s = \dfrac{3PL}{10GA}$.

Result: $U_s = \dfrac{6}{5}\displaystyle\int_0^L \dfrac{V^2}{2GA}\,dx$ for a rectangular section.

Strain energy due to bending is the energy stored in a beam because of bending (flexural) stresses.

Derivation

Consider an elemental length $dx$ of a beam under a bending moment $M$. The ends of the element rotate relative to each other by

$$d\theta = \frac{M\,dx}{EI}$$

The moment rises gradually from 0 to $M$ while the rotation rises from 0 to $d\theta$, so the work done (strain energy stored) in the element is

$$dU = \tfrac12 M\,d\theta = \frac{M^2\,dx}{2EI}$$

Alternative (stress method): the bending stress at a distance $y$ from the neutral axis is $\sigma = \dfrac{My}{I}$. Energy per unit volume $= \dfrac{\sigma^2}{2E}$:

$$dU = \int \frac{M^2y^2}{2EI^2}\,dA\,dx = \frac{M^2}{2EI^2}\left(\int y^2\,dA\right)dx = \frac{M^2\,dx}{2EI}$$

For the whole beam of length $L$:

$$U_b = \int_0^L \frac{M^2}{2EI}\,dx$$

For a rectangular beam

With $I = \dfrac{bd^3}{12}$, a simply supported beam of span $L$ carrying a central load $P$ has $M = \dfrac{P x}{2}$ for $0 \le x \le L/2$:

$$U_b = 2\int_0^{L/2}\frac{P^2x^2}{8EI}\,dx = \frac{P^2L^3}{96EI}$$

For a cantilever with end load $P$: $M = Px$, so $U_b = \dfrac{P^2L^3}{6EI}$.

Result: $U_b = \displaystyle\int_0^L \frac{M^2}{2EI}\,dx$

Total strain energy of a structural system

For a member of length $L$ with axial force $N$, bending moment $M$, shear force $V$ and torque $T$, the total strain energy is the sum:

$$U = \int_0^L \frac{N^2}{2AE}\,dx + \int_0^L \frac{M^2}{2EI}\,dx + \int_0^L \frac{\alpha\,V^2}{2GA}\,dx + \int_0^L \frac{T^2}{2GJ}\,dx$$

Here $A$ = area, $I$ = moment of inertia, $J$ = polar moment of inertia, $E$ and $G$ = Young's and shear moduli, $\alpha$ = shear form factor ($6/5$ for rectangle, $10/9$ for solid circle). For a structure, the sum is taken over all members.

Derivation of energy due to shear force

Consider an element of length $dx$ with shear force $V$. Shear stress $\tau$ at a level $y$ of the section (area element $dA$):

$$\tau = \frac{V A'\bar y}{Ib}$$

Energy per unit volume $= \dfrac{\tau^2}{2G}$, so for the element

$$dU = \int_A \frac{\tau^2}{2G}\,dA\,dx$$

If the shear stress were uniform, $\tau = V/A$, giving $dU = \dfrac{V^2}{2GA}dx$. To account for the actual non-uniform distribution a shear form factor is defined as

$$\alpha = \frac{A}{I^2}\int_A \left(\frac{A'\bar y}{b}\right)^2 dA$$

Then

$$dU = \frac{\alpha V^2}{2GA}\,dx \quad\Rightarrow\quad U_s = \int_0^L \frac{\alpha V^2}{2GA}\,dx$$

For a rectangular section $\tau = \dfrac{V}{2I}\left(\dfrac{d^2}{4}-y^2\right)$, which gives $\alpha = 1.2$.

The dynamic multiplier (impact factor) is the ratio of the maximum dynamic deflection (or stress) to the static deflection (or stress) produced by the same load:

$$\text{Dynamic multiplier} = \frac{\delta_{max}}{\delta_{st}}$$

Mass falling on the mid-span of a simply supported beam

Let a weight $W$ fall from a height $h$ on to the mid-span of a simply supported beam of span $L$ and flexural rigidity $EI$. Let $\delta$ be the maximum mid-span deflection and neglect the beam's weight and energy losses.

Static deflection of the beam under $W$ at mid-span:

$$\delta_{st} = \frac{WL^3}{48EI}$$

So the beam stiffness at mid-span is $k = \dfrac{W}{\delta_{st}} = \dfrac{48EI}{L^3}$, and the equivalent dynamic load is $P = k\delta$.

Energy balance: loss of potential energy of the falling weight = strain energy stored in the beam.

$$W(h + \delta) = \tfrac12 P\delta = \tfrac12 k\delta^2 = \tfrac12\frac{W}{\delta_{st}}\delta^2$$ $$\delta^2 - 2\delta_{st}\delta - 2\delta_{st}h = 0$$ $$\delta = \delta_{st}\left[1 + \sqrt{1 + \frac{2h}{\delta_{st}}}\right]$$

Therefore the dynamic multiplier is

$$\boxed{\frac{\delta}{\delta_{st}} = 1 + \sqrt{1+\frac{2h}{\delta_{st}}}}$$

The maximum bending moment and stress are multiplied by the same factor, since the structure is linear. If $h = 0$ (suddenly applied load) the multiplier is 2. If $h \gg \delta_{st}$, then $\delta \approx \sqrt{2h\,\delta_{st}}$.

Given data

Length $L = 8$ m, diameter $d = 20$ cm $= 0.2$ m, end load $P = 15$ kN, twisting moment $T = 8$ kNm.

Assumed: $E = 200$ GPa $= 2\times10^8$ kN/m$^2$, $G = 80$ GPa $= 8\times10^7$ kN/m$^2$.

Section properties

$$I = \frac{\pi d^4}{64} = \frac{\pi(0.2)^4}{64} = 7.854\times10^{-5}\ \text{m}^4$$ $$J = \frac{\pi d^4}{32} = 1.5708\times10^{-4}\ \text{m}^4$$

Strain energy due to bending

At distance $x$ from the free end, $M = Px$:

$$U_b = \int_0^L \frac{(Px)^2}{2EI}dx = \frac{P^2L^3}{6EI}$$ $$U_b = \frac{15^2\times 8^3}{6\times 2\times10^{8}\times 7.854\times10^{-5}} = \frac{115200}{94247.8} = 1.2223\ \text{kNm}$$

Strain energy due to torsion

Torque is constant along the length:

$$U_t = \frac{T^2L}{2GJ} = \frac{8^2\times 8}{2\times 8\times10^{7}\times 1.5708\times10^{-4}} = \frac{512}{25132.7} = 0.02037\ \text{kNm}$$

Answer: $U_b = 1.222$ kNm ($\approx 1222$ J) and $U_t = 0.0204$ kNm ($\approx 20.4$ J).

The torsional energy is small compared to the bending energy.

Given data

$L = 4$ m, $d = 15$ cm $= 0.15$ m, $P = 10$ kN, $T = 5$ kNm, $E = 200$ kN/mm$^2 = 2\times10^8$ kN/m$^2$, $G = 80$ kN/mm$^2 = 8\times10^7$ kN/m$^2$.

Section properties

$$\begin{aligned} A &= \frac{\pi d^2}{4} = 0.017671\ \text{m}^2\\ I &= \frac{\pi d^4}{64} = 2.4850\times10^{-5}\ \text{m}^4\\ J &= \frac{\pi d^4}{32} = 4.9701\times10^{-5}\ \text{m}^4 \end{aligned}$$

Bending energy

$M = Px$, so

$$U_b = \frac{P^2L^3}{6EI} = \frac{10^2\times 4^3}{6\times2\times10^8\times2.4850\times10^{-5}} = \frac{6400}{29820.6} = 0.2146\ \text{kNm}$$

Shear energy

$V = P$ constant. Shape factor for a solid circle $\alpha = 10/9$:

$$U_s = \frac{\alpha P^2 L}{2GA} = \frac{(10/9)\times 100\times 4}{2\times 8\times10^{7}\times 0.017671} = \frac{444.44}{2827433} = 1.572\times10^{-4}\ \text{kNm}$$

Torsion energy

$$U_t = \frac{T^2L}{2GJ} = \frac{5^2\times4}{2\times8\times10^7\times4.9701\times10^{-5}} = \frac{100}{7952.2} = 0.01258\ \text{kNm}$$

Answer:

Shear energy is negligible compared with bending energy.

Given data

Bar: $d = 3$ cm, $L = 130$ cm, $E = 210$ GPa. Falling mass $m = 4$ kg ($W = mg = 39.24$ N). Spring: $k = 40$ kN/m $= 40\,000$ N/m, free length 10 cm. The mass falls 1.15 m measured from the top of the collar.

The spring (on the collar) and the bar act in series: the same force $P$ passes through both, so the mass first falls freely to the top of the spring, then compresses the spring while stretching the bar.

Free fall before touching the spring:

$$h' = 1.15 - 0.10 = 1.05\ \text{m}$$

Section properties and stiffnesses

$$A = \frac{\pi d^2}{4} = 706.86\ \text{mm}^2 = 7.0686e-04\ \text{m}^2$$ $$\frac{1}{k_{bar}} = \frac{L}{AE} = \frac{1.300}{7.0686e-04\times2.100e+11} = 8.758e-09\ \text{m/N},\qquad \frac{1}{k} = 2.500e-05\ \text{m/N}$$

Total flexibility of spring + bar: $c = 2.5009e-05$ m/N. If the maximum force is $P$, the total shortening/extension is $\delta = cP$.

Energy balance

Loss of potential energy = strain energy stored in spring and bar:

$$W\,(h' + \delta) = \tfrac12 P\delta,\qquad \delta = cP$$ $$W(h' + cP) = \tfrac12 cP^2 \;\Rightarrow\; P^2 - 2WP - \frac{2Wh'}{c} = 0$$ $$P = W + \sqrt{W^2 + \frac{2Wh'}{c}} = 39.24 + \sqrt{1539.8 + 3.2950e+06} = 1854.9\ \text{N}$$

Spring compression $= P/k = 46.37$ mm; bar elongation $= PL/AE = 0.01624$ mm (very small).

Stress in the bar

$$\sigma = \frac{P}{A} = \frac{1854.9}{706.86} = 2.624\ \text{N/mm}^2$$

Answer: maximum stress in the bar $\approx 2.62$ N/mm$^2$ (max. load 1.855 kN). The spring absorbs almost all the energy, which is why the stress is small. For comparison, the stress if the same weight were applied statically would be only 0.056 N/mm$^2$.

Given data

Bar: $d = 2$ cm, $L = 125$ cm, $E = 210$ GPa. Falling mass $m = 4$ kg ($W = mg = 39.24$ N). Spring: $k = 40$ kN/m $= 40\,000$ N/m, free length 10 cm. The mass falls 1.0 m measured from the top of the collar.

The spring (on the collar) and the bar act in series: the same force $P$ passes through both, so the mass first falls freely to the top of the spring, then compresses the spring while stretching the bar.

Free fall before touching the spring:

$$h' = 1.0 - 0.10 = 0.90\ \text{m}$$

Section properties and stiffnesses

$$A = \frac{\pi d^2}{4} = 314.16\ \text{mm}^2 = 3.1416e-04\ \text{m}^2$$ $$\frac{1}{k_{bar}} = \frac{L}{AE} = \frac{1.250}{3.1416e-04\times2.100e+11} = 1.895e-08\ \text{m/N},\qquad \frac{1}{k} = 2.500e-05\ \text{m/N}$$

Total flexibility of spring + bar: $c = 2.5019e-05$ m/N. If the maximum force is $P$, the total shortening/extension is $\delta = cP$.

Energy balance

Loss of potential energy = strain energy stored in spring and bar:

$$W\,(h' + \delta) = \tfrac12 P\delta,\qquad \delta = cP$$ $$W(h' + cP) = \tfrac12 cP^2 \;\Rightarrow\; P^2 - 2WP - \frac{2Wh'}{c} = 0$$ $$P = W + \sqrt{W^2 + \frac{2Wh'}{c}} = 39.24 + \sqrt{1539.8 + 2.8231e+06} = 1719.9\ \text{N}$$

Spring compression $= P/k = 43.00$ mm; bar elongation $= PL/AE = 0.03259$ mm (very small).

Stress in the bar

$$\sigma = \frac{P}{A} = \frac{1719.9}{314.16} = 5.475\ \text{N/mm}^2$$

Answer: maximum stress in the bar $\approx 5.47$ N/mm$^2$ (max. load 1.720 kN). The spring absorbs almost all the energy, which is why the stress is small. For comparison, the stress if the same weight were applied statically would be only 0.125 N/mm$^2$.

Strain energy is the energy stored in a deformed elastic body due to the work done by external loads in deforming it. It is recovered when the load is removed (within the elastic limit). For direct load, $U = \dfrac{\sigma^2}{2E}\times\text{Volume}$.

Gradually and suddenly applied direct loads

For sudden loading, work done $P\delta_{max}$ equals strain energy $\tfrac12 k\delta_{max}^2$ , so $\delta_{max} = 2P/k$: the stress and deflection are twice those for a gradually applied load.

Strain energy due to shear force in a beam

For an element of length $dx$ with shear force $V$, shear stress at distance $y$ from the neutral axis is $\tau = \dfrac{VA'\bar y}{Ib}$, and energy per unit volume is $\dfrac{\tau^2}{2G}$. Integrating over the area, for the rectangular section:

$$dU = \int \frac{\tau^2}{2G}\,dA\,dx,\qquad \tau = \frac{V}{2I}\left(\frac{d^2}{4}-y^2\right)$$ $$dU = \frac{V^2 b}{8GI^2}\cdot\frac{d^5}{30}\,dx = \frac{6}{5}\cdot\frac{V^2}{2GA}\,dx$$

For any section:

$$U_s = \int_0^L \frac{\alpha V^2}{2GA}\,dx$$

where $\alpha$ is the shear form factor ($\alpha = 1.2$ for a rectangle, $10/9$ for a solid circle).

Strain energy

Strain energy $U$ is the energy stored in an elastic body by the work done by external forces in deforming it. It equals the area under the load-displacement curve (up to the displacement):

$$U = \int_0^\Delta P\,d\Delta$$

For linear behaviour, $U = \tfrac12 P\Delta$.

Complementary strain energy

Complementary strain energy $U^*$ is the area between the load-displacement curve and the load axis:

$$U^* = \int_0^P \Delta\,dP$$

 P |        /
   |  U*  /|
   |    /  |      U + U* = P*Delta
   |  /  U |
   +-------+---- Delta

For a linear structure, $U^* = U = \tfrac12 P\Delta$; for a non-linear one they differ. Castigliano's theorems use $U$ (first) and $U^*$ (second).

Strain energy due to bending

An element $dx$ under moment $M$ rotates by $d\theta = \dfrac{M\,dx}{EI}$. The moment grows gradually, so

$$dU = \tfrac12 M\,d\theta = \frac{M^2\,dx}{2EI}$$ $$U_b = \int_0^L \frac{M^2}{2EI}\,dx$$

Using the stress method as a check: $\sigma = \dfrac{My}{I}$, $dU = \displaystyle\int\frac{\sigma^2}{2E}dA\,dx = \frac{M^2}{2EI^2}\Big(\int y^2dA\Big)dx = \frac{M^2dx}{2EI}$.

Example: cantilever with end load $P$: $M = Px$, so $U_b = \displaystyle\int_0^L \frac{P^2x^2}{2EI}dx = \frac{P^2L^3}{6EI}$.

Strain energy is the energy stored in an elastic body when it is deformed by external loads. It equals the work done by the loads and is recovered fully on unloading. It is the area under the load-displacement curve:

$$U = \int_0^\Delta P\,d\Delta \quad(= \tfrac12 P\Delta \text{ for a linear body})$$

Complementary strain energy is the area between the load-displacement curve and the load axis:

$$U^* = \int_0^P \Delta\,dP \quad(= \tfrac12 P\Delta \text{ for a linear body})$$

The sum $U + U^* = P\Delta$. For linearly elastic materials $U^* = U$; they differ only for non-linear materials.

Strain energy is the energy stored in an elastic body due to deformation produced by loads. It equals the work done by the gradually applied loads: $U = \tfrac12 P\Delta$. For a beam,

$$U = \int \frac{M^2}{2EI}\,dx + \int \frac{\alpha V^2}{2GA}\,dx$$

with the first term due to bending and the second due to shear.

Comparison of bending and shear deflections

Take a simply supported beam of span $L$, depth $d$ and rectangular section with a central load $P$. Here $M = Px/2$ (for $x \le L/2$) and $V = P/2$.

Bending energy:

$$U_b = \frac{P^2L^3}{96EI}\ \Rightarrow\ \delta_b = \frac{\partial U_b}{\partial P} = \frac{PL^3}{48EI}$$

Shear energy:

$$U_s = \frac{\alpha (P/2)^2 L}{2GA} = \frac{\alpha P^2L}{8GA}\ \Rightarrow\ \delta_s = \frac{\partial U_s}{\partial P} = \frac{\alpha PL}{4GA}$$

Ratio:

$$\frac{\delta_s}{\delta_b} = \frac{\alpha PL}{4GA}\times\frac{48EI}{PL^3} = 12\,\alpha\,\frac{E}{G}\,\frac{I}{AL^2}$$

For a rectangular section $I/A = d^2/12$, so

$$\frac{\delta_s}{\delta_b} = \alpha\,\frac{E}{G}\left(\frac{d}{L}\right)^2$$

Substituting $\alpha = 1.2$ and $E/G = 2.4$:

$$\frac{\delta_s}{\delta_b} = 1.2\times 2.4\times\left(\frac{d}{L}\right)^2 = 2.88\left(\frac{d}{L}\right)^2$$

For an ordinary beam, $L/d$ is 10 to 20, so

Hence the shear deflection is only about 1 to 3% of the bending deflection and can be neglected in an ordinary beam. It becomes important only for deep beams ($L/d < 5$ or so).

Given data

$b = 25$ cm, $d = 50$ cm, $L = 6$ m $= 600$ cm, $P = 100$ kN, $E = 2\times10^6$ kg/cm$^2$, $G = 0.85\times10^6$ kg/cm$^2$.

Assumption: $1\ \text{kgf} = 9.81$ N, so $P = 100\,000/9.81 = 10\,193.7$ kgf. Shear form factor for rectangle $\alpha = 1.2$.

$$I = \frac{25\times50^3}{12} = 2.604\times10^5\ \text{cm}^4,\qquad A = 25\times50 = 1250\ \text{cm}^2$$

Strain energy due to bending

For a central load, $M = Px/2$ for $x \le L/2$:

$$U_b = 2\int_0^{L/2}\frac{(Px/2)^2}{2EI}dx = \frac{P^2L^3}{96EI}$$ $$U_b = \frac{(10193.7)^2\times(600)^3}{96\times2\times10^6\times2.604\times10^5} = 448.9\ \text{kg-cm}$$

Strain energy due to shear

$V = P/2$ throughout the span:

$$U_s = \frac{\alpha V^2 L}{2GA} = \frac{1.2\times(5096.8)^2\times600}{2\times0.85\times10^6\times1250} = 8.80\ \text{kg-cm}$$

Answer:

• Bending: $U_b = 448.9$ kg-cm ($\approx 44.0$ N·m)
• Shear: $U_s = 8.80$ kg-cm ($\approx 0.863$ N·m)

Shear energy is only about 2% of the bending energy.

Assume both bars are linearly elastic, have the same modulus $E$, carry the same axial force along their length, and absorb the same strain energy $U$.

Bar 1 (uniform area $A$, length $L$)

Stress $\sigma_1 = P_1/A$ throughout.

$$U_1 = \frac{\sigma_1^2}{2E}\times AL = \frac{\sigma_1^2 AL}{2E}$$

Bar 2 (upper half $2A$, lower half $A$, each $0.5L$)

The same force $P_2$ acts in both halves. Maximum stress occurs in the smaller area (lower half):

$$\sigma_2 = \frac{P_2}{A}\quad(\text{lower half}),\qquad \frac{\sigma_2}{2} = \frac{P_2}{2A}\quad(\text{upper half})$$ $$\begin{aligned} U_2 &= \frac{\sigma_2^2}{2E}\,(A)(0.5L) + \frac{(\sigma_2/2)^2}{2E}\,(2A)(0.5L)\\ &= \frac{\sigma_2^2 AL}{4E} + \frac{\sigma_2^2 AL}{8E} = \frac{3\sigma_2^2AL}{8E} \end{aligned}$$

Equal energy

$$U_1 = U_2 \;\Rightarrow\; \frac{\sigma_1^2}{2} = \frac{3\sigma_2^2}{8} \;\Rightarrow\; \sigma_2^2 = \frac{4}{3}\sigma_1^2$$ $$\sigma_2 = \frac{2}{\sqrt3}\,\sigma_1 = 1.155\,\sigma_1$$

Answer: The maximum stress in bar 2 (in its lower, smaller-area half) is $1.155\,\sigma_1$, i.e. 15.5% higher than in bar 1; the upper half of bar 2 carries only $0.577\,\sigma_1$. For the same energy absorbed, the bar of uniform section (bar 1) is stressed less, so it is the better energy absorber. Reducing the area over part of the length increases stress and wastes material in the larger portion.

Given and assumptions

Beam ABC, $EI$ constant: overhang $AB = 1.5$ m, span $BC = 3$ m. Load $P = 20$ kN (down) at free end A, hinge at B, and C supported on a spring of stiffness $k = EI/9$ (kN/m, with $EI$ in kNm$^2$). Deflection of A is found by Castigliano's theorem, $\delta_A = \partial U/\partial P$, with the total strain energy of beam bending plus spring.

Reactions

Take moments about B:

$$R_C\times3 = P\times1.5 \;\Rightarrow\; R_C = 0.5P = 10\ \text{kN (downward, spring in tension)}$$ $$R_B = P + 0.5P = 1.5P = 30\ \text{kN (upward)}$$

Bending moments (x from A, y from C)

• $AB$: $M = Px$, $0 \le x \le 1.5$
• $BC$: $M = R_C\,y = 0.5Py$, $0 \le y \le 3$

Strain energy

$$U = \int_0^{1.5}\frac{(Px)^2}{2EI}dx + \int_0^{3}\frac{(0.5Py)^2}{2EI}dy + \frac{R_C^2}{2k}$$ $$U = \frac{P^2}{2EI}\left[\frac{1.5^3}{3} + 0.25\times\frac{3^3}{3}\right] + \frac{(0.5P)^2\times 9}{2EI} = \frac{P^2}{2EI}\left[1.125 + 2.25\right] + \frac{1.125P^2}{EI}$$ $$U = \frac{1.6875P^2 + 1.125P^2}{EI} = \frac{2.8125\,P^2}{EI}$$

Deflection at A

$$\delta_A = \frac{\partial U}{\partial P} = \frac{5.625\,P}{EI} = \frac{5.625\times20}{EI}$$

Parts: beam bending $= 67.5/EI$, spring (rotation of the beam about B due to the spring) $= 45/EI$.

Answer: $\delta_A = \dfrac{112.5}{EI}$ (kNm$^3$/EI, i.e. in metres when $EI$ is in kNm$^2$), downward.

Given data and assumptions

Frame fixed at joint 0, all members of solid circular section $d = 100$ mm, $E = 2\times10^5$ N/mm$^2$ ($= 2\times10^8$ kN/m$^2$). Taking the figure as a cantilever frame with horizontal member 0-1 (3 m), vertical member 1-2 (1 m) and horizontal member 2-3 (2 m), the 500 kN load $P$ acts downward at joint 3. Shear energy is neglected (small for slender members; $G$ is not given). Vertical displacement of joint 3 is found by Castigliano's theorem: $\delta_3 = \partial U/\partial P$.

$$I = \frac{\pi d^4}{64} = 4.9087e-06\ \text{m}^4,\quad EI = 981.7\ \text{kNm}^2,\quad A = 0.00785\ \text{m}^2,\quad AE = 1.5708e+06\ \text{kN}$$

Internal forces (s measured from joint 3)

Strain energy and displacement

$$U = \int\frac{M^2}{2EI}dx + \int\frac{N^2}{2AE}dx$$ $$\delta_3 = \frac{\partial U}{\partial P} = \frac{P}{EI}\left[\int_0^2 s^2ds + (2)^2(1) + \int_0^3 (2+s)^2 ds\right] + \frac{PL_{2-1}}{AE}$$ $$\int_0^2 s^2ds = 2.667,\qquad 2^2\times1 = 4,\qquad \int_0^3(2+s)^2ds = \frac{5^3-2^3}{3} = 39$$ $$\delta_{bending} = \frac{500\times(2.667+4+39)}{981.7} = \frac{500\times45.667}{981.7} = 23.258\ \text{m}$$ $$\delta_{axial} = \frac{500\times1}{1.5708e+06} = 0.318\ \text{mm}$$

Answer: vertical displacement of the loaded joint $\approx 23.26$ m downward (bending governs; axial part is only 0.32 mm). The very large value shows that the given load is extremely heavy for a 100 mm rod; the small-deflection theory is used as asked.