Finite element method — nodes, elements, stiffness matrix: Exercises

3.6.18 · D4 · Physics › Spacecraft Structures & Systems Engineering › Finite element method — nodes, elements, stiffness matrix

Yeh page parent topic ke liye ek self-test ladder hai. Har problem mein sirf wohi ideas use hoti hain jo wahan build ki gayi hain: 1D bar element stiffness $K^{e} = \frac{E A}{L} [1 - 1 - 1 1]$ , shape functions, assembly, aur boundary conditions. Kuch bhi naya assume nahi kiya gaya — agar koi symbol aata hai, toh woh parent mein define kiya gaya tha.

Poore note mein, ek bar element sirf axial displacement carry karta hai: $1$ degree of freedom (DOF) per node. Parent se reuse hone wale symbols:

$E$ = Young's modulus (material ki stiffness, pascals mein).
$A$ = cross-section area.
$L$ = element length.
$k \equiv \frac{E A}{L}$ = ek bar ki "spring constant". Units: force / length (N/m).
$d_{i}$ = node $i$ ka axial displacement. $F_{i}$ = node $i$ par lagaya gaya axial force.

Figure — Finite element method — nodes, elements, stiffness matrix

Upar ka figure hamare reference structure hai jo zyaadatar problems mein use hoga: numbered nodes ke beech bars (springs) ki ek chain, jisme sabse baaya node wall se pin kiya gaya hai.

Level 1 — Recognition

Problem 1.1

Ek single bar ke liye element stiffness matrix likho jisme $E = 70 GPa$ (aluminium), $A = 2 \times 1 0^{- 4} m^{2}$ , $L = 0.5 m$ ho. $k$ ki numeric value do.

Recall Solution 1.1

$k = \frac{E A}{L} = \frac{( 70 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 4} )}{0.5}$ . Numerator: $70 \times 1 0^{9} \times 2 \times 1 0^{- 4} = 1.4 \times 1 0^{7}$ N. $0.5$ m se divide karo: $k = 2.8 \times 1 0^{7} N/m .$ Toh $K^{e} = 2.8 \times 1 0^{7} [1 - 1 - 1 1] N/m .$

Problem 1.2

Ek $3$ -node, $2$ -bar structure ko assemble karke yeh mila hai: $K_{global} = k 1 - 1 0 - 1 2 - 1 0 - 1 1 .$ Kaun sa node do elements se connect hota hai, aur sirf matrix padhke tum yeh kaise jaante ho?

Recall Solution 1.2

Diagonal entry $K_{ii}$ count karta hai ki node $i$ par kitni total stiffness khinchti hai. Nodes $1$ aur $3$ ka diagonal $1$ hai (ek-ek bar); node $2$ ka diagonal $2$ hai (do bars add up hote hain). Toh node 2 dono elements se connect hota hai. Off-diagonal $K_{25} = 0$ type ke zeros yeh bhi batate hain ki node $1$ aur node $3$ directly connected nahi hain (unke beech koi bar nahi).

Level 2 — Application

Problem 2.1

Ek single aluminium bar ( $k = 2.8 \times 1 0^{7}$ N/m, Problem 1.1 se) node $1$ par fixed hai aur node $2$ par $F_{2} = 5600$ N se khaincha ja raha hai. $d_{2}$ aur stress $σ$ nikalo agar $A = 2 \times 1 0^{- 4} m^{2}$ ho.

Recall Solution 2.1

Fixed node $1$ matlab $d_{1} = 0$ . Sirf free DOF (node $2$ ) rakho: $k d_{2} = F_{2} ⟹ d_{2} = \frac{5600}{2.8 \times 1 0 ^{7}} = 2 \times 1 0^{- 4} m = 0.2 mm .$ Strain (parent Step 2, $d_{1} = 0$ ke saath): $ϵ = \frac{d _{2} - d _{1}}{L} = \frac{2 \times 1 0 ^{- 4}}{0.5} = 4 \times 1 0^{- 4}$ . Stress (Hooke, parent Step 3): $σ = E ϵ = 70 \times 1 0^{9} \times 4 \times 1 0^{- 4} = 2.8 \times 1 0^{7}$ Pa $= 28$ MPa. Cross-check: $σ = F / A = 5600/ (2 \times 1 0^{- 4}) = 2.8 \times 1 0^{7}$ Pa. ✓ Dekho stress & strain.

Problem 2.2

Do identical bars series mein, har ek $k$ , nodes $1$ – $2$ – $3$ . Node $1$ fixed, load $P$ node $3$ par. Reduced system reproduce karo aur $d_{2}, d_{3}$ solve karo.

Recall Solution 2.2

Global matrix se row/column $1$ (fixed) delete karo: $k [2 - 1 - 1 1] [d_{2} d_{3}] = [0 P] .$ Row 1: $2 k d_{2} - k d_{3} = 0 \Rightarrow d_{3} = 2 d_{2}$ . Row 2: $- k d_{2} + k d_{3} = P \Rightarrow - k d_{2} + 2 k d_{2} = P \Rightarrow d_{2} = \frac{P}{k}$ , aur $d_{3} = \frac{2 P}{k}$ . Physical check: series mein do springs free end par double stretch karte hain. ✓

Level 3 — Analysis

Problem 3.1

Do bars series mein lekin alag stiffnesses ke saath: element A ( $k_{A}$ ) nodes $1$ – $2$ ke beech, element B ( $k_{B}$ ) nodes $2$ – $3$ ke beech. Node $1$ fixed, load $P$ node $3$ par. Assemble karo, phir $d_{2}, d_{3}$ solve karo.

Recall Solution 3.1

Element A DOFs (1,2) par, element B DOFs (2,3) par. Node $2$ ko dono diagonals milte hain: $K_{global} = k_{A} - k_{A} 0 - k_{A} k_{A} + k_{B} - k_{B} 0 - k_{B} k_{B} .$ Reduce karo (node 1 drop karo): $[k_{A} + k_{B} - k_{B} - k_{B} k_{B}] [d_{2} d_{3}] = [0 P] .$ Row 1: $(k_{A} + k_{B}) d_{2} = k_{B} d_{3} \Rightarrow d_{2} = \frac{k _{B}}{k _{A} + k _{B}} d_{3}$ . Row 2: $- k_{B} d_{2} + k_{B} d_{3} = P$ . Substitute karo: $k_{B} d_{3} (1 - \frac{k _{B}}{k _{A} + k _{B}}) = P \Rightarrow k_{B} d_{3} \cdot \frac{k _{A}}{k _{A} + k _{B}} = P .$ $d_{3} = P \cdot \frac{k _{A} + k _{B}}{k _{A} k _{B}} = P (\frac{1}{k _{A}} + \frac{1}{k _{B}}) .$ Yahi series-spring rule hai: total flexibility = flexibilities ka sum. Aur $d_{2} = \frac{k _{B}}{k _{A} + k _{B}} d_{3} = \frac{P}{k _{A}}$ . Sanity check: agar $k_{A} = k_{B} = k$ , toh $d_{3} = 2 P / k$ , $d_{2} = P / k$ — Problem 2.2 se match karta hai. ✓

Problem 3.2

Dikhao ki single free-free bar ka un-reduced $2 \times 2$ $K^{e}$ singular hai aur zero-determinant ko physically interpret karo.

Recall Solution 3.2

$det [1 - 1 - 1 1] = (1) (1) - (- 1) (- 1) = 1 - 1 = 0.$ Zero determinant matlab $K^{e}$ ka koi inverse nahi: $Kd = F$ ka ya toh koi solution nahi ya infinitely many hain. Physically, koi support nahi hone par, dono nodes ko same amount se slide karna ( $d_{1} = d_{2} = c$ ) zero strain produce karta hai ( $ϵ = (d_{2} - d_{1}) / L = 0$ ) aur isliye zero force — yeh ek rigid-body translation hai jisme energy nahi lagti. Vector $[11]$ null-space direction hai. Boundary conditions exactly isi mode ko khatam karne ke liye exist karti hain (parent ka doosra [!mistake]).

Level 4 — Synthesis

Problem 4.1

Ek stepped rocket strut: bar A ( $E$ , $A_{1} = 4 \times 1 0^{- 4} m^{2}$ , $L = 1 m$ ) phir bar B ( $E$ , $A_{2} = 2 \times 1 0^{- 4} m^{2}$ , $L = 1 m$ ), $E = 70$ GPa, series mein, nodes $1$ – $2$ – $3$ . Node $1$ fixed, $P = 1.4 \times 1 0^{4}$ N node $3$ par. $d_{2}$ , $d_{3}$ , aur har bar mein stress nikalo. Kaun sa bar zyada stressed hai?

Recall Solution 4.1

$k_{A} = \frac{E A _{1}}{L} = \frac{70 \times 1 0 ^{9} \cdot 4 \times 1 0 ^{- 4}}{1} = 2.8 \times 1 0^{7}$ N/m. $k_{B} = \frac{E A _{2}}{L} = \frac{70 \times 1 0 ^{9} \cdot 2 \times 1 0 ^{- 4}}{1} = 1.4 \times 1 0^{7}$ N/m. Problem 3.1 ke results use karo: $d_{2} = \frac{P}{k _{A}} = \frac{1.4 \times 1 0 ^{4}}{2.8 \times 1 0 ^{7}} = 5 \times 1 0^{- 4}$ m. $d_{3} = P (\frac{1}{k _{A}} + \frac{1}{k _{B}}) = 1.4 \times 1 0^{4} (\frac{1}{2.8 \times 1 0 ^{7}} + \frac{1}{1.4 \times 1 0 ^{7}})$ . $= 1.4 \times 1 0^{4} (3.5714 \times 1 0^{- 8} + 7.1429 \times 1 0^{- 8}) = 1.4 \times 1 0^{4} \times 1.0714 \times 1 0^{- 7} = 1.5 \times 1 0^{- 3}$ m. Bar A mein stress: $ϵ_{A} = (d_{2} - 0) / L = 5 \times 1 0^{- 4}$ , $σ_{A} = E ϵ_{A} = 70 \times 1 0^{9} \times 5 \times 1 0^{- 4} = 3.5 \times 1 0^{7}$ Pa $= 35$ MPa. Bar B mein stress: $ϵ_{B} = (d_{3} - d_{2}) / L = (1.5 \times 1 0^{- 3} - 5 \times 1 0^{- 4}) /1 = 1.0 \times 1 0^{- 3}$ , $σ_{B} = 70 \times 1 0^{9} \times 1.0 \times 1 0^{- 3} = 7.0 \times 1 0^{7}$ Pa $= 70$ MPa. $σ = F / A$ se cross-check: dono mein same internal force $P = 1.4 \times 1 0^{4}$ N flow karta hai (series). Bar A: $1.4 \times 1 0^{4} /4 \times 1 0^{- 4} = 3.5 \times 1 0^{7}$ Pa ✓. Bar B: $1.4 \times 1 0^{4} /2 \times 1 0^{- 4} = 7.0 \times 1 0^{7}$ Pa ✓. Bar B zyada stressed hai — patalee wali section. Isliye cutouts aur thin walls design drive karte hain; dekho buckling.

Problem 4.2

Ek single free-free bar of mass ke liye — isko ek element ki tarah discretise karo jisme har node par lumped mass $m /2$ ho. Bina poori modal machinery ke, singular stiffness se argue karo ki free-free bar mein zero-frequency (rigid-body) mode kyun hota hai, phir batao ki ek free unconstrained 3D structure mein kitne rigid-body modes hote hain.

Recall Solution 4.2

Natural frequency satisfy karti hai $K ϕ = ω^{2} M ϕ$ (dekho modal analysis). Rigid-translation vector $ϕ = [11]$ deta hai $K ϕ = 0$ (Problem 3.2 ke null space se), isliye $ω^{2} M ϕ = 0$ , jo force karta hai $ω = 0$ . Zero-frequency mode = motion jisme koi elastic restoring force nahi = rigid-body mode. Free 3D space mein ek unconstrained body mein 6 rigid-body modes hote hain: 3 translations + 3 rotations. Ek supported structure ka valid analysis boundary conditions ke zariye inhe sab remove karta hai.

Level 5 — Mastery

Problem 5.1

Wall se ek single free node tak teen bars ek parallel bundle ki tarah arranged hain: nodes $1, 2, 3$ sab wall se fixed hain ( $d_{1} = d_{2} = d_{3} = 0$ ) aur har ek apne bar ke zariye ek common free node $4$ se connected hai, stiffnesses $k_{1}, k_{2}, k_{3}$ same axis ke along act kar rahi hain. Node $4$ par load $P$ lagaya gaya hai. Reduced equation assemble karo aur $d_{4}$ nikalo. Phir $k_{1} = k_{2} = k_{3} = k$ ke liye evaluate karo.

Recall Solution 5.1

Har bar shared free node $4$ ke diagonal mein $+ k_{i}$ contribute karta hai (teeno node $4$ ko khinchte hain). Fixed nodes $1, 2, 3$ drop out ho jaate hain. Single free DOF: $(k_{1} + k_{2} + k_{3}) d_{4} = P ⟹ d_{4} = \frac{P}{k _{1} + k _{2} + k _{3}} .$ Parallel springs ki stiffnesses add hoti hain — compare karo Problem 3.1 ke series result se jisme flexibilities add hui thi. $k_{1} = k_{2} = k_{3} = k$ ke liye: $d_{4} = \frac{P}{3 k}$ — bundle ek bar se teen guna stiff hai, isliye teen guna kam deflect hoti hai.

Problem 5.2

Convergence reasoning. Ek tapered bar (area linearly shrink ho raha hai $A_{0} \to A_{0} /2$ length $L$ ke over) ko ek constant-area bar se model kiya gaya hai jiska area $\overset{ˉ}{A} = 0.75 A_{0}$ hai (average). End load $P$ ke under exact tip displacement hai: $d_{exact} = \frac{P L}{E A _{0}} ln 2 \approx 0.6931 \frac{P L}{E A _{0}} .$ One-element FEM estimate aur uska percent error compute karo. Phir batao ki h-refinement kya karega.

Recall Solution 5.2

One-element model: $k = \frac{E A ˉ}{L} = \frac{E ( 0.75 A _{0} )}{L}$ , toh $d_{FEM} = \frac{P}{k} = \frac{P L}{0.75 E A _{0}} = 1.3333 \frac{P L}{E A _{0}} .$ Ruko — like for like compare karo $\frac{P L}{E A _{0}}$ ko unit maanke. FEM coefficient $= 1/0.75 = 1.3333$ ; exact coefficient $= ln 2 = 0.6931$ . Percent error $= \frac{1.3333 - 0.6931}{0.6931} \times 100% \approx 92.4%$ . Yeh bahut bada hai: ek single average-area element ek strong taper ko badly mis-model karta hai kyunki hamara linear shape function constant strain force karta hai, lekin ek tapered bar mein strain patalee end ki taraf badhti hai. h-refinement (bahut chhote bars mein split karna, har ek ka area almost constant) har element ki constant-strain assumption ko locally valid banata hai; summed displacement $0.6931 P L / (E A_{0})$ par converge karta hai. Dekho numerical methods aur, temperature-driven area/property changes ke liye, thermal analysis.

Recall Quick self-quiz

Single bar stiffness $k$ barabar hoti hai? ::: $E A / L$ Assembled $K$ ka kisi node par diagonal entry kya batata hai? ::: wahan milne wali bars ki total stiffness (number/strength) Un-constrained $K$ singular kyun hoti hai? ::: rigid-body motion zero strain energy cost karti hai → zero-determinant null space Series springs combine hoti hain add karke? ::: flexibilities ( $1/ k$ ) Parallel springs combine hoti hain add karke? ::: stiffnesses ( $k$ ) Free 3D body mein kitne rigid-body modes hote hain? ::: 6 (3 translations + 3 rotations) Linear bar element sirf kaisa strain represent kar sakta hai? ::: constant strain