BE Civil Engineering (IOE, TU) Strength of Materials (IOE, CE 503) Question Paper 2078 Nepal

Given: $A_s = 600\ \text{mm}^2$, $A_c = 1000\ \text{mm}^2$, $L = 400\ \text{mm}$, $E_s = 200\ \text{GPa}$, $E_c = 100\ \text{GPa}$.

Part (a): Axial load 90 kN

For a composite bar with ends rigidly connected, both materials share the same strain (compatibility) and the loads add up (equilibrium).

Compatibility (equal strain):

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

Equilibrium:

$$P = \sigma_s A_s + \sigma_c A_c$$ $$90{,}000 = (2\sigma_c)(600) + \sigma_c(1000) = 1200\sigma_c + 1000\sigma_c = 2200\,\sigma_c$$ $$\sigma_c = \frac{90{,}000}{2200} = 40.91\ \text{N/mm}^2$$ $$\sigma_s = 2\sigma_c = 81.82\ \text{N/mm}^2$$

Both are compressive.

Shortening:

$$\delta = \varepsilon_s L = \frac{\sigma_s}{E_s}L = \frac{81.82}{200{,}000}\times 400 = 0.1636\ \text{mm}$$

• Steel stress = 81.82 N/mm² (compressive)
• Copper stress = 40.91 N/mm² (compressive)
• Shortening = 0.164 mm

Check: $\sigma_s A_s + \sigma_c A_c = 81.82(600)+40.91(1000)=49{,}092+40{,}910 \approx 90{,}000\ \text{N}$ ✓

Part (b): Temperature rise of 50 °C (no external load)

Copper has the higher $\alpha$, so it tends to expand more; the steel restrains it. The result: copper is put in compression, steel in tension, with the internal forces self-balancing.

Equilibrium of internal forces (no external load):

$$\sigma_s A_s = \sigma_c A_c \quad\Rightarrow\quad \sigma_s (600) = \sigma_c (1000) \Rightarrow \sigma_s = 1.6667\,\sigma_c$$

Compatibility (free expansions differ; final lengths equal):

$$\alpha_c\,\Delta T - \frac{\sigma_c}{E_c} = \alpha_s\,\Delta T + \frac{\sigma_s}{E_s}$$

(copper expansion reduced by its compressive strain equals steel expansion increased by its tensile strain)

$$(\alpha_c-\alpha_s)\Delta T = \frac{\sigma_s}{E_s} + \frac{\sigma_c}{E_c}$$ $$(18-12)\times10^{-6}\times 50 = \frac{1.6667\,\sigma_c}{200{,}000} + \frac{\sigma_c}{100{,}000}$$ $$300\times10^{-6} = \sigma_c\left(8.333\times10^{-6} + 10\times10^{-6}\right) = \sigma_c(18.333\times10^{-6})$$ $$\sigma_c = \frac{300\times10^{-6}}{18.333\times10^{-6}} = 16.36\ \text{N/mm}^2\ \text{(compressive)}$$ $$\sigma_s = 1.6667\times16.36 = 27.27\ \text{N/mm}^2\ \text{(tensile)}$$

• Steel stress (thermal) = 27.27 N/mm² (tensile)
• Copper stress (thermal) = 16.36 N/mm² (compressive)

Check equilibrium: $27.27\times600 = 16{,}362\ \text{N}$ and $16.36\times1000 = 16{,}360\ \text{N}$ ✓

Layout (distance from $A$): $A=0$, $B=2\ \text{m}$, $C=7\ \text{m}$, $D=8\ \text{m}$.

Loads: $20\ \text{kN}\downarrow$ at $A$; UDL $10\ \text{kN/m}$ over $BC$ (total $= 10\times5 = 50\ \text{kN}$ acting at midspan, $x=4.5\ \text{m}$); $15\ \text{kN}\downarrow$ at $D$.

Reactions

Take moments about $B$ ($x=2$). Let $R_B, R_C$ upward.

$$\sum M_B = 0:\; -20(2) + 50(2.5) + 15(6) - R_C(5) = 0$$ $$-40 + 125 + 90 = 5R_C \Rightarrow 175 = 5R_C \Rightarrow R_C = 35\ \text{kN}$$ $$\sum F_y=0:\; R_B + R_C = 20 + 50 + 15 = 85 \Rightarrow R_B = 50\ \text{kN}$$

Shear Force (from left, downward loads negative)

• Just right of $A$: $V = -20\ \text{kN}$ (constant to $B$).
• Just left of $B$: $V = -20$.
• Just right of $B$: $V = -20 + 50 = +30\ \text{kN}$.
• Across $BC$ shear drops linearly by UDL: at $C^-$: $V = 30 - 10(5) = 30-50 = -20\ \text{kN}$.
• Just right of $C$: $V = -20 + 35 = +15\ \text{kN}$.
• At $D^-$: $V = +15$; the $15\ \text{kN}$ load brings it to $0$. ✓

Zero shear in span: measuring $x'$ from $B$, $V = 30 - 10x' = 0 \Rightarrow x' = 3\ \text{m}$ (i.e. $x = 5\ \text{m}$ from $A$). Max span BM here.

Bending Moments (sagging +)

• $M_A = 0$.
• $M_B = -20(2) = -40\ \text{kN·m}$ (hogging).
• $M_D = 0$; $M_C = -15(1) = -15\ \text{kN·m}$ (hogging) (from right side).
• Max span moment at $x'=3\ \text{m}$ from $B$:

$$M = R_B(3) - 20(2+3) - 10\cdot\frac{3^2}{2}$$ $$M = 50(3) - 20(5) - 45 = 150 - 100 - 45 = +5\ \text{kN·m (sagging)}$$

Points of contraflexure (M = 0 in span $BC$)

With $x'$ from $B$: $M(x') = 50x' - 20(2+x') - 5x'^2 = -5x'^2 + 30x' - 40$. Set $=0$: $5x'^2 - 30x' + 40 = 0 \Rightarrow x'^2 - 6x' + 8 = 0 \Rightarrow (x'-2)(x'-4)=0$.

$$x' = 2\ \text{m and } 4\ \text{m from } B \;(x = 4\ \text{m and } 6\ \text{m from } A)$$

Diagrams (schematic)

SFD (kN):
  A      B          (zero @3m)        C        D
 -20 --- -20 | +30 \____ 0 ____ -20 | +15 \ 0
            (jump +50)              (jump +35)

BMD (kN·m):
   0 .... -40 (B, hog) /  +5 (x'=3) \  -15 (C, hog) .. 0
        contraflex @ x'=2 and x'=4

Summary: $R_B = 50\ \text{kN}$, $R_C = 35\ \text{kN}$; max sagging $= +5\ \text{kN·m}$ at $5\ \text{m}$ from $A$; max hogging $= -40\ \text{kN·m}$ at $B$; contraflexure at $4\ \text{m}$ and $6\ \text{m}$ from $A$.

Maximum bending moment and shear force:

$$M_{max} = \frac{WL}{4} = \frac{48\times4}{4} = 48\ \text{kN·m}; \qquad V_{max} = \frac{W}{2} = 24\ \text{kN}$$

Section properties (about neutral axis = centroid, by symmetry at mid-depth)

Overall depth $D = 240\ \text{mm}$, $y_{max} = 120\ \text{mm}$.

$$I = \underbrace{\frac{10\times200^3}{12}}_{\text{web}} + 2\Big[\underbrace{\frac{150\times20^3}{12}}_{\text{flange own}} + \underbrace{150\times20\times110^2}_{\text{transfer}}\Big]$$

Web: $\dfrac{10\times200^3}{12} = \dfrac{10\times8\times10^6}{12} = 6.667\times10^6\ \text{mm}^4$

Flange own: $\dfrac{150\times8000}{12} = 0.1\times10^6\ \text{mm}^4$ each.

Flange transfer: distance from NA to flange centroid $= 100 + 10 = 110\ \text{mm}$; $150\times20\times110^2 = 3000\times12{,}100 = 36.3\times10^6\ \text{mm}^4$ each.

$$I = 6.667\times10^6 + 2(0.1\times10^6 + 36.3\times10^6) = 6.667\times10^6 + 72.8\times10^6 = 79.467\times10^6\ \text{mm}^4$$

(a) Maximum bending stress

$$\sigma_{max} = \frac{M\,y_{max}}{I} = \frac{48\times10^6 \times 120}{79.467\times10^6} = \frac{5.76\times10^9}{79.467\times10^6} = 72.5\ \text{N/mm}^2$$

Maximum bending stress = 72.5 N/mm² (tension at bottom, compression at top).

(b) Shear stress: $\tau = \dfrac{V\,A\bar{y}}{I\,b}$

Maximum shear at neutral axis (web width $b = 10\ \text{mm}$). First moment of area above NA:

• Flange: $A\bar y = (150\times20)\times110 = 3000\times110 = 330{,}000\ \text{mm}^3$
• Half web (above NA): $(100\times10)\times50 = 1000\times50 = 50{,}000\ \text{mm}^3$
• Total $Q_{NA} = 380{,}000\ \text{mm}^3$

$$\tau_{max} = \frac{24\times10^3 \times 380{,}000}{79.467\times10^6 \times 10} = \frac{9.12\times10^9}{794.67\times10^6} = 11.48\ \text{N/mm}^2$$

At flange–web junction, $Q =$ flange only $= 330{,}000\ \text{mm}^3$.

• Using web width $b = 10\ \text{mm}$:

$$\tau = \frac{24\times10^3 \times 330{,}000}{79.467\times10^6 \times 10} = \frac{7.92\times10^9}{794.67\times10^6} = 9.97\ \text{N/mm}^2$$

• Using flange width $b = 150\ \text{mm}$ (just inside flange):

$$\tau = \frac{24\times10^3 \times 330{,}000}{79.467\times10^6 \times 150} = 0.66\ \text{N/mm}^2$$

So there is a sudden jump at the junction from $0.66$ (flange side) to $9.97\ \text{N/mm}^2$ (web side).

Shear-stress distribution (sketch)

 top flange:  small parabola, 0 -> 0.66 N/mm^2
 junction:    JUMP 0.66 -> 9.97 N/mm^2
 web:         parabola 9.97 -> 11.48 (NA, max) -> 9.97
 junction:    JUMP 9.97 -> 0.66
 bottom flg:  0.66 -> 0

Results: $\tau_{max} = 11.48\ \text{N/mm}^2$ at NA; $\tau$ at junction $= 9.97\ \text{N/mm}^2$ (web) and $0.66\ \text{N/mm}^2$ (flange).

Given: $\sigma_x = 80$, $\sigma_y = -40$, $\tau_{xy} = 30\ \text{N/mm}^2$.

(a) Analytical solution

Principal stresses:

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$ $$\frac{\sigma_x+\sigma_y}{2} = \frac{80-40}{2} = 20\ \text{N/mm}^2$$ $$\frac{\sigma_x-\sigma_y}{2} = \frac{80-(-40)}{2} = 60\ \text{N/mm}^2$$ $$R = \sqrt{60^2 + 30^2} = \sqrt{3600 + 900} = \sqrt{4500} = 67.08\ \text{N/mm}^2$$ $$\sigma_1 = 20 + 67.08 = 87.08\ \text{N/mm}^2 \;(\text{tensile})$$ $$\sigma_2 = 20 - 67.08 = -47.08\ \text{N/mm}^2 \;(\text{compressive})$$

Orientation of principal planes:

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(30)}{80-(-40)} = \frac{60}{120} = 0.5$$ $$2\theta_p = 26.57^{\circ} \Rightarrow \theta_p = 13.28^{\circ} \;(\text{and } 103.28^{\circ})$$

The plane of $\sigma_1$ is at $\theta_p = 13.28^{\circ}$ (measured anticlockwise from the plane on which $\sigma_x$ acts).

Maximum shear stress:

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

occurring on planes at $45^{\circ}$ to the principal planes, i.e. at $\theta = 13.28^{\circ}+45^{\circ} = 58.28^{\circ}$.

(b) Mohr's circle construction

1. Plot horizontal axis $\sigma$, vertical axis $\tau$ (to scale, e.g. $1\ \text{cm} = 10\ \text{N/mm}^2$).
2. Point $X = (\sigma_x, +\tau_{xy}) = (80, 30)$; point $Y = (\sigma_y, -\tau_{xy}) = (-40, -30)$.
3. Join $XY$; it crosses the $\sigma$-axis at the centre $C = \left(\frac{80-40}{2}, 0\right) = (20, 0)$.
4. Radius $R =$ distance $CX = \sqrt{(80-20)^2 + 30^2} = \sqrt{3600+900} = 67.08\ \text{N/mm}^2$.
5. Circle cuts $\sigma$-axis at $\sigma_1 = 20+67.08 = 87.08$ and $\sigma_2 = 20-67.08 = -47.08$.
6. Top of circle gives $\tau_{max} = R = 67.08\ \text{N/mm}^2$.
7. Angle $XCB$ (to $\sigma_1$) $= \arctan\frac{30}{60} = 26.57^{\circ} = 2\theta_p$, so $\theta_p = 13.28^{\circ}$ on the element.

        tau
         |        X(80,30)
         |       /
  -------C(20,0)------- sigma
  s2=-47 | \ /        s1=87
         | Y(-40,-30)
   tau_max = 67.08 at top of circle

Values read from Mohr's circle agree with analysis:

• $\sigma_1 = 87.08\ \text{N/mm}^2$, $\sigma_2 = -47.08\ \text{N/mm}^2$
• $\tau_{max} = 67.08\ \text{N/mm}^2$, $\theta_p = 13.28^{\circ}$

Torque to be transmitted

$$T = \frac{60\,P}{2\pi N} = \frac{60\times150\times10^3}{2\pi\times200} = \frac{9{,}000{,}000}{1256.64} = 7161\ \text{N·m} = 7.161\times10^6\ \text{N·mm}$$

(a) Solid shaft

Strength criterion ($\tau \le 80$): $T = \dfrac{\pi}{16}\tau d^3$

$$d^3 = \frac{16T}{\pi\tau} = \frac{16\times7.161\times10^6}{\pi\times80} = \frac{114.58\times10^6}{251.33} = 455{,}900\ \text{mm}^3$$ $$d = 76.95\ \text{mm}$$

Stiffness criterion ($\theta = 1^{\circ} = 0.01745\ \text{rad}$ over $L = 3000\ \text{mm}$): $\dfrac{T}{J} = \dfrac{G\theta}{L}$, with $J = \dfrac{\pi}{32}d^4$.

$$J = \frac{T L}{G\theta} = \frac{7.161\times10^6 \times 3000}{80{,}000\times0.01745} = \frac{21.483\times10^9}{1396} = 15.39\times10^6\ \text{mm}^4$$ $$d^4 = \frac{32 J}{\pi} = \frac{32\times15.39\times10^6}{\pi} = 156.7\times10^6 \Rightarrow d = 111.9\ \text{mm}$$

The stiffness criterion governs (larger). Adopt $d \approx 112\ \text{mm}$.

(b) Hollow shaft (same torque, same $\tau = 80$ strength design)

Let external $= D$, internal $d_i = 0.6D$, so $k = 0.6$.

$$T = \frac{\pi}{16}\tau\,\frac{D^4 - d_i^4}{D} = \frac{\pi}{16}\tau D^3 (1-k^4)$$ $$1-k^4 = 1 - 0.6^4 = 1 - 0.1296 = 0.8704$$ $$D^3 = \frac{16T}{\pi\tau(1-k^4)} = \frac{455{,}900}{0.8704} = 523{,}780\ \text{mm}^3$$ $$D = 80.6\ \text{mm}, \quad d_i = 0.6\times80.6 = 48.4\ \text{mm}$$

(Comparison for weight saving is made on the strength-designed shafts of equal $\tau$.)

Weight (∝ cross-sectional area, same material and length):

• Solid (strength): $A_s = \dfrac{\pi}{4}d^2 = \dfrac{\pi}{4}(76.95)^2 = 4650\ \text{mm}^2$
• Hollow: $A_h = \dfrac{\pi}{4}(D^2 - d_i^2) = \dfrac{\pi}{4}(80.6^2 - 48.4^2) = \dfrac{\pi}{4}(6496 - 2343) = \dfrac{\pi}{4}(4153) = 3262\ \text{mm}^2$

$$\%\text{ saving} = \frac{A_s - A_h}{A_s}\times100 = \frac{4650 - 3262}{4650}\times100 = 29.8\%$$

Results: Solid shaft diameter governed by stiffness $\approx \mathbf{112\ mm}$; hollow shaft (strength design) $D = 80.6\ \text{mm}$, $d_i = 48.4\ \text{mm}$, giving a weight saving of about 30% over the strength-designed solid shaft.

The axial force in every segment $= P = 40\ \text{kN} = 40{,}000\ \text{N}$ (series bar).

Stresses ($\sigma = P/A$)

$$\sigma_{AB} = \frac{40{,}000}{400} = 100\ \text{N/mm}^2$$ $$\sigma_{BC} = \frac{40{,}000}{800} = 50\ \text{N/mm}^2$$ $$\sigma_{CD} = \frac{40{,}000}{200} = 200\ \text{N/mm}^2$$

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

$$\delta = \frac{P}{E}\left(\frac{L_{AB}}{A_{AB}}+\frac{L_{BC}}{A_{BC}}+\frac{L_{CD}}{A_{CD}}\right)$$ $$= \frac{40{,}000}{200{,}000}\left(\frac{500}{400}+\frac{500}{800}+\frac{500}{200}\right)$$ $$= 0.2\,(1.25 + 0.625 + 2.5) = 0.2\times 4.375 = 0.875\ \text{mm}$$

Results: $\sigma_{AB}=100$, $\sigma_{BC}=50$, $\sigma_{CD}=200\ \text{N/mm}^2$; total elongation $= 0.875\ \text{mm}$.

(a) Relation $E = 2G(1+\mu)$ (outline)

Consider an element under pure shear stress $\tau$. Pure shear is equivalent to direct stresses $+\tau$ (tensile) and $-\tau$ (compressive) acting on planes at $45^{\circ}$.

The linear strain along the $45^{\circ}$ tensile diagonal due to these two direct stresses:

$$\varepsilon_{45} = \frac{\tau}{E} - \mu\frac{(-\tau)}{E} = \frac{\tau}{E}(1+\mu)$$

The shear strain $\phi = \tau/G$ produces a diagonal strain of $\phi/2 = \tau/(2G)$. Equating the two expressions for the diagonal strain:

$$\frac{\tau}{2G} = \frac{\tau}{E}(1+\mu) \;\Rightarrow\; \boxed{E = 2G(1+\mu)}$$

(b) Numerical values

$$E = 2G(1+\mu) \Rightarrow 1+\mu = \frac{E}{2G} = \frac{200}{2\times80} = \frac{200}{160} = 1.25$$ $$\mu = 0.25$$

Bulk modulus from $E = 3K(1-2\mu)$:

$$K = \frac{E}{3(1-2\mu)} = \frac{200}{3(1-0.5)} = \frac{200}{1.5} = 133.33\ \text{GPa}$$

Results: $\mu = 0.25$, $K = 133.33\ \text{GPa}$.

Given: $d = 1200\ \text{mm}$, $L = 8000\ \text{mm}$, $p = 2\ \text{N/mm}^2$, $t = 12\ \text{mm}$.

(a) Stresses

Hoop (circumferential) stress:

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

Longitudinal stress:

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

(b) Deformations

Circumferential (hoop) strain:

$$\varepsilon_h = \frac{1}{E}(\sigma_h - \mu\sigma_l) = \frac{1}{200{,}000}(100 - 0.3\times50) = \frac{85}{200{,}000} = 4.25\times10^{-4}$$ $$\Delta d = \varepsilon_h \cdot d = 4.25\times10^{-4}\times1200 = 0.51\ \text{mm}$$

Longitudinal strain:

$$\varepsilon_l = \frac{1}{E}(\sigma_l - \mu\sigma_h) = \frac{1}{200{,}000}(50 - 0.3\times100) = \frac{20}{200{,}000} = 1.0\times10^{-4}$$ $$\Delta L = \varepsilon_l \cdot L = 1.0\times10^{-4}\times8000 = 0.80\ \text{mm}$$

Results: $\sigma_h = 100\ \text{N/mm}^2$, $\sigma_l = 50\ \text{N/mm}^2$; $\Delta d = 0.51\ \text{mm}$, $\Delta L = 0.80\ \text{mm}$.

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

$$A = \frac{\pi}{4}(D^2 - d^2) = \frac{\pi}{4}(200^2 - 160^2) = \frac{\pi}{4}(40{,}000 - 25{,}600) = \frac{\pi}{4}(14{,}400) = 11{,}310\ \text{mm}^2$$ $$I = \frac{\pi}{64}(D^4 - d^4) = \frac{\pi}{64}(200^4 - 160^4) = \frac{\pi}{64}(1.6\times10^9 - 0.6554\times10^9)$$ $$I = \frac{\pi}{64}(0.9446\times10^9) = 46.37\times10^6\ \text{mm}^4$$

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

(a) Euler crippling load

$$P_E = \frac{\pi^2 E I}{L_e^2} = \frac{\pi^2 \times 100{,}000 \times 46.37\times10^6}{2000^2}$$ $$= \frac{9.8696 \times 100{,}000 \times 46.37\times10^6}{4\times10^6} = \frac{45.77\times10^{12}}{4\times10^6} = 11.44\times10^6\ \text{N}$$ $$\boxed{P_E = 11{,}443\ \text{kN} \approx 11.44\ \text{MN}}$$

(b) Rankine crippling load

Radius of gyration: $k^2 = I/A = \dfrac{46.37\times10^6}{11{,}310} = 4100\ \text{mm}^2$.

$$P_R = \frac{\sigma_c\,A}{1 + a\left(\dfrac{L_e}{k}\right)^2} = \frac{\sigma_c A}{1 + a\,\dfrac{L_e^2}{k^2}}$$ $$\frac{L_e^2}{k^2} = \frac{2000^2}{4100} = \frac{4\times10^6}{4100} = 975.6$$ $$1 + a\frac{L_e^2}{k^2} = 1 + \frac{975.6}{1600} = 1 + 0.6098 = 1.6098$$ $$P_R = \frac{550\times11{,}310}{1.6098} = \frac{6.2205\times10^6}{1.6098} = 3.864\times10^6\ \text{N}$$ $$\boxed{P_R = 3864\ \text{kN} \approx 3.86\ \text{MN}}$$

Rankine load is much lower (and more realistic for this stocky column) than the Euler load.

Cross-sectional area:

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

Direct stress:

$$\sigma = \frac{P}{A} = \frac{60{,}000}{706.86} = 84.88\ \text{N/mm}^2$$

Strain energy (gradually applied load):

$$U = \frac{\sigma^2}{2E}\times V, \qquad V = A\,L = 706.86\times2000 = 1.4137\times10^6\ \text{mm}^3$$ $$U = \frac{(84.88)^2}{2\times200{,}000}\times 1.4137\times10^6 = \frac{7204.6}{400{,}000}\times1.4137\times10^6$$ $$U = 0.018012\times1.4137\times10^6 = 25{,}460\ \text{N·mm} = 25.46\ \text{N·m (J)}$$

Check via $U = \tfrac12 P\delta$: $\delta = \dfrac{PL}{AE} = \dfrac{60{,}000\times2000}{706.86\times200{,}000} = 0.8488\ \text{mm}$; $U = \tfrac12\times60{,}000\times0.8488 = 25{,}464\ \text{N·mm}$ ✓

Strain energy stored $= 25.46\ \text{J}$.

Section modulus (rectangular):

$$Z = \frac{b d^2}{6} = \frac{150\times300^2}{6} = \frac{150\times90{,}000}{6} = 2.25\times10^6\ \text{mm}^3$$

Moment of resistance:

$$M = \sigma\,Z = 8\times2.25\times10^6 = 18\times10^6\ \text{N·mm} = 18\ \text{kN·m}$$

Maximum bending moment for UDL on simple span:

$$M_{max} = \frac{wL^2}{8} = \frac{w\times4^2}{8} = 2w \ \ (\text{kN·m, with } w \text{ in kN/m})$$

Set $M_{max} = M$:

$$2w = 18 \Rightarrow w = 9\ \text{kN/m}$$

Maximum safe UDL $w = 9\ \text{kN/m}$ (inclusive of self-weight).