--- title: "Critical position of UDL shorter than span" subject: "Theory of Structures I" chapter: 5 key: "semester_4/theory_of_structures_i:udl-shorter-than-span-critical-position" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/udl-shorter-than-span-critical-position site: BCT Engineering --- # Critical position of UDL shorter than span Theory of Structures I · Chapter 5: Influence Lines for Simple Structures - Asked in: 2079 Chaitra, 4 marks; 2077 Chaitra, 6 marks; 2071 Bhadra, 6 marks; 2071 Magh, 5 marks; 2070 Magh, 6 marks; 2066 Kartik, 6 marks (asked 6 times) - Video: [Influence Line Diagrams for Simply Supported Beams - Problem No 2 (UDL length less than beam span)](https://www.youtube.com/watch?v=vIXTV9x7lsg) — Stan Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/udl-shorter-than-span-critical-position - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5 ## Question Using influence line diagram, determine the most critical position of a stretch of uniformly distributed load shorter than the span to give maximum bending moment at a given section of a simply supported beam. (Prove that the bending moment is maximum when the section divides the span and the load in the same ratio.) ## Short answer Place the UDL so that the section divides the load in the same ratio as it divides the span: c/l = a/L, where c is load length left of the section, l is total load length, a is distance from left support. Equating end ordinates of the ILD under the load yields this critical condition. ## Answer A UDL shorter than the span gives maximum bending moment at a section when the load is placed so that the **section divides the span and the load in the same ratio**. #### ILD for bending moment at section C Simply supported beam $AB$, span $L$; section $C$ at $a$ from $A$, $b = L-a$. The ILD for $M_C$ is a triangle with peak $y_C = ab/L$ under $C$. ```text A c C (l-c) B o------+----+-------+----o ILD: y1 ab/L y2 ``` Ordinate left of $C$ at distance $x$ from $A$: $y = \dfrac{xb}{L}$. Right of $C$ at distance $x$ from $B$: $y = \dfrac{xa}{L}$. #### Proof Let a UDL of intensity $w$ and length $l$ (< $L$) have a length $c$ on the left of $C$ and $(l-c)$ on the right. Ordinates under its ends are $$ y_1 = \frac{(a-c)\,b}{L}, \qquad y_2 = \frac{\{b-(l-c)\}\,a}{L} $$ $M_C = w\times(\text{area under ILD below the load})$. Shift the load by a small distance $dx$ to the right. The left end loses a strip of height $y_1$ and the right end gains a strip of height $y_2$, so $$ \frac{dM_C}{dx} = w\,(y_2 - y_1) $$ $M_C$ is maximum when $dM_C/dx = 0$, i.e. $y_1 = y_2$: $$ \begin{aligned} (a-c)\,b &= (b-l+c)\,a \\ ab - cb &= ab - la + ca \\ c(a+b) &= la \Rightarrow \frac{c}{l} = \frac{a}{L} \end{aligned} $$ Hence the part of the load on the left of $C$ is to the whole load as the left part of the span is to the whole span: $$ \frac{c}{l} = \frac{a}{L} $$ #### Maximum moment With $c = la/L$ the end ordinates are equal: $y_1 = y_2 = \dfrac{ab}{L}\left(1-\dfrac{l}{L}\right)$. The area under the load is two trapezoids: $$ M_{max} = w\,\frac{l}{2}\left(\frac{ab}{L} + y_1\right) = w\,l\,\frac{ab}{L}\left(1 - \frac{l}{2L}\right) $$ **Result:** place the UDL so that the section divides the load in the ratio $a : b$, the same as it divides the span. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the shape of the ILD for bending moment at section C of a simply supported beam? A. A triangle with peak ab/L under C B. A rectangle of height ab/L C. A straight line from 1 at A to 0 at B D. A parabola with peak at mid-span Answer: A. M_C ILD is a triangle, zero at both supports and ab/L under the section. (Key point: ILD shape) 2. Shifting the UDL by dx changes M_C by w(y2 - y1). When is M_C maximum? A. When the ordinates under the two ends of the load are equal (y1 = y2) B. When the load is centred on the span C. When the load starts at the section C D. When y1 is zero Answer: A. dM/dx = 0 requires y2 = y1, the end ordinates being equal. (Key point: Condition for maximum) 3. What does y1 = y2 lead to? A. c/l = a/L, the section divides the load and the span in the same ratio B. c = l/2, the section is at the load centre C. c = a, the load begins at the support D. l = L - a Answer: A. Algebra of (a-c)b = (b-l+c)a gives c(a+b) = la, so c/l = a/L. (Key point: Result of proof) 4. With the load critically placed, how are the moment and ordinates described? A. The area under the load is two trapezoids with equal end ordinates ab/L(1 - l/L) B. The load area is one triangle C. The end ordinates are 0 D. The end ordinates are ab/L Answer: A. With c = la/L, y1 = y2 = (ab/L)(1 - l/L), and M_max = w l (ab/L)(1 - l/2L). (Key point: Maximum moment) ## Related in Theory of Structures I - [Moment area: mid-span deflection under UDL](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-ss-udl-midspan) — Theory of Structures I, chapter 4 - [No bending moment under UDL in parabolic arch](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-no-bm-udl) — Theory of Structures I, chapter 6, asked 2× - [Arch 160 m: rolling UDL, section 50 m from right](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-160m-udl-30knm-60m-50m) — Theory of Structures I, chapter 6 - [ILD and BM at mid span for triangular load](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-triangular-load-mid-span-c) — Theory of Structures I, chapter 5 - [Define influence line diagram and its uses](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Adefine-ild-uses) — Theory of Structures I, chapter 5, asked 5× - [Conjugate beam: free end of overhang, UDL and point load](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-udl-10kn-4m-3m-2071b) — Theory of Structures I, chapter 4 ## Related in other subjects - [Simply supported T-beam: max bending stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-6m-100kn-10kn) — Strength of Materials, chapter 7 - [I-section beam: moment of resistance and UDL](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ai-section-moment-resistance-udl-10m) — Strength of Materials, chapter 7 - [Design simply supported beam, 6 m span](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adesign-simply-supported-beam-6-m) — Design of RCC Structure, chapter 4 - [Simply supported I-beam: stress at flange-web junction](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ai-beam-flange-web-junction-stress) — Strength of Materials, chapter 7