Exercises — Buckling — Euler column buckling load derivation

KYA: pin–pin ka matlab effective-length factor $K=1$ hai, toh $L_e=KL=1.2\text{ m}$. Woh $K$ kyun: pinned ends rotate kar sakte hain lekin sideways move nahi kar sakte, isliye exactly ek half sine wave poori length mein fit hoti hai (figure s01 dekho, leftmost case). Plug in karo: $P_{cr}=\frac{{\pi }^{2}EI}{L_e^2}=\frac{{\pi }^2(70\times 10^9)(5.0\times 10^{-10})}{(1.2{)}^2}$ $P_{cr}=\frac{{\pi }^2(70\times 10^9)(5.0\times 10^{-10})}{1.44}\approx 240\text{N}$

Answer: (b) yield stress ${\sigma }_y$. KYU: Euler buckling ek stiffness-and-geometry event hai, strength event nahi. Formula $P_{cr}={\pi }^{2}EI/L_e^2$ mein sirf $E$ (stiffness), $I$ (cross-section ki geometry), aur $L_e$ (length ki geometry) hain. Ek column tab bhi buckle kar sakta hai jab har fibre perfectly elastic ho — material sense mein kuch "toota" nahi hai.

KYA: fixed–free ka $K=2$ hota hai — ise seedha figure s01 ke doosre column se padho — toh $L_e=2L=2.4\text{ m}$. Woh $K$ kyun: clamped-base/free-top column sirf ek quarter sine wave fit kar sakta hai; use mirror karo aur tum dekhoge ki yeh length $2L$ ke pin–pin column ke aadhe ki tarah behave karta hai (figure s01, doosra case). Zyada effective length $\Rightarrow$ zyada weak. $P_{cr}=\frac{{\pi }^{2}EI}{(2L{)}^2}=\frac{{\pi }^2(70\times 10^9)(5.0\times 10^{-10})}{(2.4{)}^2}\approx 60\text{N}$ Kyunki $P_{cr}\propto 1/L_e^2$ aur $L_e$ double ho gayi, load $2^2=4$ se ghata: $240/4=60\text{N}$. ✓

Step 1 — geometry $I$: $I=\frac{\pi r_0^4}{4}=\frac{\pi (0.006{)}^4}{4}=1.018\times 10^{-9}{\text{m}}^4$ Step 2 — Euler load ($K=1$, $L_e=0.9\text{ m}$): $P_{cr}=\frac{{\pi }^2(200\times 10^9)(1.018\times 10^{-9})}{(0.9{)}^2}\approx 2481\text{N}$

Pehle: $I_{\text{tube}}\approx \pi R^{3}t$ kyun hota hai (thin-wall approximation). Circular tube ka exact second moment, outer radius $R_o$ aur inner radius $R_i$ ke saath, $I=\frac{\pi }{4}(R_o^4-R_i^4)$ hai. Wall ko mean radius $R=\frac{1}{2}(R_o+R_i)$ aur thickness $t=R_o-R_i$ mein likho, toh $R_o=R+\frac{t}{2}$, $R_i=R-\frac{t}{2}$. Tab $R_o^4-R_i^4=(R_o^2-R_i^2)(R_o^2+R_i^2)=[(R_o-R_i)(R_o+R_i)](R_o^2+R_i^2)=(t)(2R)(2R^2+\frac{t^2}{2}).$ $t^2$ term kyun drop karein: thin wall ke liye $t\ll R$, toh $t^2/2$ negligible hai $2R^2$ ke comparison mein. Sirf leading term rakhne par: $I\approx \frac{\pi }{4}\cdot t\cdot 2R\cdot 2R^2=\pi R^{3}t.$ Yeh sirf isliye valid hai kyunki wall thin hai — yahi reason hai ki tube rod se better hai: uski almost saari area door wale radius $R$ par baith jaati hai.

Solid: $r_0=\sqrt{A/\pi }=\sqrt{2.83\times 10^{-4}/\pi }=9.49\times 10^{-3}\text{m}$. $I_{\text{solid}}=\frac{\pi r_0^4}{4}=\frac{\pi (9.49\times 10^{-3}{)}^4}{4}=6.37\times 10^{-9}{\text{m}}^4$ Tube: wall $t=A/(2\pi R)=2.83\times 10^{-4}/(2\pi \cdot 0.04)=1.126\times 10^{-3}\text{m}$ (check karo $t/R\approx 0.028\ll 1$, toh thin-wall safe hai). $I_{\text{tube}}\approx \pi R^{3}t=\pi (0.04{)}^3(1.126\times 10^{-3})=2.264\times 10^{-7}{\text{m}}^4$ Ratio (same $E$, same $L$, toh $P_{cr}$ ratio $=$ $I$ ratio): $\frac{P_{cr,\text{tube}}}{P_{cr,\text{solid}}}=\frac{2.264\times 10^{-7}}{6.37\times 10^{-9}}\approx 35.5$ Interpretation: identical mass, lekin tube buckling ~35× better resist karta hai kyunki $I$ axis se material ki distance ke square ke saath scale karta hai. Exactly yahi reason hai ki launch vehicles aur lander legs tubes hote hain, rods nahi.

Solid circle ke liye radius of gyration: $I=\pi r_0^4/4$, $A=\pi r_0^2$, toh $r=\sqrt{\frac{I}{A}}=\sqrt{\frac{\pi r_0^4/4}{\pi r_0^2}}=\frac{r_0}{2}=4.0\times 10^{-3}\text{m}$ Crossover: ${\sigma }_{cr}={\sigma }_y$ set karo: $\frac{{\pi }^{2}E}{{\lambda }^2}={\sigma }_y\Rightarrow \lambda =\pi \sqrt{\frac{E}{{\sigma }_y}}=\pi \sqrt{\frac{70\times 10^9}{270\times 10^6}}=50.6$ Length ($K=1$ toh $L_e=L^{\ast }$): $L^{\ast }=\lambda r=50.6\times 4.0\times 10^{-3}=0.202\text{m}$ Matlab: figure s02 par $\lambda =50.6$ locate karo — yeh exactly crossover par baith ta hai, dashed inelastic band ki boundary par. ~20 cm se chhota hone par yeh rod pehle yield karti hai (left region); usse lamba hone par Euler buckling govern karta hai. Exactly is length ka ek real rod actually dono curves se thoda neeche fail hoga, inelastic regime mein.

(a) Required load: $P_{cr}=1.5\times 2500=3750\text{ N}$. $I$ ke liye Euler rearrange karo ($L_e=L=0.75\text{ m}$): $I\ge \frac{P_{cr}{L}_e^2}{{\pi }^{2}E}=\frac{3750(0.75{)}^2}{{\pi }^2(200\times 10^9)}=1.069\times 10^{-9}{\text{m}}^4$ (b) Wall thickness (Exercise 3.1 se $I\approx \pi R^{3}t$ use karke): $t\ge \frac{I}{\pi R^3}=\frac{1.069\times 10^{-9}}{\pi (0.025{)}^3}=2.177\times 10^{-5}\text{m}\approx 0.022\text{ mm}$ Reality check: itni thin wall global Euler buckling se pehle local wall wrinkling se fail ho jaayegi — ek reminder ki Euler ek necessary lekin sufficient nahi check hai. Phir bhi, yeh global-buckling requirement ko sahi size karta hai.

Euler load: $I=\pi r_0^4/4=\pi (0.005{)}^4/4=4.909\times 10^{-11}{\text{m}}^4$ (exponent note karo: $(0.005{)}^4=6.25\times 10^{-10}$, $\pi /4$ se multiply karne par $4.909\times 10^{-11}$ milta hai). $P_{cr}=\frac{{\pi }^2(70\times 10^9)(4.909\times 10^{-11})}{(0.30{)}^2}\approx 377\text{N}$ Yield load: $A=\pi r_0^2=7.854\times 10^{-5}{\text{m}}^2$, $P_y={\sigma }_{y}A=(270\times 10^6)(7.854\times 10^{-5})\approx 21206\text{N}$ Compare karo: $P_{cr}=377\text{ N}\ll P_y=21206\text{ N}$, toh Euler buckling govern karta hai; strut $\approx 377\text{ N}$ par fail hoti hai. Uska slenderness $\lambda =L_e/r=0.30/(0.005/2)=120$ ise figure s02 ke right (slender) branch par far out rakhta hai, inelastic band se kaafi door — toh yahan pure Euler trustworthy hai.

(a) Fixed–fixed ka $K=0.5$ hota hai (figure s01 ka teesra column), toh $L_e=0.5L$. Kyunki $P_{cr}\propto 1/L_e^2$: $P_{cr,\text{new}}=\frac{{\pi }^{2}EI}{(0.5L{)}^2}=\frac{1}{0.25}\cdot \frac{{\pi }^{2}EI}{L^2}=4\times 800=3200\text{N}$ (b) Factor $=1/K^2=1/0.25=4$. Clamping buckling load ko chaar guna kar deta hai — geometry se free strength milti hai. (c) Pin–pin column ko chhota karke same load paana: $L_e^{\prime 2}=L^2/4$ chahiye, yaani $L^{\prime }=L/2$. Toh length half karna identical $4\times$ gain deta hai — dono routes ($K$ aur $L$) formula mein same tarike se enter karte hain, $L_e^2$ ke through.

Effective length: $L_e=0.7\times 1.1=0.77\text{ m}$. Required $P_{cr}=2.0\times 1800=3600\text{ N}$. (a) $I\ge \frac{P_{cr}{L}_e^2}{{\pi }^{2}E}=\frac{3600(0.77{)}^2}{{\pi }^2(70\times 10^9)}=3.090\times 10^{-9}{\text{m}}^4$ (b) $t\ge \frac{I}{\pi R^3}=\frac{3.090\times 10^{-9}}{\pi (0.02{)}^3}=1.229\times 10^{-4}\text{m}\approx 0.123\text{ mm}$ (c) $A\approx 2\pi Rt$ kyun: thin ring ke liye, wall ek strip hai jiska length mean radius par circumference $2\pi R$ ke barabar hai aur width $t$ hai; kyunki $t\ll R$ toh inner aur outer radii dono $\approx R$ hain, toh exact area $\pi (R_o^2-R_i^2)=\pi (R_o+R_i)(R_o-R_i)=\pi (2R)(t)=2\pi Rt$ hoti hai. Tab: $A\approx 2\pi Rt=2\pi (0.02)(1.229\times 10^{-4})=1.545\times 10^{-5}{\text{m}}^2$ $P_y={\sigma }_{y}A=(270\times 10^6)(1.545\times 10^{-5})\approx 4171\text{N}$ Kyunki $P_y\approx 4171\text{ N}>P_{cr}=3600\text{ N}$ hai, tube apni Euler limit yield se pehle reach kar leta hai — buckling correctly govern karta hai, exactly jaisa ek slender strut ke liye intended tha. ✓ (Practically thin-wall local buckling aur figure s02 ka inelastic band bhi check kiya jaayega.)