BE Civil Engineering (IOE, TU) Strength of Materials (IOE, CE 503) Question Paper 2079 Nepal

Part (a): Load sharing under mechanical load only

Since the rigid bar stays horizontal, both wires elongate by the same amount:

$$\delta_s = \delta_c \implies \frac{P_s L}{A_s E_s} = \frac{P_c L}{A_c E_c}$$

With equal $L$:

$$\frac{P_s}{A_s E_s} = \frac{P_c}{A_c E_c}$$

Compute axial stiffness $AE$ (in N):

• Steel: $A_s E_s = 120\times10^{-6}\times200\times10^9 = 24\times10^{6}\ \text{N}$
• Copper: $A_c E_c = 200\times10^{-6}\times100\times10^9 = 20\times10^{6}\ \text{N}$

So $\dfrac{P_s}{P_c} = \dfrac{A_s E_s}{A_c E_c} = \dfrac{24}{20} = 1.2$.

Equilibrium: $P_s + P_c = 15\ \text{kN}$.

$$P_s = 1.2 P_c \implies 1.2P_c + P_c = 15 \implies P_c = \frac{15}{2.2} = 6.818\ \text{kN}$$ $$P_s = 15 - 6.818 = 8.182\ \text{kN}$$

Stresses:

$$\sigma_s = \frac{P_s}{A_s} = \frac{8.182\times10^3}{120\times10^{-6}} = 68.18\times10^{6}\ \text{Pa} = \mathbf{68.18\ MPa}$$ $$\sigma_c = \frac{P_c}{A_c} = \frac{6.818\times10^3}{200\times10^{-6}} = 34.09\times10^{6}\ \text{Pa} = \mathbf{34.09\ MPa}$$

Part (b): Add a uniform temperature rise of $40\ ^{\circ}\text{C}$

The load equilibrium is unchanged: $P_s + P_c = 15\ \text{kN}$. The compatibility condition is that total elongations remain equal:

$$\underbrace{\alpha_s\,\Delta T\,L + \frac{P_s L}{A_s E_s}}_{\text{steel}} = \underbrace{\alpha_c\,\Delta T\,L + \frac{P_c L}{A_c E_c}}_{\text{copper}}$$

Divide through by $L$:

$$\alpha_s\Delta T + \frac{P_s}{A_s E_s} = \alpha_c\Delta T + \frac{P_c}{A_c E_c}$$ $$\frac{P_s}{A_sE_s} - \frac{P_c}{A_cE_c} = (\alpha_c-\alpha_s)\Delta T$$

RHS: $(18-12)\times10^{-6}\times40 = 6\times10^{-6}\times40 = 240\times10^{-6} = 2.4\times10^{-4}$.

LHS coefficients:

• $\dfrac{1}{A_sE_s} = \dfrac{1}{24\times10^{6}} = 4.1667\times10^{-8}\ \text{N}^{-1}$
• $\dfrac{1}{A_cE_c} = \dfrac{1}{20\times10^{6}} = 5.0000\times10^{-8}\ \text{N}^{-1}$

So:

$$4.1667\times10^{-8}P_s - 5.0000\times10^{-8}P_c = 2.4\times10^{-4} \quad (1)$$ $$P_s + P_c = 15000 \quad (2)$$

From (2): $P_s = 15000 - P_c$. Substitute into (1):

$$4.1667\times10^{-8}(15000 - P_c) - 5.0000\times10^{-8}P_c = 2.4\times10^{-4}$$ $$6.250\times10^{-4} - 4.1667\times10^{-8}P_c - 5.0000\times10^{-8}P_c = 2.4\times10^{-4}$$ $$-9.1667\times10^{-8}P_c = 2.4\times10^{-4} - 6.250\times10^{-4} = -3.850\times10^{-4}$$ $$P_c = \frac{3.850\times10^{-4}}{9.1667\times10^{-8}} = 4200\ \text{N} = 4.20\ \text{kN}$$ $$P_s = 15000 - 4200 = 10800\ \text{N} = 10.80\ \text{kN}$$

New stresses:

$$\sigma_s = \frac{10800}{120\times10^{-6}} = 90.0\times10^{6}\ \text{Pa} = \mathbf{90.0\ MPa\ (tension)}$$ $$\sigma_c = \frac{4200}{200\times10^{-6}} = 21.0\times10^{6}\ \text{Pa} = \mathbf{21.0\ MPa\ (tension)}$$

Interpretation: Because copper expands more with temperature, to keep the bar horizontal the copper wire must shed load to the steel wire; hence steel stress rises (68.18 → 90.0 MPa) and copper stress falls (34.09 → 21.0 MPa).

Coordinates (measure $x$ from $A$): $A=0$, $B=2$, $C=6$, $D=8\ \text{m}$.

Part (a): Reactions

Total UDL on $BC$ = $5\times4 = 20\ \text{kN}$ acting at midspan $x=4\ \text{m}$ (i.e. $2\ \text{m}$ right of $B$).

Take moments about $B$ (CW positive), with $R_B, R_C$ upward:

$$\sum M_B = 0:\; -10(2) + 20(2) + 8(6) - R_C(4) = 0$$ $$-20 + 40 + 48 - 4R_C = 0 \implies 4R_C = 68 \implies R_C = 17\ \text{kN}$$

Vertical equilibrium:

$$R_B + R_C = 10 + 20 + 8 = 38 \implies R_B = 38 - 17 = \mathbf{21\ kN}$$ $$\boxed{R_B = 21\ \text{kN},\quad R_C = 17\ \text{kN}}$$

Part (b): Shear Force (taking upward forces left of section as +)

• Just right of $A$: $V = -10\ \text{kN}$ (constant to $B$).
• Just left of $B$: $V = -10\ \text{kN}$.
• Just right of $B$: $V = -10 + 21 = +11\ \text{kN}$.
• Across $BC$ the UDL reduces $V$ at $5\ \text{kN/m}$. At $C^-$: $V = 11 - 5(4) = 11 - 20 = -9\ \text{kN}$.
• Just right of $C$: $V = -9 + 17 = +8\ \text{kN}$.
• At $D^-$: $V = +8\ \text{kN}$; the $8\ \text{kN}$ load at $D$ brings it to $0$. ✓

Shear is zero within $BC$ where $11 - 5(x-2) = 0 \Rightarrow x-2 = 2.2 \Rightarrow x = 4.2\ \text{m}$ (from $A$).

SFD (kN), x from A:
 A      B-     B+              x=4.2          C-    C+      D
-10 ---(-10) | +11 \___ ramp down ___ 0 ___\ -9 | +8 ---- +8 -> 0

Bending Moments (sagging +):

• At $A$: $M_A = 0$.
• At $B$: $M_B = -10\times2 = -20\ \text{kN·m}$ (hogging).
• Within $BC$, at distance $x$ from $A$ ($2\le x \le 6$):

$$M(x) = -10x + 21(x-2) - 5\frac{(x-2)^2}{2}$$

• At $C$ ($x=6$): $M_C = -10(6) + 21(4) - 5\frac{(4)^2}{2} = -60 + 84 - 40 = -16\ \text{kN·m}$ (hogging). Check from right: $M_C = -8\times2 = -16\ \text{kN·m}$ ✓.
• At $D$: $M_D = 0$.

Part (c): Extreme moment within $BC$ occurs where $V=0$, i.e. $x = 4.2\ \text{m}$:

$$M(4.2) = -10(4.2) + 21(2.2) - 5\frac{(2.2)^2}{2}$$ $$= -42 + 46.2 - 5\times2.42 = -42 + 46.2 - 12.1 = -7.9\ \text{kN·m}$$ $$\boxed{M_{extreme} = -7.9\ \text{kN·m at }x = 4.2\ \text{m from }A\ \text{(hogging)}}$$

Points of contraflexure (where $M=0$ inside $BC$): set $M(x)=0$. Writing $M(x) = -10x +21(x-2) -2.5(x-2)^2$ and letting $u = x-2$:

$$-10(u+2) + 21u - 2.5u^2 = 0 \Rightarrow -2.5u^2 + 11u - 20 = 0 \Rightarrow 2.5u^2 - 11u + 20 = 0$$ $$u = \frac{11 \pm \sqrt{121 - 200}}{5},\quad \text{discriminant} = 121 - 200 = -79 < 0$$

There is no real root, so the bending moment never reaches zero inside $BC$. This is consistent with the salient values $M_B=-20$, interior extreme $M(4.2)=-7.9$, and $M_C=-16$ — all negative. The moment rises from $-20$ at $B$ to a peak (least-negative) value of $-7.9\ \text{kN·m}$ at $x=4.2\ \text{m}$, then falls to $-16$ at $C$.

Thus the entire span $BC$ is in hogging; there are no points of contraflexure. The numerically largest bending moment in the beam is the hogging moment $-20\ \text{kN·m}$ at support $B$.

Geometry & symmetry: The neutral axis (NA) is at mid-depth, $120\ \text{mm}$ from each outer edge.

Part (a): Second moment of area $I$ about NA

Full rectangle $150\times240$ minus the two side cut-outs (each $70\times200$):

$$I = \frac{150\times240^3}{12} - 2\times\frac{70\times200^3}{12}$$ $$\frac{150\times240^3}{12} = \frac{150\times13.824\times10^{6}}{12} = \frac{2.0736\times10^{9}}{12} = 172.8\times10^{6}\ \text{mm}^4$$ $$2\times\frac{70\times200^3}{12} = 2\times\frac{70\times8\times10^{6}}{12} = 2\times\frac{560\times10^{6}}{12} = 2\times46.667\times10^{6} = 93.333\times10^{6}\ \text{mm}^4$$ $$\boxed{I = 172.8\times10^{6} - 93.333\times10^{6} = 79.467\times10^{6}\ \text{mm}^4}$$

Part (b): Maximum bending stress at extreme fibre $y_{max}=120\ \text{mm}$:

$$\sigma_{max} = \frac{M\,y_{max}}{I} = \frac{60\times10^{6}\ \text{N·mm}\times120\ \text{mm}}{79.467\times10^{6}\ \text{mm}^4}$$ $$= \frac{7.2\times10^{9}}{79.467\times10^{6}} = 90.6\ \text{N/mm}^2 = \mathbf{90.6\ MPa}$$

Tension at the bottom fibre, compression at the top fibre (sagging).

Part (c): Shear stress using $\tau = \dfrac{V Q}{I\,b}$.

At the NA (max shear), $Q$ = first moment of all area above NA about NA.

• Top flange ($150\times20$), centroid at $120-10 = 110\ \text{mm}$ from NA: $Q_f = 150\times20\times110 = 330{,}000\ \text{mm}^3$.
• Upper half of web ($10\times100$), centroid at $50\ \text{mm}$ from NA: $Q_w = 10\times100\times50 = 50{,}000\ \text{mm}^3$.
• $Q_{NA} = 330{,}000 + 50{,}000 = 380{,}000\ \text{mm}^3$.

Width at NA $b = 10\ \text{mm}$ (web):

$$\tau_{max} = \frac{80\times10^{3}\times380{,}000}{79.467\times10^{6}\times10} = \frac{3.04\times10^{10}}{7.9467\times10^{8}} = 38.3\ \text{N/mm}^2 = \mathbf{38.3\ MPa}$$

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

• Using web width $b=10\ \text{mm}$ (just below junction):

$$\tau_{web,jn} = \frac{80\times10^{3}\times330{,}000}{79.467\times10^{6}\times10} = \frac{2.64\times10^{10}}{7.9467\times10^{8}} = 33.2\ \text{MPa}$$

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

$$\tau_{flange,jn} = \frac{33.2}{15} = 2.22\ \text{MPa}$$

Shear stress distribution (description):

 top edge:        tau = 0
 flange (parabolic, small):  rises to 2.22 MPa at junction
  --- jump at junction (width change 150->10) ---
 web just below junction:    33.2 MPa
 web (parabolic):            rises to 38.3 MPa at NA
 web (symmetric):            falls back to 33.2 MPa at lower junction
  --- jump back to 2.22 MPa ---
 bottom flange -> 0 at edge

The web carries almost all the shear; the discontinuity at the junction is caused by the abrupt change in width $b$.

Part (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 stress (centre):

$$\frac{\sigma_x+\sigma_y}{2} = \frac{80 + (-40)}{2} = 20\ \text{MPa}$$

Half-difference:

$$\frac{\sigma_x-\sigma_y}{2} = \frac{80-(-40)}{2} = 60\ \text{MPa}$$

Radius:

$$R = \sqrt{60^2 + 30^2} = \sqrt{3600 + 900} = \sqrt{4500} = 67.08\ \text{MPa}$$

Principal stresses:

$$\sigma_1 = 20 + 67.08 = \mathbf{87.08\ MPa\ (tensile)}$$ $$\sigma_2 = 20 - 67.08 = \mathbf{-47.08\ MPa\ (compressive)}$$

Orientation of principal planes:

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y} = \frac{2\times30}{80-(-40)} = \frac{60}{120} = 0.5$$ $$2\theta_p = 26.57^{\circ} \implies \theta_p = 13.28^{\circ}\ \text{(and }13.28^{\circ}+90^{\circ}=103.28^{\circ})$$

The plane of $\sigma_1$ is at $\theta_p = \mathbf{13.28^{\circ}}$ (measured CCW from the $x$-face), since $\sigma_x>\sigma_y$.

Part (b): Maximum in-plane shear stress

$$\tau_{max} = R = \mathbf{67.08\ MPa}$$

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

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

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

Part (c): Mohr's circle verification

Plot points (using convention $\tau$ positive downward for the $x$-face so the circle is traversed correctly):

• $X = (\sigma_x,\ \tau_{xy}) = (80,\ 30)$
• $Y = (\sigma_y,\ -\tau_{xy}) = (-40,\ -30)$

Centre $C$ is on the $\sigma$-axis at the average:

$$C = \left(\frac{80+(-40)}{2},\ 0\right) = (20,\ 0)\ \text{MPa}$$

Radius is the distance from $C$ to $X$:

$$R = \sqrt{(80-20)^2 + 30^2} = \sqrt{60^2 + 30^2} = \sqrt{4500} = 67.08\ \text{MPa}$$

Principal stresses are where the circle crosses the $\sigma$-axis:

$$\sigma_1 = C + R = 20 + 67.08 = 87.08\ \text{MPa}$$ $$\sigma_2 = C - R = 20 - 67.08 = -47.08\ \text{MPa}$$

The top/bottom of the circle give $\tau_{max}=R=67.08\ \text{MPa}$ at $\sigma=20\ \text{MPa}$. The angle $2\theta_p$ from $CX$ to the $+\sigma$-axis is $\arctan(30/60)=26.57^{\circ}$, i.e. $\theta_p=13.28^{\circ}$ on the element. All results agree with Part (a). ✓

Part (a): Section properties

Area:

$$A = \frac{\pi}{4}(D^2 - d^2) = \frac{\pi}{4}(200^2 - 160^2) = \frac{\pi}{4}(40000 - 25600) = \frac{\pi}{4}(14400)$$ $$A = 11309.7\ \text{mm}^2$$

Second moment of area:

$$I = \frac{\pi}{64}(D^4 - d^4) = \frac{\pi}{64}(200^4 - 160^4)$$ $$200^4 = 1.6\times10^{9},\quad 160^4 = 6.5536\times10^{8}$$ $$I = \frac{\pi}{64}(1.6\times10^{9} - 0.65536\times10^{9}) = \frac{\pi}{64}(0.94464\times10^{9}) = 46.36\times10^{6}\ \text{mm}^4$$

Least radius of gyration:

$$k = \sqrt{\frac{I}{A}} = \sqrt{\frac{46.36\times10^{6}}{11309.7}} = \sqrt{4099.6} = 64.03\ \text{mm}$$

Effective length (both ends fixed): $L_e = \dfrac{L}{2} = \dfrac{4500}{2} = 2250\ \text{mm}$.

Slenderness ratio: $\dfrac{L_e}{k} = \dfrac{2250}{64.03} = 35.14$.

Part (b): Euler crippling load

$$P_E = \frac{\pi^2 E I}{L_e^2} = \frac{\pi^2\times100\times10^{3}\ \text{N/mm}^2\times46.36\times10^{6}\ \text{mm}^4}{(2250)^2\ \text{mm}^2}$$

Numerator $= 9.8696\times100\times10^{3}\times46.36\times10^{6} = 4.5757\times10^{13}$. Denominator $= 5.0625\times10^{6}$.

$$P_E = \frac{4.5757\times10^{13}}{5.0625\times10^{6}} = 9.038\times10^{6}\ \text{N} = \mathbf{9038\ kN}$$

Part (c): Rankine crippling load

$$P_R = \frac{\sigma_c\,A}{1 + a\left(\dfrac{L_e}{k}\right)^2}$$ $$\left(\frac{L_e}{k}\right)^2 = 35.14^2 = 1234.8$$ $$a\left(\frac{L_e}{k}\right)^2 = \frac{1234.8}{1600} = 0.7717$$

Numerator $= \sigma_c A = 550\times11309.7 = 6.2203\times10^{6}\ \text{N}$.

$$P_R = \frac{6.2203\times10^{6}}{1 + 0.7717} = \frac{6.2203\times10^{6}}{1.7717} = 3.5110\times10^{6}\ \text{N} = \mathbf{3511\ kN}$$

Part (d): Which governs

The slenderness ratio is low ($L_e/k = 35.1$), so the column is a short/intermediate column. Euler's formula is unconservative here because it ignores direct crushing; it predicts a very high load (Euler stress $= P_E/A = 799\ \text{MPa}$, which exceeds the crushing stress of $550\ \text{MPa}$ and is therefore physically impossible). The Rankine load (3511 kN) governs because it accounts for both crushing and buckling and gives a realistic value below the crushing capacity ($\sigma_c A = 6220\ \text{kN}$).

$$\boxed{\text{Safe crippling load} = P_R = 3511\ \text{kN (Rankine governs)}}$$

Cross-sectional area:

$$A = \frac{\pi}{4}(50)^2 = \frac{\pi}{4}\times2500 = 1963.5\ \text{mm}^2$$

Stress:

$$\sigma = \frac{P}{A} = \frac{150\times10^{3}}{1963.5} = 76.39\ \text{N/mm}^2$$

Axial (longitudinal) strain:

$$\varepsilon_L = \frac{\delta L}{L} = \frac{0.12}{300} = 4.0\times10^{-4}$$

(a) Young's modulus:

$$E = \frac{\sigma}{\varepsilon_L} = \frac{76.39}{4.0\times10^{-4}} = 190{,}980\ \text{N/mm}^2 \approx \mathbf{191\ GPa}$$

Lateral strain:

$$\varepsilon_t = \frac{\delta d}{d} = \frac{0.0066}{50} = 1.32\times10^{-4}$$

(b) Poisson's ratio:

$$\nu = \frac{\varepsilon_t}{\varepsilon_L} = \frac{1.32\times10^{-4}}{4.0\times10^{-4}} = \mathbf{0.33}$$

(c) Bulk modulus and modulus of rigidity:

$$K = \frac{E}{3(1-2\nu)} = \frac{190.98\times10^{3}}{3(1-0.66)} = \frac{190.98\times10^{3}}{3\times0.34} = \frac{190.98\times10^{3}}{1.02} = 187{,}235\ \text{N/mm}^2 \approx \mathbf{187\ GPa}$$ $$G = \frac{E}{2(1+\nu)} = \frac{190.98\times10^{3}}{2(1.33)} = \frac{190.98\times10^{3}}{2.66} = 71{,}797\ \text{N/mm}^2 \approx \mathbf{71.8\ GPa}$$

Torque transmitted:

$$T = \frac{60\,P}{2\pi N} = \frac{60\times90\times10^{3}}{2\pi\times200} = \frac{5.4\times10^{6}}{1256.64} = 4297.2\ \text{N·m} = 4.2972\times10^{6}\ \text{N·mm}$$

Condition 1 — Strength (shear stress):

$$\tau = \frac{16T}{\pi d^3} \le 50 \implies d^3 \ge \frac{16T}{\pi\times50} = \frac{16\times4.2972\times10^{6}}{157.08}$$ $$d^3 \ge \frac{6.8755\times10^{7}}{157.08} = 4.3771\times10^{5}\ \text{mm}^3 \implies d \ge 75.93\ \text{mm}$$

Condition 2 — Stiffness (angle of twist):

$$\theta = \frac{T L}{G J},\quad \theta = 1^{\circ} = \frac{\pi}{180} = 0.017453\ \text{rad},\quad J = \frac{\pi d^4}{32}$$ $$\theta = \frac{32\,T L}{G\,\pi d^4} \le 0.017453$$ $$d^4 \ge \frac{32 T L}{G\,\pi\,\theta} = \frac{32\times4.2972\times10^{6}\times2000}{80\times10^{3}\times\pi\times0.017453}$$

Numerator $= 32\times4.2972\times10^{6}\times2000 = 2.7502\times10^{11}$. Denominator $= 80\times10^{3}\times3.14159\times0.017453 = 4386.9$.

$$d^4 \ge \frac{2.7502\times10^{11}}{4386.9} = 6.2691\times10^{7}\ \text{mm}^4 \implies d \ge (6.2691\times10^{7})^{1/4}$$ $$d \ge 88.99\ \text{mm}$$

Governing diameter: the stiffness condition demands the larger diameter, so

$$\boxed{d_{min} \approx 89\ \text{mm (use }90\ \text{mm)}}$$

The angle-of-twist (stiffness) criterion governs; a shaft sized only for strength ($\approx76\ \text{mm}$) would twist more than $1^{\circ}$.

Given: $d = 1200\ \text{mm}$, $t = 10\ \text{mm}$, $L = 3000\ \text{mm}$, $p = 2.0\ \text{N/mm}^2$.

(a) Stresses

Hoop stress:

$$\sigma_h = \frac{p d}{2t} = \frac{2.0\times1200}{2\times10} = \frac{2400}{20} = \mathbf{120\ MPa}$$

Longitudinal stress:

$$\sigma_l = \frac{p d}{4t} = \frac{2.0\times1200}{4\times10} = 60\ \text{N/mm}^2 = \mathbf{60\ MPa}$$

(b) Change in diameter and length

Circumferential (hoop) strain:

$$\varepsilon_h = \frac{1}{E}(\sigma_h - \nu\sigma_l) = \frac{1}{200\times10^{3}}(120 - 0.3\times60) = \frac{120 - 18}{200\times10^{3}} = \frac{102}{200\times10^{3}} = 5.10\times10^{-4}$$ $$\delta d = \varepsilon_h\,d = 5.10\times10^{-4}\times1200 = \mathbf{0.612\ mm}$$

Longitudinal strain:

$$\varepsilon_l = \frac{1}{E}(\sigma_l - \nu\sigma_h) = \frac{1}{200\times10^{3}}(60 - 0.3\times120) = \frac{60 - 36}{200\times10^{3}} = \frac{24}{200\times10^{3}} = 1.20\times10^{-4}$$ $$\delta L = \varepsilon_l\,L = 1.20\times10^{-4}\times3000 = \mathbf{0.36\ mm}$$

(c) Change in internal volume

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

$$\varepsilon_V = 2(5.10\times10^{-4}) + 1.20\times10^{-4} = 10.20\times10^{-4} + 1.20\times10^{-4} = 11.40\times10^{-4}$$

Original internal volume:

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

Change in volume:

$$\delta V = \varepsilon_V\,V = 11.40\times10^{-4}\times3.3929\times10^{9} = 3.868\times10^{6}\ \text{mm}^3$$ $$\boxed{\delta V \approx 3.87\times10^{6}\ \text{mm}^3 = 3.87\ \text{litres}}$$

Section properties:

$$A = \frac{\pi}{4}(25)^2 = 490.87\ \text{mm}^2,\qquad L = 2500\ \text{mm}$$

Static stress (for comparison):

$$\sigma_{st} = \frac{W}{A} = \frac{400}{490.87} = 0.8149\ \text{N/mm}^2$$

(a) Impact stress — from energy balance, the instantaneous stress for a falling weight is:

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

Compute the term under the root:

$$\frac{2 h A E}{W L} = \frac{2\times80\times490.87\times200\times10^{3}}{400\times2500}$$

Numerator $= 2\times80\times490.87\times200\times10^{3} = 1.5708\times10^{10}$. Denominator $= 400\times2500 = 1.0\times10^{6}$.

$$\frac{2 h A E}{W L} = \frac{1.5708\times10^{10}}{1.0\times10^{6}} = 15708$$ $$\sqrt{1 + 15708} = \sqrt{15709} = 125.34$$ $$\sigma = 0.8149\,[1 + 125.34] = 0.8149\times126.34 = \mathbf{102.96\ N/mm^2 \approx 103\ MPa}$$

(b) Instantaneous elongation:

$$\delta = \frac{\sigma L}{E} = \frac{102.96\times2500}{200\times10^{3}} = \frac{257400}{200000} = 1.287\ \text{mm}$$ $$\boxed{\delta \approx 1.29\ \text{mm}}$$

(c) Strain energy stored:

$$U = \frac{\sigma^2}{2E}\times(\text{volume}) = \frac{\sigma^2}{2E}\,A L$$ $$U = \frac{(102.96)^2}{2\times200\times10^{3}}\times490.87\times2500$$ $$\frac{(102.96)^2}{4\times10^{5}} = \frac{10600.8}{4\times10^{5}} = 0.026502\ \text{N/mm}^2$$ $$U = 0.026502\times(490.87\times2500) = 0.026502\times1.2272\times10^{6} = 32522\ \text{N·mm} \approx \mathbf{32.5\ J}$$

Check by work done by falling weight: $W(h+\delta) = 400(80 + 1.287) = 400\times81.287 = 32515\ \text{N·mm} \approx 32.5\ \text{J}$ ✓.

Comparison: Impact factor $= \dfrac{\sigma}{\sigma_{st}} = \dfrac{102.96}{0.8149} \approx 126$. The suddenly-applied falling load produces a stress about 126 times the static stress, illustrating the severe effect of impact loading.

Internal axial forces (given, all tensile):

• $AB$: $P_{AB} = 80\ \text{kN}$
• $BC$: $P_{BC} = 60\ \text{kN}$
• $CD$: $P_{CD} = 30\ \text{kN}$

Stress in each segment $\sigma = P/A$:

$$\sigma_{AB} = \frac{80\times10^{3}}{600} = 133.3\ \text{N/mm}^2 = \mathbf{133.3\ MPa}$$ $$\sigma_{BC} = \frac{60\times10^{3}}{400} = 150.0\ \text{N/mm}^2 = \mathbf{150.0\ MPa}$$ $$\sigma_{CD} = \frac{30\times10^{3}}{250} = 120.0\ \text{N/mm}^2 = \mathbf{120.0\ MPa}$$

Elongation of each segment $\delta = \dfrac{P L}{A E} = \dfrac{\sigma L}{E}$:

$$\delta_{AB} = \frac{133.3\times1000}{200\times10^{3}} = 0.6667\ \text{mm}$$ $$\delta_{BC} = \frac{150.0\times1500}{200\times10^{3}} = 1.1250\ \text{mm}$$ $$\delta_{CD} = \frac{120.0\times1200}{200\times10^{3}} = 0.7200\ \text{mm}$$

Total elongation:

$$\delta_{total} = 0.6667 + 1.1250 + 0.7200 = \mathbf{2.512\ mm}$$ $$\boxed{\sigma_{AB}=133.3,\ \sigma_{BC}=150.0,\ \sigma_{CD}=120.0\ \text{MPa};\quad \delta_{total}=2.51\ \text{mm}}$$

(a) Maximum bending moment (simply supported, UDL over full span):

$$M_{max} = \frac{w L^2}{8} = \frac{6\times(4)^2}{8} = \frac{6\times16}{8} = 12\ \text{kN·m}$$ $$M_{max} = 12\times10^{6}\ \text{N·mm}$$

(b) Required section modulus from $\sigma = M/Z$:

$$Z_{req} = \frac{M_{max}}{\sigma_{perm}} = \frac{12\times10^{6}}{8} = 1.5\times10^{6}\ \text{mm}^3$$

(c) Dimensions — for a rectangle, $Z = \dfrac{b d^2}{6}$. With $d = 2b$:

$$Z = \frac{b(2b)^2}{6} = \frac{b\times4b^2}{6} = \frac{4b^3}{6} = \frac{2b^3}{3}$$

Set $Z = Z_{req}$:

$$\frac{2b^3}{3} = 1.5\times10^{6} \implies b^3 = \frac{3\times1.5\times10^{6}}{2} = 2.25\times10^{6}\ \text{mm}^3$$ $$b = (2.25\times10^{6})^{1/3} = 131.0\ \text{mm}$$ $$d = 2b = 262.1\ \text{mm}$$

Adopt $b = 135\ \text{mm}$, $d = 270\ \text{mm}$ (rounded up for safety). Check: $Z = \dfrac{135\times270^2}{6} = \dfrac{135\times72900}{6} = \dfrac{9.8415\times10^{6}}{6} = 1.640\times10^{6}\ \text{mm}^3 > 1.5\times10^{6}\ \text{mm}^3$ ✓.

$$\boxed{b = 135\ \text{mm},\quad d = 270\ \text{mm}}$$