--- title: "Truss ILDs for DK, DJ, DE" subject: "Theory of Structures I" chapter: 5 key: "semester_4/theory_of_structures_i:truss-ild-dk-dj-de" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-dk-dj-de site: BCT Engineering --- # Truss ILDs for DK, DJ, DE Theory of Structures I · Chapter 5: Influence Lines for Simple Structures - Asked in: 2077 Chaitra, 6 marks - 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/truss-ild-dk-dj-de - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5 ## Question Draw influence line diagram for member DK, DJ, and DE of the truss as shown in figure below. Assume the unit load moves along the bottom chord. [Figure: truss with bottom joints A, B, C, D, E, F, G (6 panels of 6 m = 36 m), top joints L, K, J, I, H, height 8 m; A hinge, G roller.] ## Short answer DK: tension for load right of D, compression left of C; max +0.625 at D, min −0.417 at C. DJ: zero for all load positions (unstressed vertical). DE: always tension, peak ordinate +1.0 at E. ## Answer **Assumed truss (Pratt type, figure not available):** bottom chord $A$–$G$, 6 panels of 6 m (span 36 m); top joints $L, K, J, I, H$ at 8 m height above $B, C, D, E, F$; inclined end posts $AL$ and $GH$; verticals above $B$ to $F$; diagonals $LC$, $KD$, $ID$, $HE$ (sloping down towards the centre). $A$ hinge, $G$ roller. The unit load moves along the bottom chord; the ILD is straight between panel points. Reactions: $R_A = \dfrac{36-x}{36}$, $R_G = \dfrac{x}{36}$. #### Member DK (diagonal in panel $CD$) Section through $KJ$, $KD$ and $CD$. Diagonal $KD$: $\tan\theta = 8/6$, $\sin\theta = 0.8$. - Load to the right of $D$ (left part considered): $F_{DK}\sin\theta = R_A \Rightarrow F_{DK} = 1.25\,R_A$ (tension). - Load to the left of $C$ (right part considered): $F_{DK} = -1.25\,R_G$ (compression). - Between $C$ and $D$ the ILD is the straight line joining the two ordinates, passing through zero. #### Member DJ (vertical) At the unloaded top joint $J$, three members meet: two collinear top-chord members ($KJ$ and $JI$) and the vertical $DJ$, so $F_{DJ} = 0$. For bottom-chord loading the ILD of $DJ$ is zero throughout. (The load at $D$ itself is carried by the diagonals $KD$ and $ID$ which are symmetric, so the vertical is still unstressed.) #### Member DE (bottom chord) Section through $JI$, $ID$ and $DE$ (panel $DE$). $JI$ and $ID$ meet at $I$ ($x = 24$ m, $y = 8$ m), so take moments about $I$ (lever arm of $DE$ = 8 m): - Load to the left of $D$ (right part): $F_{DE}\times 8 = R_G\times 12 \Rightarrow F_{DE} = 1.5\,R_G$ (tension). - Load to the right of $E$ (left part): $F_{DE}\times 8 = R_A\times 24 \Rightarrow F_{DE} = 3\,R_A$ (tension). #### Ordinates (per unit load; + tension) | Member | A | B | C | D | E | F | G | |---|---|---|---|---|---|---|---| | DK | 0.000 | -0.208 | -0.417 | 0.625 | 0.417 | 0.208 | 0.000 | | DJ | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | | DE | 0.000 | 0.250 | 0.500 | 0.750 | 1.000 | 0.500 | 0.000 | #### Shapes - **DK:** negative ordinates for load on $A$–$C$ (maximum $-0.417$ at $C$), positive for load on $D$–$G$ (maximum $+0.625$ at $D$); the zero crossing lies between $C$ and $D$ at $x = 12 + 6\times\frac{0.417}{0.417+0.625} = 14.4$ m from $A$. - **DJ:** zero for all positions of the load. - **DE:** tension everywhere, rising from 0 at $A$ to a peak $+1.0$ under $E$ ($x = 24$ m), then falling to 0 at $G$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What gives the force in diagonal DK for a load right of D? A. F = 1.25 R_A tension, since sin theta = 0.8 B. F = 0.8 R_A C. F = R_A compression D. F = 1.5 R_G Answer: A. Vertical equilibrium F sin theta = R_A gives 1.25 R_A. (Key point: DK method) 2. Why is the ILD of DJ zero? A. It carries all the shear B. At the unloaded top joint J, two collinear chords and the vertical meet, so DJ carries no force C. The vertical is in compression D. The load is on the top chord Answer: B. A joint of two collinear members and a third member with no load forces the third member to be zero. (Key point: DJ) 3. What does the DE ILD look like? A. Compression everywhere B. Peak at D C. Tension everywhere, peak +1.0 under E D. Sign change at C Answer: C. Moments about I give 1.5 R_G or 3 R_A, with maximum 1.0 at E. (Key point: DE) 4. Where does the ILD of DK change sign? A. At B B. At F C. At mid-span 18 m D. About 14.4 m from A, between C and D Answer: D. Ordinates are -0.417 at C and +0.625 at D, so zero is at 12 + 6(0.417/1.042) = 14.4 m. 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