--- title: "Stiffness matrix: element matrices, member forces, BMD" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:stiffness-element-matrices-600kn-bmd" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-element-matrices-600kn-bmd site: BCT Engineering --- # Stiffness matrix: element matrices, member forces, BMD Theory of Structures II · Chapter 3: Displacement method - Asked in: 2069 Chaitra, 18 marks - Video: [Continuous Beam Analysis By Stiffness Matrix Method। Lecture 9। Beam Numerical।Structural Analysis 3](https://www.youtube.com/watch?v=IrSidZTsssY) — Civil Gyani - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-element-matrices-600kn-bmd - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Determine element stiffness matrices, deformations at joints and member forces. Also draw bending moment diagram, using the stiffness matrix method. $E = 2\times10^{5}$ MPa, $A = 200\ \text{cm}^2$, $I = 2.5\times10^{5}\ \text{mm}^4$ [as printed]. [Figure: frame ABCD; beam AB-C of 4 m (2 m + 2 m) with 600 kN downward at B (mid-span); column CD of 5 m, D fixed, A roller/hinge; coordinates 1 (horizontal), 2 (vertical) and 3 (rotation) at C.] ## Short answer Deformations at C: u_C=-0.06148 mm, v_C=-0.4398 mm, θ_C=0.000537 rad. Bending moments: M_A=0, M_C=207.41 kNm, max under load 496.30 kNm. ## Answer **Data and assumptions:** $E=2\times10^5$ MPa $=2\times10^8$ kN/m², $A=200$ cm² $=0.02$ m². The printed $I=2.5\times10^5$ mm⁴ would give $EI=50$ kNm² (deflections of several metres), so $I=2.5\times10^5$ cm⁴ $=2.5\times10^{-3}$ m⁴ is used: $EA=4\times10^6$ kN, $EI=5\times10^5$ kNm². Geometry: beam AC = 4 m (2 m + 2 m) with the 600 kN load downward at B (mid-span); A is a hinge (translations fixed, rotation free); column CD = 5 m, fixed at D. Coordinates at C: **1 = horizontal (right +), 2 = vertical (up +), 3 = rotation (anticlockwise +)**. B is only a load point, so AC is one element and the load is replaced by fixed-end actions. #### Step 1: Element stiffness matrices (global axes, kN and m) Member constants: $EA/L$, $12EI/L^3$, $6EI/L^2$, $4EI/L$, $2EI/L$. | Element | L (m) | $EA/L$ | $12EI/L^3$ | $6EI/L^2$ | $4EI/L$ | $2EI/L$ | |---|---|---|---|---|---|---| | 1: AC (horizontal) | 4 | 1000000 | 93750 | 187500 | 500000 | 250000 | | 2: CD (vertical) | 5 | 800000 | 48000 | 120000 | 400000 | 200000 | **Element 1** (A: dofs $u,v,\theta$; then C: $u,v,\theta$), member axis along global $x$: $$ [k_1]=\begin{bmatrix}1000000 & 0 & 0 & -1000000 & 0 & 0 \\ 0 & 93750 & 187500 & 0 & -93750 & 187500 \\ 0 & 187500 & 500000 & 0 & -187500 & 250000 \\ -1000000 & 0 & 0 & 1000000 & 0 & 0 \\ 0 & -93750 & -187500 & 0 & 93750 & -187500 \\ 0 & 187500 & 250000 & 0 & -187500 & 500000\end{bmatrix} $$ **Element 2** (C: $u,v,\theta$; then D: $u,v,\theta$), member axis along global $-y$ (axial terms act in $v$, bending terms in $u$): $$ [k_2]=\begin{bmatrix}48000 & 0 & 120000 & -48000 & 0 & 120000 \\ 0 & 800000 & 0 & 0 & -800000 & 0 \\ 120000 & 0 & 400000 & -120000 & 0 & 200000 \\ -48000 & 0 & -120000 & 48000 & 0 & -120000 \\ 0 & -800000 & 0 & 0 & 800000 & 0 \\ 120000 & 0 & 200000 & -120000 & 0 & 400000\end{bmatrix} $$ #### Step 2: Fixed-end actions of element 1 600 kN at mid-span: $FEM=\mp\dfrac{PL}{8}=\mp300$ kNm and end shears 300 kN. The fixed-end actions are reversed to give the equivalent joint loads. #### Step 3: Structure stiffness matrix at C (hinge A condensed) and solution Assembling $[k_1]$ and $[k_2]$ and eliminating the free rotation at A ($\theta_A$, using $M_A=0$): $$ [K]_{CC}=\begin{bmatrix}1048000 & 0 & 120000 \\ 0 & 823438 & -93750 \\ 120000 & -93750 & 775000\end{bmatrix},\qquad \{P\}=\begin{Bmatrix}0.00\\-412.50\\450.00\end{Bmatrix}\ (\text{kN, kN, kNm}) $$ $$ \{d\}=[K]^{-1}\{P\}:\quad u_C=-0.06148\ \text{mm},\ v_C=-0.4398\ \text{mm},\ \theta_C=0.000537\ \text{rad},\ \theta_A=-0.001033\ \text{rad} $$ #### Step 4: Member end forces (local axes, kN, kNm) - Element 1 (A to C): axial -61.48 kN (tension +), shear at A 248.15 kN, $M_A=0$; at C: shear 351.85 kN, $M_C=207.41$ kNm (clockwise on the member end). - Element 2 (C to D): axial -351.85 kN (tension +), shear at C 61.48 kN, $M_C=-207.41$ kNm; at D shear -61.48 kN, $M_D=-100.01$ kNm. Joint C equilibrium of moments: $207.41+(-207.41)=0$ ✓. #### Step 5: Bending moment diagram (kNm) | Member | BM at first end | BM at second end | Max within member | |---|---|---|---| | AC | 0.00 | -207.41 | 496.30 at 2.00 m | | CD | -207.41 | 100.01 | - | Under the 600 kN load the sagging moment is 496.30 kNm. The BMD of AC consists of two straight lines peaking under the load, zero at A and 207.41 kNm (hogging) at C; CD has a straight-line diagram from the value at C to 100.01 kNm at the fixed end D. Reactions: A: $H=61.48$, $V=248.15$, $M=0.00$; D: $H=-61.48$, $V=351.85$, $M=100.01$. **Answer: $u_C=-0.0615$ mm, $v_C=-0.440$ mm, $\theta_C=0.000537$ rad; $M_C=207.41$ kNm, $M_D=-100.01$ kNm, BM under the load 496.30 kNm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the answer use I = 2.5x10^5 cm^4 instead of the printed 2.5x10^5 mm^4? A. mm^4 gives EI = 50 kNm^2, which would produce absurd deflections of several metres B. Because cm^4 is always the unit for I in IOE problems C. Because mm^4 cannot be converted to m^4 D. Because A is given in cm^2 Answer: A. With mm^4, EI is only 50 kNm^2 and deflections would be unrealistically large, so the intended unit is cm^4, giving EI = 5x10^5 kNm^2. (Key point: Unit conversion of I) 2. B is only a load point on beam AC. How is the 600 kN load handled? A. AC is split into two elements at B B. AC is kept as one element and the load is replaced by fixed-end actions (PL/8 = 300 kNm, shears 300 kN) C. The load is applied directly as a moment at C D. The load is ignored because B has no coordinate Answer: B. No joint is needed at B, so one element AC is used and the load becomes fixed-end actions that are reversed to give equivalent joint loads. (Key point: Treatment of the load at B) 3. For element CD (L = 5 m, EI = 5x10^5), which values are right for 12EI/L^3 and 4EI/L? A. 93750 and 500000 B. 48000 and 400000 C. 120000 and 200000 D. 800000 and 48000 Answer: B. 12EI/L^3 = 12x5x10^5/125 = 48000 and 4EI/L = 4x5x10^5/5 = 400000. (Key point: Element stiffness terms) 4. How is the hinge at A (free rotation) accounted for when forming the structure stiffness at C? A. Its rotation is set to zero B. Its rotation theta_A is eliminated using the condition M_A = 0 (static condensation) C. A is treated as fixed and the moment released afterwards D. The column stiffness is doubled Answer: B. A hinge carries no moment, so M_A = 0 is used to eliminate theta_A and give the 3x3 matrix at C. 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