--- title: "Beam reactions, deflection, BMD (30 kN, 20 kN/m)" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:beam-reactions-midspan-bmd-a2" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/beam-reactions-midspan-bmd-a2 site: BCT Engineering --- # Beam reactions, deflection, BMD (30 kN, 20 kN/m) Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2072 Asoj, 10 marks - Video: [Method of superposition for beams explained (slope & deflection with tables)](https://www.youtube.com/watch?v=DdZc7kLa4iI) — Engineer4Free - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/beam-reactions-midspan-bmd-a2 - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question Determine the support reactions and deflections at mid-span for the given structure. Also draw the bending moment diagram. $E = 20\times10^5$ MPa, $I = 5\times10^6$ mm$^4$. [Figure: beam fixed at the left end; 30 kN point load 3 m from the fixed end; intermediate support at 8 m from the fixed end carrying an applied moment of 40 kNm; 20 kN/m UDL over the final 6 m ending in a roller support] ## Short answer Direct stiffness with three beam elements gives RA=10.298 kN up, MA=7.930 kNm anticlockwise, RB=85.627 kN up, RC=54.076 kN up. Midspan deflection is 4.141 mm upward in A-B and 25.752 mm downward in B-C. Key BMD values include -7.930 at A, +22.964 at 3 m, -75.547 left of B, and +73.104 max sagging. ## Answer The beam is solved by the direct stiffness method with three beam elements. #### Data and idealisation $EI=(20\times10^{5}\ \text{N/mm}^2)(5\times10^{6}\ \text{mm}^4)=1\times10^{13}\ \text{N·mm}^2=1\times10^{4}\ \text{kN·m}^2$ (since $1\ \text{kN·m}^2=10^{9}\ \text{N·mm}^2$). ```text 30 kN 40 kNm 20 kN/m | (clockwise) |||||||||||| ##=======o=========o==============o A 2 B C fixed |<- 5 m ->| |<- 6 m ->| |<- 3 m->| ``` - Nodes: 1 = A (x = 0, fixed), 2 (x = 3 m, point load), 3 = B (x = 8 m, support, $M=40$ kNm taken clockwise), 4 = C (x = 14 m, roller). - Elements: 1 (3 m), 2 (5 m), 3 (6 m, carries the UDL from B to C). - Dofs per node: (deflection $v$ up, rotation $\theta$ anticlockwise). Known: $v_1=\theta_1=0$, $v_3=0$, $v_4=0$. Unknown: $v_2,\theta_2,\theta_3,\theta_4$. #### Element stiffness (each element) $$ [k]=\frac{EI}{L^3}\begin{bmatrix}12&6L&-12&6L\\6L&4L^2&-6L&2L^2\\-12&-6L&12&-6L\\6L&2L^2&-6L&4L^2\end{bmatrix} $$ #### Equivalent nodal loads Element 3: $wL/2=60$ kN down at each end, $wL^2/12=60$ kNm (clockwise at B, anticlockwise at C). Loads on the structure for $(v_2,\theta_2,\theta_3,\theta_4)$: $$ \{F\}=\begin{bmatrix}-30\\0\\-40-60\\+60\end{bmatrix}=\begin{bmatrix}-30\\0\\-100\\60\end{bmatrix} $$ #### Reduced equations (after assembly) $$ EI\begin{bmatrix} 0.54044 & -0.42667 & 0.24 & 0 \\ -0.42667 & 2.1333 & 0.4 & 0 \\ 0.24 & 0.4 & 1.4667 & 0.33333 \\ 0 & 0 & 0.33333 & 0.66667 \end{bmatrix}\begin{bmatrix}v_2\\\theta_2\\\theta_3\\\theta_4\end{bmatrix}=\begin{bmatrix}-30\\0\\-100\\60\end{bmatrix} $$ Solution: $$ \begin{aligned}v_2&=\frac{10.657}{EI}=1.066\times10^{-3}\ \text{m}=1.066\ \text{mm},&\theta_2&=\frac{22.551}{EI}=2.255\times10^{-3}\ \text{rad}\\ \theta_3&=-\frac{108.906}{EI}=-10.891\times10^{-3}\ \text{rad},&\theta_4&=\frac{144.453}{EI}=14.445\times10^{-3}\ \text{rad}\end{aligned} $$ (with $EI=10^4$ kN·m$^2$; $v_2$ is upward). #### Support reactions ($\{R\}=[K]\{d\}-\{F\}$) $$ R_A=10.298\ \text{kN}\uparrow,\quad M_A=7.930\ \text{kNm (anticlockwise)},\quad R_B=85.627\ \text{kN}\uparrow,\quad R_C=54.076\ \text{kN}\uparrow $$ Check: $\Sigma F_y$: $10.298+85.627+54.076=150.000=30+20\times6$. $\Sigma M_A=0$ also checks. #### Deflection at mid-span Using the element shape functions (and adding the UDL term $wx^2(L-x)^2/24EI$ inside the loaded element): - Mid-span of A-B ($x=4$ m): $v=4.141\times10^{-3}$ m $=4.141$ mm (upward). - Mid-span of B-C ($x=11$ m): $v=-25.752$ mm (downward). Without the UDL term the cubic interpolation alone would give $-19.002$ mm. #### Bending moment (sagging +), from the reactions | Section | M (kNm) | |---|---| | A (fixed end) | $-7.930$ (hogging) | | 3 m (under 30 kN) | $+22.964$ | | B, left | $-75.547$ | | B, right (after the 40 kNm) | $-35.547$ | | In B-C, max sagging at 3.296 m from B | $+73.104$ | | C | 0 | ```text BMD, sagging up (1 row = 12 kNm, 1 column = 0.35 m) 84 | 72 | ***** 60 | * ** 48 | * * 36 | * * 24 | ** * 12 | **** ** * * 0 |-***--------*---------------------------* -12 |* ** * -24 | ** -36 | * * -48 | ** -60 | ** -72 | * -84 | ``` The diagram runs from A at the left edge to C at the right edge: hogging at A, sagging under the 30 kN load, deep hogging just left of B, a jump of 40 kNm at B, then sagging in B-C with the maximum at 3.30 m from B, and zero at C. **Answer:** $R_A=10.298$, $R_B=85.627$, $R_C=54.076$ kN; $M_A=7.930$ kNm; mid-span deflections $4.141$ mm upward (AB) and $25.752$ mm downward (BC). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which DOFs are unknown in the three-element model? A. v2, theta2, theta3, theta4 B. v1, theta1, v2, theta2 C. theta2, theta3 only D. v3, v4 Answer: A. A is fixed, and supports B and C prevent deflection, so v2, theta2 (under 30 kN load), theta3 and theta4 remain. (Key point: Unknown DOFs) 2. For the 20 kN/m UDL on the 6 m element B-C, what are the equivalent loads? A. 60 kN down and 60 kN m at each end (clockwise at B, anticlockwise at C) B. 120 kN at B C. 60 kN and 90 kN m D. 30 kN and 30 kN m Answer: A. wL/2 = 60 kN and wL^2/12 = 20 x 36/12 = 60 kN m. (Key point: UDL equivalent loads) 3. Why does the load at theta3 become -100 kN m? A. The 40 kN m clockwise moment (-40) adds to the clockwise UDL end moment (-60) at B B. The two moments cancel C. Only the UDL moment acts D. It is the sum of both vertical forces Answer: A. Clockwise moments are negative, so -40 - 60 = -100. (Key point: Applied moment in load vector) 4. Which statements are correct? A. R_A + R_B + R_C = 150 kN = 30 + 20 x 6, and the BMD jumps by 40 kN m at B B. R_A + R_B + R_C = 30 kN C. The BMD has no discontinuity at B D. The reactions must sum to 20 kN Answer: A. Vertical equilibrium requires the total load of 150 kN, and an applied moment causes a jump in the BMD at its point. (Key point: Reaction check and BMD jump) ## Close variants set in other exams - [Beam reactions and mid-span deflection (2EI)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-support-reactions-midspan-k8) — Computational Techniques in Civil Engineering, chapter 4, asked 2× ## Related in Computational Techniques in Civil Engineering - [Beam BMD, SFD and mid-span deflection (FEM)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-bmd-sfd-midspan-concrete) — Computational Techniques in Civil Engineering, chapter 4 - [Beam: deflection, slope under load, BMD](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-fem-s79) — Computational Techniques in Civil Engineering, chapter 4 - [Rotation and deflection at free end, BMD](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-fem-rotation-deflection-free-end) — Computational Techniques in Civil Engineering, chapter 4 ## Related in other subjects - [Slope deflection: beam, 360 kN load, fixed end](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-10kn-20knm-360kn) — Theory of Structures II, chapter 3 - [Mid-span deflection of beam, 10 kN load](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ass-beam-deflection-mid-span-4m) — Strength of Materials, chapter 7 - [ILD: bending moment at fixed support, 10 m beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-fixed-support-10m-1) — Theory of Structures II, chapter 4 - [Beam mid-span deflection with temperature gradient](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Abeam-midspan-defl-temperature) — Theory of Structures I, chapter 3