BE Civil Engineering (IOE, TU) Strength of Materials (IOE, CE 503) Question Paper 2076 Nepal

(a) Load sharing under external load $P = 120\ \text{kN}$

Compatibility: all rods have the same length $L$ and equal elongation $\delta$, so equal strain $\varepsilon = \delta/L$.

$$\varepsilon = \frac{\sigma_s}{E_s} = \frac{\sigma_c}{E_c} \implies \sigma_s = \sigma_c\,\frac{E_s}{E_c} = 2\sigma_c$$

Equilibrium: two steel rods + one copper rod carry $P$.

$$P = 2\sigma_s A_s + \sigma_c A_c$$

Substitute $\sigma_s = 2\sigma_c$, $A_s = 300\ \text{mm}^2$, $A_c = 600\ \text{mm}^2$:

$$120\times10^3 = 2(2\sigma_c)(300) + \sigma_c(600) = 1200\,\sigma_c + 600\,\sigma_c = 1800\,\sigma_c$$ $$\sigma_c = \frac{120\times10^3}{1800} = 66.67\ \text{N/mm}^2 = 66.67\ \text{MPa}$$ $$\sigma_s = 2\sigma_c = 133.33\ \text{MPa}$$

Loads:

• Copper rod: $F_c = \sigma_c A_c = 66.67\times600 = 40{,}000\ \text{N} = \mathbf{40\ kN}$
• Each steel rod: $F_s = \sigma_s A_s = 133.33\times300 = 40{,}000\ \text{N} = \mathbf{40\ kN}$

Check: $40 + 2(40) = 120\ \text{kN}$ ✓

Answers: $\boxed{\sigma_c = 66.67\ \text{MPa},\ \sigma_s = 133.33\ \text{MPa};\ F_c = 40\ \text{kN},\ F_{s,\text{each}} = 40\ \text{kN}}$

(b) Thermal stress on heating ($\Delta T = 40\ ^{\circ}\text{C}$, no external load)

Free expansions differ; the rigid bar forces equal final elongation. Copper wants to expand more than steel ($\alpha_c > \alpha_s$). The bar restrains copper (compression in copper) and drags steel (tension in steel).

Equilibrium (no external load): internal forces self-balance.

$$2\sigma_s A_s + \sigma_c A_c = 0$$

Using tension positive in steel and compression in copper, let copper force be $\sigma_c$ (negative = compression):

$$2\sigma_s A_s = -\sigma_c A_c \implies 2\sigma_s(300) = -\sigma_c(600) \implies \sigma_s = -\sigma_c$$

So $|\sigma_s| = |\sigma_c|$ (magnitudes equal); steel in tension, copper in compression.

Compatibility (equal final elongation per unit length):

$$\alpha_s \Delta T + \frac{\sigma_s}{E_s} = \alpha_c \Delta T - \frac{\sigma_c'}{E_c}$$

where $\sigma_s$ is tensile in steel and $\sigma_c' = |\sigma_c|$ is the compressive magnitude in copper. Rearranging:

$$\frac{\sigma_s}{E_s} + \frac{\sigma_c'}{E_c} = (\alpha_c - \alpha_s)\Delta T$$

With $\sigma_s = \sigma_c'$ (from equilibrium magnitudes):

$$\sigma_s\left(\frac{1}{E_s} + \frac{1}{E_c}\right) = (\alpha_c - \alpha_s)\Delta T$$ $$\sigma_s\left(\frac{1}{200{,}000} + \frac{1}{100{,}000}\right) = (18-12)\times10^{-6}\times40$$ $$\sigma_s\left(5\times10^{-6} + 10\times10^{-6}\right) = 6\times10^{-6}\times40 = 240\times10^{-6}$$ $$\sigma_s\,(15\times10^{-6}) = 240\times10^{-6}$$ $$\sigma_s = \frac{240}{15} = 16\ \text{N/mm}^2 = 16\ \text{MPa}$$

Thermal stresses:

• Steel: $\boxed{16\ \text{MPa tensile}}$
• Copper: $\boxed{16\ \text{MPa compressive}}$

Force check: steel $2(16)(300)=9600\ \text{N}$ tension; copper $16(600)=9600\ \text{N}$ compression — balanced ✓.

Geometry (measure $x$ from $A$)

A at 0, B at 1.5 m, C at 6.5 m, D at 8.0 m.

Loads: $20\ \text{kN}\downarrow$ at A; UDL $10\ \text{kN/m}$ over BC (total $10\times5 = 50\ \text{kN}$ acting at midspan, $x = 4.0\ \text{m}$); $30\ \text{kN}\downarrow$ at D.

Reactions

Take moments about B ($x=1.5$). Distances from B: A is $-1.5$, UDL resultant at $4.0-1.5=2.5$, C at $5.0$, D at $6.5$.

$$\sum M_B = 0:\quad R_C(5) = 20(-1.5)\cdot(-1)\ldots$$

Work directly:

$$R_C \times 5 - 50\times2.5 - 30\times6.5 + 20\times1.5 = 0$$

(20 kN at A is on the opposite side of B, giving a CCW moment $+20\times1.5$.)

$$5R_C = 50\times2.5 + 30\times6.5 - 20\times1.5 = 125 + 195 - 30 = 290$$ $$R_C = 58\ \text{kN}$$ $$\sum F_y = 0:\quad R_B + R_C = 20 + 50 + 30 = 100 \implies R_B = 100 - 58 = 42\ \text{kN}$$

$R_B = 42\ \text{kN}$, $R_C = 58\ \text{kN}$.

Shear force (just to the right of each point), N taken upward positive

• At A: $V=0$, drop of 20 → just right of A, $V=-20\ \text{kN}$.
• A→B: no load, constant $V=-20$. Just left of B: $-20$.
• At B: $+R_B=+42$ → just right of B: $-20+42 = +22\ \text{kN}$.
• B→C: UDL 10 kN/m over 5 m reduces V linearly: at C (left) $V = 22 - 10\times5 = 22-50 = -28\ \text{kN}$.
• At C: $+R_C=+58$ → just right of C: $-28+58 = +30\ \text{kN}$.
• C→D: no load, constant $+30$. At D: $+30$, then $-30$ point load → $0$. ✓ closes.

SFD salient values (kN): $0,\ -20\ (\text{A}^+),\ -20\ (\text{B}^-),\ +22\ (\text{B}^+),\ -28\ (\text{C}^-),\ +30\ (\text{C}^+),\ 0\ (\text{D}^+).$

Zero shear in span BC (max sagging moment)

From B, $V(x') = 22 - 10x' = 0 \implies x' = 2.2\ \text{m}$ from B (i.e. $x = 3.7\ \text{m}$ from A).

Bending moments

• At A: $M_A = 0$.
• At B: $M_B = -20\times1.5 = -30\ \text{kN·m}$ (hogging).
• Max in span, $2.2\ \text{m}$ right of B:

$$M = -20(1.5+2.2) + R_B(2.2) - 10\frac{(2.2)^2}{2}$$ $$= -20(3.7) + 42(2.2) - 10(2.42) = -74 + 92.4 - 24.2 = -5.8\ \text{kN·m}$$

Alternatively from B reference: $M = M_B + 22(2.2) - 10\frac{2.2^2}{2} = -30 + 48.4 - 24.2 = -5.8\ \text{kN·m}$. So the "maximum" (zero-shear) moment is $\mathbf{-5.8\ kN\cdot m}$ — still hogging; the heavy overhang at D dominates.

• At C: $M_C = -30\times1.5 = -45\ \text{kN·m}$ (hogging, from the D-side cantilever).
• At D: $M_D = 0$.

BMD salient values (kN·m): $M_A=0,\ M_B=-30,\ M_{x=3.7}=-5.8,\ M_C=-45,\ M_D=0.$

The whole span BC is in hogging here (all values negative). The local extremum at zero shear ($-5.8\ \text{kN·m}$) is the least-hogging (peak) point of the parabola in BC; the largest hogging moment is at C, $\mathbf{-45\ kN\cdot m}$.

Points of contraflexure

The moment stays negative throughout BC (it never crosses zero between B and C since the peak $-5.8$ is still negative). Therefore there are no points of contraflexure in this beam — bending is hogging over the supports and zero only at the free ends A and D.

ASCII sketch

Load:  20kN        UDL 10 kN/m              30kN
        |    ^B   |||||||||||||   ^C        |
        A----B---------------------C--------D
 SFD: -20 |  +22\__________ -28 | +30 ___ 0
         (jump at B)        (jump at C)
 BMD:  0   -30  ...peak -5.8...  -45    0

Section properties

Rectangle $b=150\ \text{mm}$, $d=300\ \text{mm}$.

$$I = \frac{bd^3}{12} = \frac{150\times300^3}{12} = \frac{150\times2.7\times10^7}{12} = 3.375\times10^8\ \text{mm}^4$$ $$Z = \frac{I}{d/2} = \frac{bd^2}{6} = \frac{150\times300^2}{6} = \frac{150\times90{,}000}{6} = 2.25\times10^6\ \text{mm}^3$$

(a) Maximum UDL from bending

For a simply supported beam with UDL over span $L=4\ \text{m}$:

$$M_{max} = \frac{wL^2}{8}$$

Limiting: $M_{max} = \sigma_b Z = 8\ \text{N/mm}^2 \times 2.25\times10^6\ \text{mm}^3 = 18\times10^6\ \text{N·mm} = 18\ \text{kN·m}$

$$\frac{w(4)^2}{8} = 18 \implies 2w = 18 \implies w = 9\ \text{kN/m}$$

Maximum safe UDL (bending) $= \boxed{9\ \text{kN/m}}$.

(b) Maximum shear stress at this load

Max shear force at support: $V_{max} = \dfrac{wL}{2} = \dfrac{9\times4}{2} = 18\ \text{kN} = 18{,}000\ \text{N}$.

For a rectangle, max shear stress (at neutral axis):

$$\tau_{max} = \frac{3}{2}\cdot\frac{V}{A},\quad A = b d = 150\times300 = 45{,}000\ \text{mm}^2$$ $$\tau_{max} = 1.5\times\frac{18{,}000}{45{,}000} = 1.5\times0.4 = 0.6\ \text{N/mm}^2 = 0.6\ \text{MPa}$$

Since $0.6\ \text{MPa} < 1.0\ \text{MPa}$ permissible, the section is safe in shear. Bending governs the design.

(c) Shear stress distribution

For a rectangular section the transverse shear stress varies parabolically: zero at top and bottom fibres, maximum at the neutral axis.

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

 top    τ=0
        \
         \___ parabola
  N.A. -- τ_max = 0.6 MPa
         /
        /
bottom  τ=0

Average shear stress $\tau_{avg} = V/A = 18{,}000/45{,}000 = 0.4\ \text{MPa}$.

$$\frac{\tau_{max}}{\tau_{avg}} = \frac{0.6}{0.4} = 1.5 = \frac{3}{2}$$

Ratio $\tau_{max}/\tau_{avg} = 1.5$ for a rectangular section.

Given

$\sigma_x = 80\ \text{MPa},\ \sigma_y = -40\ \text{MPa},\ \tau_{xy} = 30\ \text{MPa}.$

(a) Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

Mean: $\dfrac{80+(-40)}{2} = \dfrac{40}{2} = 20\ \text{MPa}$.

Half-difference: $\dfrac{80-(-40)}{2} = \dfrac{120}{2} = 60\ \text{MPa}$.

Radius term:

$$R = \sqrt{60^2 + 30^2} = \sqrt{3600 + 900} = \sqrt{4500} = 67.08\ \text{MPa}$$ $$\sigma_1 = 20 + 67.08 = 87.08\ \text{MPa (tensile)}$$ $$\sigma_2 = 20 - 67.08 = -47.08\ \text{MPa (compressive)}$$

Principal stresses: $\boxed{\sigma_1 = 87.08\ \text{MPa},\ \sigma_2 = -47.08\ \text{MPa}}$.

Orientation of principal planes:

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y} = \frac{2\times30}{80-(-40)} = \frac{60}{120} = 0.5$$ $$2\theta_p = 26.57^{\circ} \implies \theta_p = 13.28^{\circ}$$

The second principal plane is at $13.28^{\circ} + 90^{\circ} = 103.28^{\circ}$.

(b) Maximum in-plane shear stress

$$\tau_{max} = R = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2} = 67.08\ \text{MPa}$$

Also $\tau_{max} = \dfrac{\sigma_1-\sigma_2}{2} = \dfrac{87.08-(-47.08)}{2} = \dfrac{134.16}{2} = 67.08\ \text{MPa}$ ✓.

It acts on planes at $45^{\circ}$ to the principal planes:

$$\theta_s = \theta_p + 45^{\circ} = 13.28 + 45 = 58.28^{\circ}$$

On these planes the normal stress equals the mean $= 20\ \text{MPa}$.

$\boxed{\tau_{max} = 67.08\ \text{MPa at}\ 58.28^{\circ}}$.

(c) Mohr's circle check

• Centre: $C = \left(\dfrac{\sigma_x+\sigma_y}{2},\ 0\right) = (20,\ 0)\ \text{MPa}$.
• Radius: $R = 67.08\ \text{MPa}$.
• Principal stresses = $C \pm R = 20 \pm 67.08 = 87.08$ and $-47.08\ \text{MPa}$ ✓.
• Top of circle gives $\tau_{max} = R = 67.08\ \text{MPa}$ ✓.

        τ
        |    *  (top: τmax=67.08)
   ------C(20,0)------ σ
  σ2=-47.08    σ1=87.08

All analytical values are confirmed by Mohr's circle.

(a) Torque transmitted

$$P = \frac{2\pi N T}{60} \implies T = \frac{60 P}{2\pi N} = \frac{60\times150{,}000}{2\pi\times200}$$ $$T = \frac{9{,}000{,}000}{1256.64} = 7161.97\ \text{N·m} \approx 7162\ \text{N·m}$$

$T \approx 7.162\ \text{kN·m} = 7.162\times10^6\ \text{N·mm}$.

(b) Diameter required

Strength criterion (shear stress)

For a solid shaft: $\tau = \dfrac{16T}{\pi d^3}$.

$$d^3 = \frac{16T}{\pi\tau} = \frac{16\times7.162\times10^6}{\pi\times60} = \frac{1.14592\times10^8}{188.50} = 6.0792\times10^5\ \text{mm}^3$$ $$d = (6.0792\times10^5)^{1/3} = 84.75\ \text{mm}$$

So $d_{strength} \approx 84.8\ \text{mm}$.

Stiffness criterion (angle of twist)

$$\theta = \frac{T L}{G J},\quad J = \frac{\pi d^4}{32}$$

Allowable $\theta = 1^{\circ} = \dfrac{\pi}{180} = 0.017453\ \text{rad}$, $L = 2000\ \text{mm}$, $G = 80{,}000\ \text{N/mm}^2$.

$$J = \frac{T L}{G\theta} = \frac{7.162\times10^6 \times 2000}{80{,}000\times0.017453} = \frac{1.4324\times10^{10}}{1396.26} = 1.02588\times10^7\ \text{mm}^4$$ $$d^4 = \frac{32 J}{\pi} = \frac{32\times1.02588\times10^7}{\pi} = \frac{3.28282\times10^8}{3.14159} = 1.04494\times10^8\ \text{mm}^4$$ $$d = (1.04494\times10^8)^{1/4} = 101.1\ \text{mm}$$

So $d_{stiffness} \approx 101.1\ \text{mm}$.

Governing diameter

$$d = \max(84.8,\ 101.1) = 101.1\ \text{mm}$$

The stiffness (angle-of-twist) criterion governs; minimum diameter $\boxed{d \approx 102\ \text{mm}}$ (round up to a convenient size).

Check at $d=102\ \text{mm}$: $\tau = 16T/(\pi d^3) = 16\times7.162\times10^6/(\pi\times102^3) = 1.14592\times10^8/3.334\times10^6 = 34.4\ \text{MPa} < 60$ ✓; twist $\le 1^{\circ}$ ✓.

Stress and strains

Area: $A = \dfrac{\pi}{4}d^2 = \dfrac{\pi}{4}(40)^2 = 1256.64\ \text{mm}^2$.

Axial stress: $\sigma = \dfrac{P}{A} = \dfrac{90{,}000}{1256.64} = 71.62\ \text{N/mm}^2$.

Longitudinal strain: $\varepsilon_L = \dfrac{\delta L}{L} = \dfrac{1.43}{2500} = 5.72\times10^{-4}$.

Lateral strain: $\varepsilon_d = \dfrac{\delta d}{d} = \dfrac{0.0091}{40} = 2.275\times10^{-4}$.

Young's modulus

$$E = \frac{\sigma}{\varepsilon_L} = \frac{71.62}{5.72\times10^{-4}} = 1.2521\times10^5\ \text{N/mm}^2 \approx \boxed{125.2\ \text{GPa}}$$

Poisson's ratio

$$\nu = \frac{\varepsilon_d}{\varepsilon_L} = \frac{2.275\times10^{-4}}{5.72\times10^{-4}} = 0.3977 \approx \boxed{0.40}$$

Bulk modulus

$$K = \frac{E}{3(1-2\nu)} = \frac{125.2}{3(1-0.80)} = \frac{125.2}{3(0.20)} = \frac{125.2}{0.60} = \boxed{208.7\ \text{GPa}}$$

Modulus of rigidity

$$G = \frac{E}{2(1+\nu)} = \frac{125.2}{2(1.40)} = \frac{125.2}{2.80} = \boxed{44.7\ \text{GPa}}$$

Summary: $E \approx 125.2\ \text{GPa}$, $\nu \approx 0.40$, $K \approx 208.7\ \text{GPa}$, $G \approx 44.7\ \text{GPa}$.

Given

$d = 1200\ \text{mm}$, $t = 10\ \text{mm}$, $p = 2.0\ \text{N/mm}^2$, $E = 2\times10^5\ \text{N/mm}^2$, $\nu = 0.3$.

(a) Stresses

Hoop (circumferential) stress:

$$\sigma_h = \frac{pd}{2t} = \frac{2.0\times1200}{2\times10} = \frac{2400}{20} = 120\ \text{N/mm}^2 = \boxed{120\ \text{MPa}}$$

Longitudinal stress:

$$\sigma_l = \frac{pd}{4t} = \frac{2.0\times1200}{4\times10} = \frac{2400}{40} = 60\ \text{N/mm}^2 = \boxed{60\ \text{MPa}}$$

(b) Hoop strain and change in diameter

Hoop strain:

$$\varepsilon_h = \frac{1}{E}\left(\sigma_h - \nu\,\sigma_l\right) = \frac{1}{2\times10^5}\left(120 - 0.3\times60\right)$$ $$= \frac{120 - 18}{2\times10^5} = \frac{102}{2\times10^5} = 5.1\times10^{-4}$$

Hoop strain $\varepsilon_h = 5.1\times10^{-4}$.

Change in diameter $= \varepsilon_h \times d$ (since circumferential strain equals diametral strain):

$$\delta d = 5.1\times10^{-4}\times1200 = 0.612\ \text{mm}$$

$\boxed{\delta d = 0.612\ \text{mm (increase)},\ \varepsilon_h = 5.1\times10^{-4}}$.

Section properties

$D = 200\ \text{mm}$, $d = 160\ \text{mm}$.

$$A = \frac{\pi}{4}(D^2 - d^2) = \frac{\pi}{4}(200^2 - 160^2) = \frac{\pi}{4}(40{,}000 - 25{,}600) = \frac{\pi}{4}(14{,}400) = 11{,}309.7\ \text{mm}^2$$ $$I = \frac{\pi}{64}(D^4 - d^4) = \frac{\pi}{64}(200^4 - 160^4) = \frac{\pi}{64}(1.6\times10^9 - 6.5536\times10^8)$$ $$= \frac{\pi}{64}(9.4464\times10^8) = 4.6373\times10^7\ \text{mm}^4$$

Radius of gyration: $k = \sqrt{I/A} = \sqrt{4.6373\times10^7/11{,}309.7} = \sqrt{4100.3} = 64.03\ \text{mm}$.

(a) Euler load (both ends fixed)

Effective length $L_e = L/2 = 4000/2 = 2000\ \text{mm}$.

$$P_E = \frac{\pi^2 E I}{L_e^2} = \frac{\pi^2\times100{,}000\times4.6373\times10^7}{2000^2}$$ $$= \frac{9.8696\times100{,}000\times4.6373\times10^7}{4\times10^6} = \frac{4.5768\times10^{13}}{4\times10^6} = 1.1442\times10^7\ \text{N}$$

$P_E \approx 11{,}442\ \text{kN} \approx \boxed{11.44\ \text{MN}}$.

(b) Rankine load (both ends fixed)

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

Slenderness: $\dfrac{L_e}{k} = \dfrac{2000}{64.03} = 31.24$, so $\left(\dfrac{L_e}{k}\right)^2 = 975.9$.

$$a\left(\frac{L_e}{k}\right)^2 = \frac{975.9}{1600} = 0.6099$$ $$P_R = \frac{550\times11{,}309.7}{1 + 0.6099} = \frac{6{,}220{,}335}{1.6099} = 3.8638\times10^6\ \text{N}$$

$P_R \approx 3864\ \text{kN} \approx \boxed{3.86\ \text{MN}}$.

The Rankine load is much lower than Euler, reflecting the stocky (low-slenderness) column where crushing-type failure dominates; Rankine governs the practical design.

Energy balance (impact / suddenly applied falling load)

Let $\sigma$ = instantaneous (impact) stress, $L=3000\ \text{mm}$, $A=500\ \text{mm}^2$, $E=2\times10^5\ \text{N/mm}^2$, $W=2000\ \text{N}$, $h=120\ \text{mm}$.

Work done by falling weight $=$ strain energy stored in rod:

$$W(h + \delta) = \frac{\sigma^2}{2E}\,(A L),\qquad \delta = \frac{\sigma L}{E}$$

This gives the impact-stress quadratic:

$$\sigma = \frac{W}{A}\left[1 + \sqrt{1 + \frac{2 h A E}{W L}}\right]$$

Substitute values

Static stress: $\dfrac{W}{A} = \dfrac{2000}{500} = 4\ \text{N/mm}^2$.

Inside the root:

$$\frac{2 h A E}{W L} = \frac{2\times120\times500\times2\times10^5}{2000\times3000} = \frac{2.4\times10^{10}}{6\times10^6} = 4000$$ $$\sqrt{1 + 4000} = \sqrt{4001} = 63.25$$ $$\sigma = 4\left[1 + 63.25\right] = 4\times64.25 = 257.0\ \text{N/mm}^2$$

(a) Maximum impact stress $\boxed{\sigma \approx 257\ \text{MPa}}$.

(b) Instantaneous elongation

$$\delta = \frac{\sigma L}{E} = \frac{257.0\times3000}{2\times10^5} = \frac{771{,}000}{200{,}000} = 3.855\ \text{mm}$$

(b) Instantaneous elongation $\boxed{\delta \approx 3.86\ \text{mm}}$.

Check via energy: $W(h+\delta) = 2000(120+3.855) = 247{,}710\ \text{N·mm}$; strain energy $= \dfrac{\sigma^2 AL}{2E} = \dfrac{257^2\times500\times3000}{2\times2\times10^5} = \dfrac{9.907\times10^{10}}{4\times10^5} = 247{,}680\ \text{N·mm}$ ✓ (agree to rounding).

Maximum bending moment

For a cantilever with an end point load, the maximum moment is at the fixed (built-in) end:

$$M_{max} = W\,L = 12\ \text{kN}\times2.5\ \text{m} = 30\ \text{kN·m} = 30\times10^6\ \text{N·mm}$$

Maximum bending stress

Extreme-fibre distance: $y_{max} = \dfrac{200}{2} = 100\ \text{mm}$.

$$\sigma_{max} = \frac{M_{max}\,y_{max}}{I} = \frac{30\times10^6\times100}{4.0\times10^7}$$ $$= \frac{3.0\times10^9}{4.0\times10^7} = 75\ \text{N/mm}^2$$

$\boxed{\sigma_{max} = 75\ \text{MPa}}$, occurring at the extreme top and bottom fibres at the fixed end (tension on the top fibre, compression on the bottom fibre for a downward end load).

Equivalently, section modulus $Z = I/y_{max} = 4.0\times10^7/100 = 4.0\times10^5\ \text{mm}^3$, and $\sigma_{max} = M/Z = 30\times10^6/4.0\times10^5 = 75\ \text{MPa}$ ✓.

Areas

$$A_1 = \frac{\pi}{4}(25)^2 = 490.87\ \text{mm}^2$$ $$A_2 = \frac{\pi}{4}(40)^2 = 1256.64\ \text{mm}^2$$ $$A_3 = \frac{\pi}{4}(20)^2 = 314.16\ \text{mm}^2$$

Force $P = 50{,}000\ \text{N}$ throughout (series).

Stress in each segment

$$\sigma_1 = \frac{P}{A_1} = \frac{50{,}000}{490.87} = 101.86\ \text{N/mm}^2 = \mathbf{101.9\ MPa}$$ $$\sigma_2 = \frac{P}{A_2} = \frac{50{,}000}{1256.64} = 39.79\ \text{N/mm}^2 = \mathbf{39.8\ MPa}$$ $$\sigma_3 = \frac{P}{A_3} = \frac{50{,}000}{314.16} = 159.15\ \text{N/mm}^2 = \mathbf{159.2\ MPa}$$

Total elongation

$$\delta = \sum \frac{P L_i}{A_i E} = \frac{P}{E}\sum \frac{L_i}{A_i}$$ $$\frac{L_1}{A_1} = \frac{400}{490.87} = 0.8149$$ $$\frac{L_2}{A_2} = \frac{500}{1256.64} = 0.3979$$ $$\frac{L_3}{A_3} = \frac{300}{314.16} = 0.9549$$ $$\sum \frac{L_i}{A_i} = 0.8149 + 0.3979 + 0.9549 = 2.1677\ \text{mm}^{-1}\cdot\text{mm} = 2.1677\ \text{(1/mm)}$$ $$\delta = \frac{50{,}000}{200{,}000}\times2.1677 = 0.25\times2.1677 = 0.5419\ \text{mm}$$

Total elongation $\boxed{\delta \approx 0.542\ \text{mm}}$.

Individual extensions (check): $\delta_1 = 0.25\times0.8149 = 0.2037$, $\delta_2 = 0.25\times0.3979 = 0.0995$, $\delta_3 = 0.25\times0.9549 = 0.2387\ \text{mm}$; sum $= 0.5419\ \text{mm}$ ✓.