BE Civil Engineering (IOE, TU) Strength of Materials (IOE, CE 503) Question Paper 2077 Nepal

Cross-sectional areas

Steel rod ($d=30\ \text{mm}$):

$$A_s=\frac{\pi}{4}(30)^2 = 706.86\ \text{mm}^2$$

Copper tube ($d_o=60,\ d_i=40\ \text{mm}$):

$$A_c=\frac{\pi}{4}(60^2-40^2)=\frac{\pi}{4}(3600-1600)=\frac{\pi}{4}(2000)=1570.80\ \text{mm}^2$$

(a) Stresses under axial load

Compatibility (equal strain, rigid plates): $\varepsilon_s=\varepsilon_c$

$$\frac{\sigma_s}{E_s}=\frac{\sigma_c}{E_c}\ \Rightarrow\ \sigma_s=\sigma_c\frac{E_s}{E_c}=2\sigma_c$$

Equilibrium: $P=\sigma_s A_s+\sigma_c A_c$

$$90\times10^3 = 2\sigma_c(706.86)+\sigma_c(1570.80)=\sigma_c(1413.72+1570.80)=2984.52\,\sigma_c$$ $$\boxed{\sigma_c = 30.16\ \text{N/mm}^2\ \text{(compressive)}}$$ $$\boxed{\sigma_s = 2\sigma_c = 60.31\ \text{N/mm}^2\ \text{(compressive)}}$$

Overall shortening (using steel strain):

$$\delta=\frac{\sigma_s}{E_s}L=\frac{60.31}{200\times10^3}\times1500 = 0.452\ \text{mm}$$ $$\boxed{\delta \approx 0.45\ \text{mm (shortening)}}$$

(Check via copper: $\frac{30.16}{100\times10^3}\times1500=0.452\ \text{mm}$ — consistent.)

(b) Additional thermal stresses (ΔT = +50°C)

Copper has the higher $\alpha$, so on heating it tends to expand more than steel. Being tied together, copper is restrained (put in compression) and steel is dragged out (put in tension).

Let $\sigma_s'$ = tensile in steel, $\sigma_c'$ = compressive in copper.

Equilibrium of internal thermal forces (no external load):

$$\sigma_s' A_s = \sigma_c' A_c \ \Rightarrow\ \sigma_s'(706.86)=\sigma_c'(1570.80)\ \Rightarrow\ \sigma_s'=2.2222\,\sigma_c'$$

Compatibility (free differential expansion absorbed by the two strains):

$$\frac{\sigma_s'}{E_s}+\frac{\sigma_c'}{E_c}=(\alpha_c-\alpha_s)\Delta T$$ $$\frac{\sigma_s'}{200\times10^3}+\frac{\sigma_c'}{100\times10^3}=(18-12)\times10^{-6}\times50 = 3.0\times10^{-4}$$

Substitute $\sigma_s'=2.2222\,\sigma_c'$:

$$\frac{2.2222\,\sigma_c'}{200000}+\frac{\sigma_c'}{100000}=3.0\times10^{-4}$$ $$\sigma_c'(1.1111\times10^{-5}+1.0\times10^{-5})=3.0\times10^{-4}$$ $$\sigma_c'(2.1111\times10^{-5})=3.0\times10^{-4}\ \Rightarrow\ \sigma_c'=14.21\ \text{N/mm}^2$$ $$\boxed{\sigma_c'=14.21\ \text{N/mm}^2\ \text{(compressive, thermal)}}$$ $$\sigma_s'=2.2222\times14.21=31.58\ \text{N/mm}^2$$ $$\boxed{\sigma_s'=31.58\ \text{N/mm}^2\ \text{(tensile, thermal)}}$$

These add algebraically to the load stresses: steel net $=60.31-31.58=28.73$ N/mm² compressive; copper net $=30.16+14.21=44.37$ N/mm² compressive.

Geometry (measuring $x$ from A)

A: x=0, B: x=2 m, C: x=6 m, D: x=8 m. Supports at B and D. Loads: 20 kN ↓ at A; UDL 10 kN/m over B→C (x=2 to 6, total $10\times4=40$ kN acting at x=4 m); 30 kN ↓ at C (x=6).

Reactions

Take moments about B (CW +), span B→D = 6 m:

$$\sum M_B=0:\ -20(2) + 40(2) + 30(4) - R_D(6)=0$$

(20 kN at A is 2 m left of B; UDL resultant 40 kN at 2 m right of B; 30 kN at C is 4 m right of B; $R_D$ is 6 m right of B.)

$$-40 + 80 + 120 - 6R_D = 0 \Rightarrow 6R_D=160 \Rightarrow R_D=26.67\ \text{kN}$$ $$\sum F_y=0:\ R_B+R_D = 20+40+30 = 90 \Rightarrow R_B = 90-26.67 = 63.33\ \text{kN}$$

Shear Force (kN), left to right

• Just right of A: $V=-20$
• Just left of B: $-20$
• Just right of B: $-20+63.33 = +43.33$
• Over BC, UDL reduces V by 10 kN/m. At C (just left): $43.33 - 10(4) = +3.33$
• Just right of C (after 30 kN): $3.33 - 30 = -26.67$
• Over CD (no load): constant $-26.67$ up to just left of D
• Just right of D: $-26.67 + 26.67 = 0$ ✓ (closes)

SF changes sign within BC (from +43.33 to +3.33 it stays positive; then jumps to −26.67 at C). So the SF crosses zero at C (drops from +3.33 to −26.67). Max BM occurs at C.

Bending Moments (kN·m), sagging +

• At A: 0
• At B: $M_B = -20(2) = -40$ (hogging)
• At C ($x=6$): take left side: $-20(6) + 63.33(4) - 10\times4\times2$

$$M_C = -120 + 253.33 - 80 = +53.33\ \text{kN·m (sagging)}$$

• At D: 0 (free... support end, moment = 0 since end)

(Check from right at C: $M_C = R_D\times2 = 26.67\times2 = 53.33$ ✓)

Max sagging BM $= +53.33$ kN·m at C; max hogging $= -40$ kN·m at B.

Point of contraflexure (M = 0) in span BC

For $2\le x\le6$, with origin at A:

$$M(x) = -20x + 63.33(x-2) - \frac{10(x-2)^2}{2}$$

Set $M=0$:

$$-20x + 63.33x -126.67 -5(x-2)^2 = 0$$ $$43.33x -126.67 -5(x^2-4x+4)=0$$ $$-5x^2 +63.33x -146.67 = 0 \Rightarrow 5x^2 -63.33x +146.67=0$$ $$x=\frac{63.33\pm\sqrt{63.33^2-4(5)(146.67)}}{10}=\frac{63.33\pm\sqrt{4010.7-2933.4}}{10}=\frac{63.33\pm32.82}{10}$$ $$x=3.05\ \text{m or } 9.6\ \text{m}.$$

Only $x=3.05$ m lies in BC. Point of contraflexure at $x\approx3.05$ m from A (≈1.05 m right of B).

Diagram summary

SFD (kN):  -20 ──── [B] jump to +43.33 ─\(slope -10)─ +3.33 [C] drop to -26.67 ──── 0 [D]
BMD (kN·m): 0 [A] ↓ -40 [B] rising, =0 at x≈3.05, peak +53.33 [C] ↓ 0 [D]

Results: $R_B=63.33$ kN, $R_D=26.67$ kN; $M_{max,sag}=+53.33$ kN·m at C; $M_{max,hog}=-40$ kN·m at B; contraflexure at 3.05 m from A.

(a) Derivation of the flexure formula

Assumptions of simple bending theory:

1. Material is homogeneous, isotropic and obeys Hooke's law.
2. The beam is initially straight and bends in a circular arc.
3. Plane sections normal to the axis remain plane after bending.
4. Young's modulus is the same in tension and compression.
5. Transverse sections have an axis of symmetry in the plane of bending; loads act in that plane (pure bending).
6. Radius of curvature $R$ is large compared with section dimensions.

Strain–curvature: Consider a fibre at distance $y$ from the neutral axis (NA). Original length $=R\,d\theta$; after bending its length $=(R+y)d\theta$.

$$\varepsilon=\frac{(R+y)d\theta - R\,d\theta}{R\,d\theta}=\frac{y}{R}$$

Stress: By Hooke's law $\sigma=E\varepsilon=\dfrac{E}{R}y$, hence $\dfrac{\sigma}{y}=\dfrac{E}{R}.$ (i)

Moment of resistance: Force on element $dA$ is $\sigma\,dA=\dfrac{E}{R}y\,dA$. Its moment about NA $=\dfrac{E}{R}y^2 dA$. Total internal moment:

$$M=\frac{E}{R}\int y^2\,dA=\frac{E}{R}I \ \Rightarrow\ \frac{M}{I}=\frac{E}{R}.$$

(ii)

Combining (i) and (ii):

$$\boxed{\dfrac{M}{I}=\dfrac{\sigma}{y}=\dfrac{E}{R}}$$

(The net axial force is zero, which forces the NA to pass through the centroid.)

(b) Section design

Maximum bending moment for a simply supported beam with UDL:

$$M_{max}=\frac{wL^2}{8}=\frac{6\times4^2}{8}=12\ \text{kN·m}=12\times10^6\ \text{N·mm}$$

Let width $=b$, depth $d=2b$. Section modulus:

$$Z=\frac{bd^2}{6}=\frac{b(2b)^2}{6}=\frac{4b^3}{6}=\frac{2b^3}{3}$$

Required $Z=\dfrac{M_{max}}{\sigma_{perm}}=\dfrac{12\times10^6}{8}=1.5\times10^6\ \text{mm}^3$

$$\frac{2b^3}{3}=1.5\times10^6 \Rightarrow b^3=2.25\times10^6 \Rightarrow b=130.8\ \text{mm}$$ $$d=2b=261.5\ \text{mm}$$

Adopt practical sizes:

$$\boxed{b \approx 131\ \text{mm},\quad d \approx 262\ \text{mm}}$$

(Round up to e.g. $b=135\ \text{mm}$, $d=270\ \text{mm}$ for safety.)

Given

$\sigma_x=80,\ \sigma_y=-40,\ \tau_{xy}=30$ (N/mm²).

(a) Analytical principal stresses

Mean (centre) stress:

$$\sigma_{avg}=\frac{\sigma_x+\sigma_y}{2}=\frac{80+(-40)}{2}=20\ \text{N/mm}^2$$

Radius term:

$$R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}=\sqrt{\left(\frac{80-(-40)}{2}\right)^2+30^2}=\sqrt{60^2+30^2}=\sqrt{3600+900}=\sqrt{4500}=67.08$$

Principal stresses:

$$\sigma_1=\sigma_{avg}+R=20+67.08=87.08\ \text{N/mm}^2$$ $$\sigma_2=\sigma_{avg}-R=20-67.08=-47.08\ \text{N/mm}^2$$ $$\boxed{\sigma_1=87.08\ \text{N/mm}^2\ (T),\quad \sigma_2=-47.08\ \text{N/mm}^2\ (C)}$$

Maximum in-plane shear:

$$\tau_{max}=R=67.08\ \text{N/mm}^2$$ $$\boxed{\tau_{max}=67.08\ \text{N/mm}^2}$$

Orientation of principal planes:

$$\tan 2\theta_p=\frac{2\tau_{xy}}{\sigma_x-\sigma_y}=\frac{2(30)}{80-(-40)}=\frac{60}{120}=0.5$$ $$2\theta_p=26.57^{\circ}\Rightarrow \theta_p=13.28^{\circ}\ \text{(to plane of }\sigma_1\text{; the other principal plane at }103.28^{\circ}).$$

Plane of $\tau_{max}$ is at $\theta_p+45^{\circ}=58.28^{\circ}$.

(b) Mohr's circle verification

• Plot point $X=(\sigma_x,\,-\tau_{xy})=(80,\,-30)$ and $Y=(\sigma_y,\,+\tau_{xy})=(-40,\,30)$ (using the convention that gives consistent rotation).
• Centre $C$ on the $\sigma$-axis $=\left(\dfrac{\sigma_x+\sigma_y}{2},0\right)=(20,0)$.
• Radius $=$ distance $CX=\sqrt{(80-20)^2+30^2}=\sqrt{3600+900}=67.08$.
• Principal stresses are where the circle cuts the $\sigma$-axis: $\sigma=20\pm67.08$, i.e. $87.08$ and $-47.08$ N/mm² — matching part (a).
• The top of the circle gives $\tau_{max}=R=67.08$ N/mm² at the centre stress $20$.
• The angle $\angle XCA_1$ on the circle $=2\theta_p=26.57^{\circ}$; on the physical element this halves to $13.28^{\circ}$, agreeing with the analytical orientation.

        τ
        |     o (top, τmax=67.08 at σ=20)
   Y(-40,30)
  ------C(20,0)------------------ σ
   σ2=-47.08      σ1=87.08
        X(80,-30)

Verification complete — both methods give $\sigma_1=87.08$, $\sigma_2=-47.08$ N/mm².

(a) Torsion equation derivation

Assumptions:

1. Material is homogeneous, isotropic and linearly elastic (Hooke's law in shear).
2. The shaft is circular and remains circular after twisting.
3. Plane cross-sections remain plane and do not warp.
4. Radii remain straight after twisting (no distortion of cross-section).
5. Twist is uniform along the length; stress proportional to distance from axis.

Shear strain–twist: A line on the surface at radius $r$ rotates by angle $\gamma$ (shear strain) over length $L$, while the far end rotates through angle $\theta$. Arc compatibility: $r\theta = L\gamma$, so

$$\gamma=\frac{r\theta}{L}.$$

By Hooke's law in shear, $\tau=G\gamma=\dfrac{G\theta}{L}r$, hence

$$\frac{\tau}{r}=\frac{G\theta}{L}.$$

(i)

Torque–stress: Shear force on ring $dA$ at radius $r$ is $\tau\,dA=\dfrac{G\theta}{L}r\,dA$; its moment about the axis is $\dfrac{G\theta}{L}r^2\,dA$. Total torque:

$$T=\frac{G\theta}{L}\int r^2\,dA=\frac{G\theta}{L}J\ \Rightarrow\ \frac{T}{J}=\frac{G\theta}{L}.$$

(ii)

Combining (i) & (ii):

$$\boxed{\dfrac{T}{J}=\dfrac{\tau}{r}=\dfrac{G\theta}{L}}$$

(b) Shaft diameter

Torque from power:

$$T=\frac{60P}{2\pi N}=\frac{60\times120\times10^3}{2\pi\times200}=\frac{7.2\times10^6}{1256.6}=5729.6\ \text{N·m}=5.7296\times10^6\ \text{N·mm}$$

(i) Strength criterion ($\tau\le60$ N/mm²), using $T=\dfrac{\pi}{16}\tau d^3$:

$$d^3=\frac{16T}{\pi\tau}=\frac{16\times5.7296\times10^6}{\pi\times60}=\frac{9.167\times10^7}{188.50}=486{,}400\ \text{mm}^3$$ $$d=78.65\ \text{mm}$$

(ii) Stiffness criterion ($\theta\le1^{\circ}=0.017453$ rad), using $\theta=\dfrac{TL}{GJ}$ with $J=\dfrac{\pi}{32}d^4$:

$$J_{req}=\frac{TL}{G\theta}=\frac{5.7296\times10^6\times2000}{80\times10^3\times0.017453}=\frac{1.14592\times10^{10}}{1396.3}=8.207\times10^{6}\ \text{mm}^4$$ $$d^4=\frac{32 J_{req}}{\pi}=\frac{32\times8.207\times10^6}{\pi}=8.357\times10^{7}\ \text{mm}^4$$ $$d=(8.357\times10^7)^{1/4}=95.6\ \text{mm}$$

Governing diameter = larger of the two (stiffness governs):

$$\boxed{d \approx 96\ \text{mm (adopt }100\ \text{mm)}}$$

Given

$A=50\times50=2500\ \text{mm}^2$, $L=250\ \text{mm}$, $P=200\ \text{kN}=200000\ \text{N}$, $\delta L=0.125\ \text{mm}$, lateral contraction $=0.00625$ mm on the 50 mm side.

(a) Young's modulus

$$\sigma=\frac{P}{A}=\frac{200000}{2500}=80\ \text{N/mm}^2$$ $$\varepsilon_{long}=\frac{\delta L}{L}=\frac{0.125}{250}=5\times10^{-4}$$ $$E=\frac{\sigma}{\varepsilon_{long}}=\frac{80}{5\times10^{-4}}=1.6\times10^{5}\ \text{N/mm}^2$$ $$\boxed{E=160\ \text{GPa}}$$

(b) Poisson's ratio

$$\varepsilon_{lat}=\frac{0.00625}{50}=1.25\times10^{-4}$$ $$\nu=\frac{\varepsilon_{lat}}{\varepsilon_{long}}=\frac{1.25\times10^{-4}}{5\times10^{-4}}=0.25$$ $$\boxed{\nu=0.25}$$

(c) Bulk modulus

$$K=\frac{E}{3(1-2\nu)}=\frac{160}{3(1-0.5)}=\frac{160}{1.5}=106.67\ \text{GPa}$$ $$\boxed{K=106.67\ \text{GPa}}$$

(d) Modulus of rigidity

$$G=\frac{E}{2(1+\nu)}=\frac{160}{2(1.25)}=\frac{160}{2.5}=64\ \text{GPa}$$ $$\boxed{G=64\ \text{GPa}}$$

Section properties (symmetric I, NA at mid-depth)

Overall depth $=200$ mm, so NA is $100$ mm from top/bottom.

Moment of inertia about NA:

$$I=\frac{B D^3}{12}-\frac{(B-t)\,d^3}{12}=\frac{150\times200^3}{12}-\frac{(150-10)\times160^3}{12}$$ $$=\frac{150\times8\times10^6}{12}-\frac{140\times4.096\times10^6}{12}=100\times10^6 - 47.79\times10^6 = 52.21\times10^6\ \text{mm}^4$$

Shear formula: $\tau=\dfrac{V\,A\bar{y}}{I\,b}$, with $V=80\times10^3$ N.

Shear stress at NA (max), $b=t=10$ mm

First moment $Q$ of area above NA about NA:

• Flange ($150\times20$), centroid at $100-10=90$ mm: $A\bar y = 150\times20\times90 = 270000\ \text{mm}^3$
• Half web ($10\times80$), centroid at $40$ mm: $A\bar y = 10\times80\times40 = 32000\ \text{mm}^3$

$$Q_{NA}=270000+32000=302000\ \text{mm}^3$$ $$\tau_{max}=\frac{80\times10^3\times302000}{52.21\times10^6\times10}=\frac{2.416\times10^{10}}{5.221\times10^{8}}=46.3\ \text{N/mm}^2$$ $$\boxed{\tau_{max}=46.3\ \text{N/mm}^2 \text{ (at NA)}}$$

At flange–web junction

$Q$ = flange only $=270000\ \text{mm}^3$.

In the web ($b=10$ mm):

$$\tau_{web}=\frac{80\times10^3\times270000}{52.21\times10^6\times10}=\frac{2.16\times10^{10}}{5.221\times10^8}=41.4\ \text{N/mm}^2$$

In the flange ($b=150$ mm) at the same level:

$$\tau_{flange}=\frac{80\times10^3\times270000}{52.21\times10^6\times150}=2.76\ \text{N/mm}^2$$ $$\boxed{\tau_{junction}=41.4\ \text{N/mm}^2 \text{ (web side)},\ 2.76\ \text{N/mm}^2 \text{ (flange side)}}$$

Distribution sketch

  top  0
 flange  small parabola → 2.76 (flange side at junction)
  ┌── jump (b: 150→10) ──┐
  web   41.4 → rises to 46.3 at NA → 41.4
  └── jump (b: 10→150) ──┘
 flange  2.76 → 0 at bottom

The distribution is parabolic in each part with abrupt jumps at the flange–web junctions due to the sudden change in width; maximum at NA = 46.3 N/mm².

Given

$d=1200\ \text{mm}$, $t=12\ \text{mm}$, $p=2\ \text{N/mm}^2$, $E=200000\ \text{N/mm}^2$, $\nu=0.3$.

(a) Hoop (circumferential) stress

$$\sigma_h=\frac{p d}{2t}=\frac{2\times1200}{2\times12}=\frac{2400}{24}=100\ \text{N/mm}^2$$ $$\boxed{\sigma_h=100\ \text{N/mm}^2}$$

(b) Longitudinal stress

$$\sigma_l=\frac{p d}{4t}=\frac{2\times1200}{4\times12}=\frac{2400}{48}=50\ \text{N/mm}^2$$ $$\boxed{\sigma_l=50\ \text{N/mm}^2}$$

(c) Change in diameter

Circumferential (hoop) strain:

$$\varepsilon_h=\frac{1}{E}(\sigma_h-\nu\sigma_l)=\frac{1}{200000}(100-0.3\times50)=\frac{100-15}{200000}=\frac{85}{200000}=4.25\times10^{-4}$$ $$\delta d=\varepsilon_h\,d=4.25\times10^{-4}\times1200=0.51\ \text{mm}$$ $$\boxed{\delta d=0.51\ \text{mm (increase)}}$$

(d) Volumetric strain

Longitudinal strain:

$$\varepsilon_l=\frac{1}{E}(\sigma_l-\nu\sigma_h)=\frac{1}{200000}(50-0.3\times100)=\frac{50-30}{200000}=\frac{20}{200000}=1.0\times10^{-4}$$ $$\varepsilon_v=2\varepsilon_h+\varepsilon_l=2(4.25\times10^{-4})+1.0\times10^{-4}=8.5\times10^{-4}+1.0\times10^{-4}=9.5\times10^{-4}$$ $$\boxed{\varepsilon_v=9.5\times10^{-4}=0.00095}$$

Section properties

$$A=\frac{\pi}{4}(200^2-160^2)=\frac{\pi}{4}(40000-25600)=\frac{\pi}{4}(14400)=11309.7\ \text{mm}^2$$ $$I=\frac{\pi}{64}(200^4-160^4)=\frac{\pi}{64}(1.6\times10^9-6.5536\times10^8)=\frac{\pi}{64}(9.4464\times10^8)=4.6378\times10^{7}\ \text{mm}^4$$

Radius of gyration:

$$k=\sqrt{I/A}=\sqrt{\frac{4.6378\times10^7}{11309.7}}=\sqrt{4101.2}=64.04\ \text{mm}$$

Both ends fixed: effective length $L_e=\dfrac{L}{2}=\dfrac{4000}{2}=2000\ \text{mm}$.

(a) Euler's crippling load

$$P_E=\frac{\pi^2 E I}{L_e^2}=\frac{\pi^2\times100\times10^3\times4.6378\times10^7}{2000^2}=\frac{9.8696\times4.6378\times10^{12}}{4\times10^6}$$ $$=\frac{4.5773\times10^{13}}{4\times10^6}=1.1443\times10^{7}\ \text{N}$$ $$\boxed{P_E\approx11{,}443\ \text{kN}\approx11.44\ \text{MN}}$$

(b) Rankine's load

$$P_R=\frac{\sigma_c\,A}{1+a\left(\dfrac{L_e}{k}\right)^2}$$

Slenderness ratio: $\dfrac{L_e}{k}=\dfrac{2000}{64.04}=31.23$, so $\left(\dfrac{L_e}{k}\right)^2=975.5$.

$$\text{Denominator}=1+\frac{1}{1600}\times975.5=1+0.6097=1.6097$$ $$\text{Numerator}=550\times11309.7=6.2203\times10^{6}\ \text{N}$$ $$P_R=\frac{6.2203\times10^6}{1.6097}=3.865\times10^{6}\ \text{N}$$ $$\boxed{P_R\approx3865\ \text{kN}}$$

Comparison

Euler gives $\approx11{,}443$ kN whereas Rankine gives $\approx3865$ kN. Euler over-predicts because for this relatively stocky column (slenderness only $\approx31$) the failure is dominated by crushing, not pure elastic buckling. Rankine's load (≈3865 kN) is the realistic/safe estimate.

Given

$d=25\ \text{mm}$, $L=2000\ \text{mm}$, $W=500\ \text{N}$, $h=80\ \text{mm}$, $E=200000\ \text{N/mm}^2$.

$$A=\frac{\pi}{4}(25)^2=490.87\ \text{mm}^2$$

(a) Impact stress

For a falling weight, the instantaneous stress $\sigma$ satisfies the energy balance $W(h+\delta)=\tfrac{1}{2}\sigma\varepsilon\,(AL)$, giving the standard result:

$$\sigma=\frac{W}{A}\left[1+\sqrt{1+\frac{2hAE}{WL}}\right]$$

Static stress $\dfrac{W}{A}=\dfrac{500}{490.87}=1.0186\ \text{N/mm}^2$.

Inner term:

$$\frac{2hAE}{WL}=\frac{2\times80\times490.87\times200000}{500\times2000}=\frac{1.5708\times10^{10}}{1.0\times10^{6}}=15707.96$$ $$\sqrt{1+15707.96}=\sqrt{15708.96}=125.34$$ $$\sigma=1.0186\,(1+125.34)=1.0186\times126.34=128.7\ \text{N/mm}^2$$ $$\boxed{\sigma\approx128.7\ \text{N/mm}^2}$$

(b) Instantaneous elongation

$$\delta=\frac{\sigma L}{E}=\frac{128.7\times2000}{200000}=1.287\ \text{mm}$$ $$\boxed{\delta\approx1.29\ \text{mm}}$$

(c) Strain energy stored

$$U=\frac{\sigma^2}{2E}\times(\text{volume})=\frac{\sigma^2}{2E}\,A L$$ $$U=\frac{128.7^2}{2\times200000}\times490.87\times2000=\frac{16563.7}{400000}\times9.8174\times10^{5}$$ $$=0.041409\times9.8174\times10^5=40655\ \text{N·mm}\approx40.7\ \text{J}$$ $$\boxed{U\approx40.7\ \text{J}}$$

(Check by work-energy: $W(h+\delta)=500\,(80+1.287)=500\times81.287=40643\ \text{N·mm}$ ≈ U ✓.)

Given

$P=40\ \text{kN}=40000\ \text{N}$, $E=200000\ \text{N/mm}^2$. Same axial force passes through all segments.

Areas

$$A_{AB}=\frac{\pi}{4}(20)^2=314.16\ \text{mm}^2$$ $$A_{BC}=\frac{\pi}{4}(30)^2=706.86\ \text{mm}^2$$ $$A_{CD}=\frac{\pi}{4}(25)^2=490.87\ \text{mm}^2$$

Stresses ($\sigma=P/A$)

$$\sigma_{AB}=\frac{40000}{314.16}=127.32\ \text{N/mm}^2$$ $$\sigma_{BC}=\frac{40000}{706.86}=56.59\ \text{N/mm}^2$$ $$\sigma_{CD}=\frac{40000}{490.87}=81.49\ \text{N/mm}^2$$ $$\boxed{\sigma_{AB}=127.3,\ \sigma_{BC}=56.6,\ \sigma_{CD}=81.5\ \text{N/mm}^2}$$

Total elongation ($\delta=\sum \dfrac{PL}{AE}$)

$$\delta_{AB}=\frac{40000\times400}{314.16\times200000}=\frac{1.6\times10^{7}}{6.2832\times10^{7}}=0.2546\ \text{mm}$$ $$\delta_{BC}=\frac{40000\times500}{706.86\times200000}=\frac{2.0\times10^{7}}{1.41372\times10^{8}}=0.1415\ \text{mm}$$ $$\delta_{CD}=\frac{40000\times600}{490.87\times200000}=\frac{2.4\times10^{7}}{9.8174\times10^{7}}=0.2445\ \text{mm}$$ $$\delta_{total}=0.2546+0.1415+0.2445=0.6406\ \text{mm}$$ $$\boxed{\delta_{total}\approx0.64\ \text{mm}}$$

Alternatively each $\delta_i=\dfrac{\sigma_i L_i}{E}$ gives the same values (e.g. $\dfrac{127.32\times400}{200000}=0.2546$ mm ✓).