BE Civil Engineering (IOE, TU) Strength of Materials (IOE, CE 503) Question Paper 2080 Nepal

Cross-sectional areas

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

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

Copper tube ($D_o=50\,\text{mm},\ D_i=35\,\text{mm}$):

$$A_c = \frac{\pi}{4}(50^2 - 35^2) = \frac{\pi}{4}(2500-1225) = \frac{\pi}{4}(1275) = 1001.41\,\text{mm}^2$$

(a) Stresses under the 200 kN load

Compatibility (equal strain, both length $L$):

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

Equilibrium:

$$P = \sigma_s A_s + \sigma_c A_c$$ $$200\times10^3 = 2\sigma_c(706.86) + \sigma_c(1001.41) = \sigma_c(1413.72 + 1001.41) = \sigma_c(2415.13)$$ $$\sigma_c = \frac{200\times10^3}{2415.13} = 82.81\,\text{MPa (compressive)}$$ $$\sigma_s = 2\sigma_c = 165.62\,\text{MPa (compressive)}$$

(b) Shortening of the assembly

$$\delta = \varepsilon_s L = \frac{\sigma_s}{E_s}L = \frac{165.62}{200\times10^3}\times 1500 = 1.242\,\text{mm}$$

(Check via copper: $\delta = \frac{82.81}{100\times10^3}\times1500 = 1.242\,\text{mm}$ — consistent.)

$$\boxed{\sigma_s = 165.6\,\text{MPa},\quad \sigma_c = 82.8\,\text{MPa},\quad \delta = 1.24\,\text{mm (shortening)}}$$

(c) Thermal stresses for $\Delta T = 40^{\circ}\text{C}$, $P=0$

Since $\alpha_c > \alpha_s$, copper wants to expand more but is restrained by steel. Copper goes into compression, steel into tension. The total internal force is self-balancing:

$$\sigma_s A_s = \sigma_c A_c \quad\text{(tension in steel = compression in copper)}$$

Compatibility — the two end up with the same final length:

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

From force balance: $\sigma_c = \sigma_s\dfrac{A_s}{A_c} = \sigma_s\dfrac{706.86}{1001.41} = 0.70586\,\sigma_s$.

Substitute:

$$\frac{\sigma_s}{200\times10^3} + \frac{0.70586\,\sigma_s}{100\times10^3} = (17-12)\times10^{-6}\times 40$$ $$\sigma_s\left(5.0\times10^{-6} + 7.0586\times10^{-6}\right) = 200\times10^{-6}$$ $$\sigma_s(12.0586\times10^{-6}) = 200\times10^{-6}$$ $$\sigma_s = 16.59\,\text{MPa (tensile)}$$ $$\sigma_c = 0.70586\times16.59 = 11.71\,\text{MPa (compressive)}$$

Check force balance: $16.59\times706.86 = 11{,}727\,\text{N}$; $11.71\times1001.41 = 11{,}726\,\text{N}$ — balanced.

$$\boxed{\sigma_{s,\text{thermal}} = 16.6\,\text{MPa (tension)},\quad \sigma_{c,\text{thermal}} = 11.7\,\text{MPa (compression)}}$$

Set-up. Measure $x$ from $A$. Positions: $A=0$, $B=2\,\text{m}$, $C=7\,\text{m}$, $D=8\,\text{m}$.

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

Reactions. $\sum M_B = 0$ (taking moments about $B$, clockwise +):

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

Shear Force (taking left of section, upward +).

• Just right of $A$: $V = -20\,\text{kN}$ (the $20$ kN down). Constant from $A$ to $B$: $V_{B^-} = -20\,\text{kN}$.
• At $B$, add $R_B = +50$: $V_{B^+} = -20 + 50 = +30\,\text{kN}$.
• Across $BC$ the UDL reduces $V$ linearly: $V(x) = 30 - 10(x-2)$ for $2\le x \le 7$.
  • At $C^-$: $V = 30 - 10(5) = -20\,\text{kN}$.
• At $C$, add $R_C=+35$: $V_{C^+} = -20 + 35 = +15\,\text{kN}$.
• From $C$ to $D$: constant $+15\,\text{kN}$; at $D$ the $15$ kN load brings it to $0$. ✓ (closes)

Zero shear (location of max BM in span). In $BC$: $30 - 10(x-2) = 0 \Rightarrow x-2 = 3 \Rightarrow x = 5\,\text{m}$ (i.e. $3\,\text{m}$ right of $B$).

Bending Moments (sagging +), computed from the left:

• At $A$: $M_A = 0$.
• At $B$: $M_B = -20(2) = -40\,\text{kN·m}$ (hogging).
• At $D$: $M_D = 0$; at $C$ (from right): $M_C = -15(1) = -15\,\text{kN·m}$ (hogging).
• At max-shear point $x=5\,\text{m}$ (from left):

$$M = R_B(x-2) - 20\,x - \tfrac{10(x-2)^2}{2}\Big|_{x=5}$$

Using left segment: contributions = $-20(5)$ [load at A, arm 5] $+\,50(3)$ [$R_B$, arm 3] $-\,10\cdot3\cdot(3/2)$ [UDL over 3 m]

$$M_{x=5} = -100 + 150 - 45 = +5\,\text{kN·m (sagging)}$$

Max sagging in span $= +5\,\text{kN·m}$ at $3\,\text{m}$ right of $B$. Max hogging $= -40\,\text{kN·m}$ at $B$.

Points of contraflexure (where $M=0$ inside $BC$). General moment in $BC$ ($2\le x\le 7$):

$$M(x) = -20x + 50(x-2) - 5(x-2)^2$$

Let $u=x-2$: $M = -20(u+2) + 50u - 5u^2 = -40 + 30u - 5u^2$. Set $=0$:

$$5u^2 - 30u + 40 = 0 \Rightarrow u^2 - 6u + 8 = 0 \Rightarrow u = 2\ \text{or}\ 4$$

So contraflexure at $x = 4\,\text{m}$ and $x = 6\,\text{m}$ (i.e. $2\,\text{m}$ and $4\,\text{m}$ right of $B$).

SFD (kN) salient values:

A: 0 -> -20 (const to B)
B+: +30, falling linearly to -20 at C-
C+: +15, const to D, then 0

BMD (kN·m) salient values:

A: 0   B: -40   x=4m: 0   x=5m: +5 (peak)   x=6m: 0   C: -15   D: 0

$$\boxed{R_B = 50\,\text{kN},\ R_C = 35\,\text{kN};\ M_{\max,\text{sag}}=+5\,\text{kN·m},\ M_{\max,\text{hog}}=-40\,\text{kN·m at }B}$$

Section properties. Symmetric I-section: depth $D=200$, flange $120\times15$, web $10\times(200-2\times15)=10\times170$.

Moment of inertia about the neutral (centroidal horizontal) axis:

$$I = \frac{120\times200^3}{12} - \frac{(120-10)\times170^3}{12}$$ $$= \frac{120\times8\times10^6}{12} - \frac{110\times4.913\times10^6}{12}$$ $$= 80.0\times10^6 - \frac{110\times4{,}913{,}000}{12} = 80.0\times10^6 - 45.036\times10^6 = 34.96\times10^6\,\text{mm}^4$$

($170^3 = 4{,}913{,}000$.)

Internal forces (cantilever, load at tip). Max bending moment at fixed end:

$$M = W\cdot L = 40\times10^3 \times 2000 = 80\times10^6\,\text{N·mm}$$

Shear force (constant): $F = 40\times10^3\,\text{N} = 40\,\text{kN}$.

(a) Maximum bending stress at extreme fibre $y_{\max}=100\,\text{mm}$:

$$\sigma_{\max} = \frac{M\,y_{\max}}{I} = \frac{80\times10^6 \times 100}{34.96\times10^6} = 228.8\,\text{MPa}$$ $$\boxed{\sigma_{\max} \approx 228.8\,\text{MPa}}$$

(b) Shear stress $\tau = \dfrac{F\,Q}{I\,b}$, where $Q = A\bar{y}$ is the first moment of area above the level considered.

Maximum shear at the neutral axis (use full area above NA, $b = t_{web}=10\,\text{mm}$):

• Flange: $A_f = 120\times15 = 1800\,\text{mm}^2$, centroid at $y = 100 - 7.5 = 92.5\,\text{mm}$.
• Half web (above NA): $A_w = 10\times85 = 850\,\text{mm}^2$, centroid at $y = 85/2 = 42.5\,\text{mm}$.

$$Q_{NA} = 1800(92.5) + 850(42.5) = 166{,}500 + 36{,}125 = 202{,}625\,\text{mm}^3$$ $$\tau_{\max} = \frac{40\times10^3 \times 202{,}625}{34.96\times10^6 \times 10} = \frac{8.105\times10^9}{3.496\times10^8} = 23.18\,\text{MPa}$$ $$\boxed{\tau_{\max} \approx 23.2\,\text{MPa at the neutral axis}}$$

At the flange–web junction (just inside the web, $b=10\,\text{mm}$; only the flange contributes to $Q$):

$$Q_{j} = A_f\,\bar{y}_f = 1800\times92.5 = 166{,}500\,\text{mm}^3$$ $$\tau_{web} = \frac{40\times10^3 \times 166{,}500}{34.96\times10^6 \times 10} = \frac{6.66\times10^9}{3.496\times10^8} = 19.05\,\text{MPa}$$

Just inside the flange ($b=120\,\text{mm}$) the same $Q$ gives

$$\tau_{flange} = \frac{40\times10^3 \times 166{,}500}{34.96\times10^6 \times 120} = 1.59\,\text{MPa}$$

So there is a sudden jump from $1.59$ MPa (flange side) to $19.05$ MPa (web side) at the junction.

Shear-stress distribution (description). Parabolic in each flange (small, max $1.59$ MPa at junction); abrupt rise at the flange–web junction to $19.05$ MPa due to width change $120\to10$ mm; then parabolic over the web rising to the maximum $23.2$ MPa at the neutral axis, symmetric about NA.

  flange: ~1.6 MPa (small parabola)
  -- jump --
  web top: 19.05 MPa
  web mid (NA): 23.18 MPa  <- max
  web bottom: 19.05 MPa
  -- jump --
  bottom flange: ~1.6 MPa

Given: $\sigma_x = +120\,\text{MPa}$, $\sigma_y = -40\,\text{MPa}$, $\tau_{xy} = 50\,\text{MPa}$.

(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: $\dfrac{120 + (-40)}{2} = 40\,\text{MPa}$. Radius term: $\dfrac{\sigma_x-\sigma_y}{2} = \dfrac{120-(-40)}{2} = 80\,\text{MPa}$.

$$R = \sqrt{80^2 + 50^2} = \sqrt{6400 + 2500} = \sqrt{8900} = 94.34\,\text{MPa}$$ $$\sigma_1 = 40 + 94.34 = 134.34\,\text{MPa (tensile)}$$ $$\sigma_2 = 40 - 94.34 = -54.34\,\text{MPa (compressive)}$$

Orientation of principal planes:

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(50)}{120-(-40)} = \frac{100}{160} = 0.625$$ $$2\theta_p = 32.0^{\circ} \Rightarrow \theta_p = 16.0^{\circ}\ \text{(and } 106.0^{\circ})$$

The plane $\theta_p = 16.0^{\circ}$ carries $\sigma_1 = 134.34\,\text{MPa}$.

$$\boxed{\sigma_1 = 134.3\,\text{MPa},\ \sigma_2 = -54.3\,\text{MPa},\ \theta_p = 16.0^{\circ}}$$

(b) Maximum in-plane shear stress.

$$\tau_{\max} = R = 94.34\,\text{MPa}$$

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

$$\theta_s = \theta_p + 45^{\circ} = 16.0 + 45 = 61.0^{\circ}$$

The normal stress on those planes equals the mean, $40\,\text{MPa}$.

$$\boxed{\tau_{\max} = 94.3\,\text{MPa at }\theta_s = 61.0^{\circ}}$$

(c) Mohr's circle verification.

• Centre $C = \left(\dfrac{\sigma_x+\sigma_y}{2},\,0\right) = (40,\,0)\,\text{MPa}$.
• Radius $R = 94.34\,\text{MPa}$.
• Plot point $X=(\sigma_x, -\tau_{xy}) = (120, -50)$ and $Y=(\sigma_y, +\tau_{xy}) = (-40, +50)$; line $XY$ is a diameter through $C$. Distance $C$ to $X = \sqrt{(120-40)^2 + 50^2} = \sqrt{80^2+50^2} = 94.34 = R$. ✓
• $\sigma_1 = C + R = 40 + 94.34 = 134.34$; $\sigma_2 = C - R = -54.34$. ✓
• Top of circle gives $\tau_{\max} = R = 94.34$ at the centre abscissa $40$. ✓
• Angle on circle from $X$ to $\sigma_1$ axis $= 2\theta_p$ where $\tan2\theta_p = 50/80 = 0.625 \Rightarrow 2\theta_p = 32^{\circ}$. ✓

All analytical results are confirmed by Mohr's circle.

Torque transmitted.

$$T = \frac{60\,P}{2\pi N} = \frac{60 \times 120\times10^3}{2\pi \times 200} = \frac{7.2\times10^6}{1256.64} = 5729.6\,\text{N·m}$$ $$T = 5.7296\times10^6\,\text{N·mm}$$

(a) Diameter — strength criterion. For a solid shaft, $T = \dfrac{\pi}{16}\tau d^3$:

$$d^3 = \frac{16T}{\pi\tau} = \frac{16\times5.7296\times10^6}{\pi\times60} = \frac{9.1674\times10^7}{188.496} = 4.8635\times10^5\,\text{mm}^3$$ $$d = (4.8635\times10^5)^{1/3} = 78.66\,\text{mm}$$

Stiffness criterion. $\dfrac{T}{J} = \dfrac{G\theta}{L}$, with $J = \dfrac{\pi}{32}d^4$ and $\theta = 1^{\circ} = \dfrac{\pi}{180} = 0.017453\,\text{rad}$:

$$J = \frac{T L}{G\theta} = \frac{5.7296\times10^6 \times 2000}{80\times10^3 \times 0.017453} = \frac{1.14592\times10^{10}}{1396.24} = 8.2071\times10^6\,\text{mm}^4$$ $$d^4 = \frac{32 J}{\pi} = \frac{32\times8.2071\times10^6}{\pi} = 8.3576\times10^7 \Rightarrow d = (8.3576\times10^7)^{1/4} = 95.6\,\text{mm}$$

The stiffness criterion governs (larger). Adopt:

$$\boxed{d \approx 96\,\text{mm}}$$

(b) Hollow shaft, same torque & same max shear stress. Let external diameter $D$, internal $d_i = 0.6D$, ratio $k=0.6$.

$$T = \frac{\pi}{16}\tau\,\frac{D^4 - d_i^4}{D} = \frac{\pi}{16}\tau D^3(1-k^4)$$ $$1 - k^4 = 1 - 0.6^4 = 1 - 0.1296 = 0.8704$$ $$D^3 = \frac{16T}{\pi\tau(1-k^4)} = \frac{4.8635\times10^5}{0.8704} = 5.5877\times10^5\,\text{mm}^3$$ $$D = (5.5877\times10^5)^{1/3} = 82.36\,\text{mm}; \quad d_i = 0.6D = 49.4\,\text{mm}$$ $$\boxed{D \approx 82.4\,\text{mm},\ d_i \approx 49.4\,\text{mm}}$$

Percentage saving in weight (same length & material ⇒ compare cross-sectional areas; compare with the strength-designed solid shaft $d_s = 78.66\,\text{mm}$, the basis used for both strength designs):

• Solid area: $A_s = \dfrac{\pi}{4}(78.66)^2 = 4859.7\,\text{mm}^2$.
• Hollow area: $A_h = \dfrac{\pi}{4}(D^2 - d_i^2) = \dfrac{\pi}{4}D^2(1-k^2) = \dfrac{\pi}{4}(82.36)^2(1-0.36)$

$$= \frac{\pi}{4}(6783.2)(0.64) = 5328.6\times0.64\ \Rightarrow\ A_h = \frac{\pi}{4}\times6783.2\times0.64 = 3409.4\,\text{mm}^2$$ $$\text{Saving} = \frac{A_s - A_h}{A_s}\times100 = \frac{4859.7 - 3409.4}{4859.7}\times100 = \frac{1450.3}{4859.7}\times100 = 29.8\%$$ $$\boxed{\text{Weight saving} \approx 29.8\%}$$

(a) Definitions.

• Young's modulus $E$: ratio of direct (normal) stress to direct strain within the elastic limit, $E = \sigma/\varepsilon$.
• Modulus of rigidity $G$ (shear modulus): ratio of shear stress to shear strain, $G = \tau/\gamma$.
• Bulk modulus $K$: ratio of volumetric (hydrostatic) stress to volumetric strain, $K = \sigma_v/(\Delta V/V)$.

Relations with Poisson's ratio $\nu$:

$$E = 2G(1+\nu), \qquad E = 3K(1-2\nu)$$

(b) Computations. Area: $A = \dfrac{\pi}{4}(50)^2 = 1963.50\,\text{mm}^2$. Axial stress: $\sigma = \dfrac{150\times10^3}{1963.50} = 76.39\,\text{MPa}$. Longitudinal strain: $\varepsilon_L = \dfrac{0.095}{250} = 3.8\times10^{-4}$. Lateral strain: $\varepsilon_t = \dfrac{0.0058}{50} = 1.16\times10^{-4}$.

Young's modulus:

$$E = \frac{\sigma}{\varepsilon_L} = \frac{76.39}{3.8\times10^{-4}} = 2.010\times10^5\,\text{MPa} = 201.0\,\text{GPa}$$

Poisson's ratio:

$$\nu = \frac{\varepsilon_t}{\varepsilon_L} = \frac{1.16\times10^{-4}}{3.8\times10^{-4}} = 0.3053$$

Modulus of rigidity:

$$G = \frac{E}{2(1+\nu)} = \frac{201.0}{2(1.3053)} = \frac{201.0}{2.6105} = 77.0\,\text{GPa}$$

Bulk modulus:

$$K = \frac{E}{3(1-2\nu)} = \frac{201.0}{3(1-0.6105)} = \frac{201.0}{3(0.3895)} = \frac{201.0}{1.1684} = 172.0\,\text{GPa}$$ $$\boxed{E = 201.0\,\text{GPa},\ \nu = 0.305,\ G = 77.0\,\text{GPa},\ K = 172.0\,\text{GPa}}$$

Given: $d = 1200\,\text{mm}$, $L = 3000\,\text{mm}$, $t = 12\,\text{mm}$, $p = 2.5\,\text{N/mm}^2$, $E = 200{,}000\,\text{MPa}$, $\nu = 0.3$.

(a) Stresses. Hoop (circumferential) stress:

$$\sigma_h = \frac{p d}{2t} = \frac{2.5\times1200}{2\times12} = \frac{3000}{24} = 125\,\text{MPa}$$

Longitudinal stress:

$$\sigma_L = \frac{p d}{4t} = \frac{125}{2} = 62.5\,\text{MPa}$$ $$\boxed{\sigma_h = 125\,\text{MPa},\ \sigma_L = 62.5\,\text{MPa}}$$

(b) Deformations.

Circumferential (hoop) strain:

$$\varepsilon_h = \frac{1}{E}(\sigma_h - \nu\sigma_L) = \frac{1}{200000}(125 - 0.3\times62.5) = \frac{125 - 18.75}{200000} = \frac{106.25}{200000} = 5.3125\times10^{-4}$$

Change in diameter $\delta d = \varepsilon_h\, d = 5.3125\times10^{-4}\times1200 = 0.6375\,\text{mm}$.

Longitudinal strain:

$$\varepsilon_L = \frac{1}{E}(\sigma_L - \nu\sigma_h) = \frac{1}{200000}(62.5 - 0.3\times125) = \frac{62.5 - 37.5}{200000} = \frac{25}{200000} = 1.25\times10^{-4}$$

Change in length $\delta L = \varepsilon_L\, L = 1.25\times10^{-4}\times3000 = 0.375\,\text{mm}$.

Volumetric strain $= 2\varepsilon_h + \varepsilon_L$:

$$\varepsilon_v = 2(5.3125\times10^{-4}) + 1.25\times10^{-4} = 10.625\times10^{-4} + 1.25\times10^{-4} = 11.875\times10^{-4}$$

Original internal volume:

$$V = \frac{\pi}{4}d^2 L = \frac{\pi}{4}(1200)^2(3000) = \frac{\pi}{4}\times1.44\times10^6\times3000 = 3.3929\times10^9\,\text{mm}^3$$

Change in volume:

$$\delta V = \varepsilon_v V = 11.875\times10^{-4}\times3.3929\times10^9 = 4.029\times10^6\,\text{mm}^3$$ $$\boxed{\delta d = 0.638\,\text{mm},\ \delta L = 0.375\,\text{mm},\ \delta V \approx 4.03\times10^6\,\text{mm}^3\ (\approx 4.03\times10^6\,\text{mm}^3 = 4.03\,\text{litre})}$$

Section properties. $D_o = 200\,\text{mm}$, $D_i = 160\,\text{mm}$.

$$A = \frac{\pi}{4}(200^2 - 160^2) = \frac{\pi}{4}(40000 - 25600) = \frac{\pi}{4}(14400) = 11{,}309.7\,\text{mm}^2$$ $$I = \frac{\pi}{64}(200^4 - 160^4) = \frac{\pi}{64}(1.6\times10^9 - 6.5536\times10^8) = \frac{\pi}{64}(9.4464\times10^8) = 4.6370\times10^7\,\text{mm}^4$$

Radius of gyration:

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

Effective length (both ends fixed): $L_e = L/2 = 4000/2 = 2000\,\text{mm}$.

(a) Euler's crippling load.

$$P_E = \frac{\pi^2 E I}{L_e^2} = \frac{\pi^2 \times 100\times10^3 \times 4.6378\times10^7}{(2000)^2}$$ $$= \frac{9.8696 \times 100000 \times 4.6378\times10^7}{4\times10^6} = \frac{4.5772\times10^{13}}{4\times10^6} = 1.1443\times10^7\,\text{N}$$ $$\boxed{P_E \approx 11{,}443\,\text{kN} \approx 11.44\,\text{MN}}$$

(b) Rankine's load.

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

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

$$1 + a\left(\frac{L_e}{k}\right)^2 = 1 + \frac{975.5}{1600} = 1 + 0.6097 = 1.6097$$ $$P_R = \frac{550 \times 11309.7}{1.6097} = \frac{6.2203\times10^6}{1.6097} = 3.8643\times10^6\,\text{N}$$ $$\boxed{P_R \approx 3864\,\text{kN} \approx 3.86\,\text{MN}}$$

Comparison. Euler predicts $\approx 11.44\,\text{MN}$, far higher than Rankine's $\approx 3.86\,\text{MN}$. The column is fairly short/stocky ($L_e/k \approx 31$), so Euler's formula greatly over-estimates capacity; Rankine's formula, which accounts for crushing, gives the realistic governing load $\approx 3.86\,\text{MN}$.

(a) Derivation. Let the instantaneous (maximum) stress be $\sigma$ and the corresponding extension $\delta = \dfrac{\sigma L}{E}$. Work done by the falling weight as it descends $(h + \delta)$ equals the strain energy stored in the bar:

$$W(h + \delta) = \frac{\sigma^2}{2E}\times (A L)$$ $$W\left(h + \frac{\sigma L}{E}\right) = \frac{\sigma^2 A L}{2E}$$

Multiplying through and rearranging into a quadratic in $\sigma$:

$$\frac{A L}{2E}\sigma^2 - \frac{W L}{E}\sigma - W h = 0$$

Solving (taking the positive root) gives the standard result:

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

For $h=0$ this reduces to $\sigma = 2W/A$ (suddenly applied load).

(b) Numerical. Area: $A = \dfrac{\pi}{4}(25)^2 = 490.87\,\text{mm}^2$. Static stress term: $\dfrac{W}{A} = \dfrac{1500}{490.87} = 3.0558\,\text{MPa}$.

Compute the radical:

$$\frac{2 h A E}{W L} = \frac{2\times80\times490.87\times200000}{1500\times2500} = \frac{2\times80\times490.87\times200000}{3.75\times10^6}$$

Numerator $= 80\times490.87 = 39269.6$; $\times2 = 78539.2$; $\times200000 = 1.57078\times10^{10}$.

$$\frac{1.57078\times10^{10}}{3.75\times10^6} = 4188.8$$ $$\sqrt{1 + 4188.8} = \sqrt{4189.8} = 64.73$$

Maximum instantaneous stress:

$$\sigma = 3.0558\,(1 + 64.73) = 3.0558\times65.73 = 200.9\,\text{MPa}$$ $$\boxed{\sigma_{\max} \approx 200.9\,\text{MPa}}$$

Instantaneous elongation:

$$\delta = \frac{\sigma L}{E} = \frac{200.9\times2500}{200000} = 2.511\,\text{mm}$$ $$\boxed{\delta \approx 2.51\,\text{mm}}$$

Strain energy stored (= work done by weight $= W(h+\delta)$):

$$U = W(h+\delta) = 1500\,(80 + 2.511) = 1500\times82.511 = 123{,}767\,\text{N·mm} \approx 123.8\,\text{J}$$

(Check via $U=\frac{\sigma^2}{2E}AL = \frac{200.9^2}{2\times200000}\times490.87\times2500 = \frac{40360.8}{400000}\times1.2272\times10^6 = 0.10090\times1.2272\times10^6 = 1.238\times10^5\,\text{N·mm}$ — consistent.)

$$\boxed{U \approx 123.8\,\text{J}}$$

Maximum bending moment (simply supported, central point load $W$ plus UDL $w$ over span $\ell$):

$$M_{\max} = \frac{W\ell}{4} + \frac{w\ell^2}{8}$$ $$= \frac{10\times4}{4} + \frac{6\times4^2}{8} = 10 + \frac{96}{8} = 10 + 12 = 22\,\text{kN·m}$$ $$M_{\max} = 22\times10^6\,\text{N·mm}$$

Required section modulus. From $\sigma = M/Z$:

$$Z_{req} = \frac{M_{\max}}{\sigma_{perm}} = \frac{22\times10^6}{8} = 2.75\times10^6\,\text{mm}^3$$

Section modulus of rectangle with $d = 2b$:

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

Set $Z = Z_{req}$:

$$\frac{2b^3}{3} = 2.75\times10^6 \Rightarrow b^3 = \frac{3\times2.75\times10^6}{2} = 4.125\times10^6\,\text{mm}^3$$ $$b = (4.125\times10^6)^{1/3} = 160.4\,\text{mm}$$ $$d = 2b = 320.8\,\text{mm}$$

Adopt practical sizes:

$$\boxed{b = 165\,\text{mm},\quad d = 330\,\text{mm}}$$

Check with $b=165$, $d=330$: $Z = \dfrac{165\times330^2}{6} = \dfrac{165\times108900}{6} = \dfrac{1.79685\times10^7}{6} = 2.995\times10^6\,\text{mm}^3 > 2.75\times10^6$. Actual stress $= \dfrac{22\times10^6}{2.995\times10^6} = 7.35\,\text{MPa} < 8\,\text{MPa}$. Safe. ✓

Given: $d = 80\,\text{mm}$, $M = 3\,\text{kN·m} = 3\times10^6\,\text{N·mm}$, $T = 4\,\text{kN·m} = 4\times10^6\,\text{N·mm}$.

Section modulus (bending) for a solid circular shaft:

$$Z = \frac{\pi d^3}{32} = \frac{\pi (80)^3}{32} = \frac{\pi\times512000}{32} = 50{,}265\,\text{mm}^3$$

Direct bending stress:

$$\sigma_b = \frac{M}{Z} = \frac{3\times10^6}{50265} = 59.68\,\text{MPa}$$

Shear stress due to torsion ($Z_p = 2Z$):

$$\tau = \frac{16T}{\pi d^3} = \frac{T}{2Z} = \frac{4\times10^6}{2\times50265} = \frac{4\times10^6}{100530} = 39.79\,\text{MPa}$$

(a) Principal & maximum shear stresses. For a point with normal stress $\sigma_b$ and shear $\tau$ (the other normal stress $=0$):

$$\sigma_{1} = \frac{\sigma_b}{2} + \sqrt{\left(\frac{\sigma_b}{2}\right)^2 + \tau^2}$$ $$\frac{\sigma_b}{2} = 29.84,\quad \sqrt{29.84^2 + 39.79^2} = \sqrt{890.5 + 1583.2} = \sqrt{2473.7} = 49.74\,\text{MPa}$$ $$\sigma_1 = 29.84 + 49.74 = 79.58\,\text{MPa (max principal, tensile)}$$ $$\sigma_2 = 29.84 - 49.74 = -19.90\,\text{MPa (compressive)}$$

Maximum shear stress:

$$\tau_{\max} = \sqrt{\left(\frac{\sigma_b}{2}\right)^2 + \tau^2} = 49.74\,\text{MPa}$$ $$\boxed{\sigma_{1,\max} = 79.6\,\text{MPa},\quad \tau_{\max} = 49.7\,\text{MPa}}$$

(b) Equivalent moments. Equivalent twisting moment:

$$T_e = \sqrt{M^2 + T^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = 5\,\text{kN·m}$$

Equivalent bending moment:

$$M_e = \frac{1}{2}\left(M + \sqrt{M^2 + T^2}\right) = \frac{1}{2}(3 + 5) = 4\,\text{kN·m}$$ $$\boxed{T_e = 5\,\text{kN·m},\quad M_e = 4\,\text{kN·m}}$$

Cross-check: $\sigma_1 = \dfrac{M_e}{Z} = \dfrac{4\times10^6}{50265} = 79.6\,\text{MPa}$ ✓ and $\tau_{\max} = \dfrac{T_e}{2Z} = \dfrac{5\times10^6}{100530} = 49.7\,\text{MPa}$ ✓ — consistent with part (a).