--- title: "Truss ILDs for AB, CD, EJ, FH" subject: "Theory of Structures I" chapter: 5 key: "semester_4/theory_of_structures_i:ild-truss-ab-cd-ej-fh" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/ild-truss-ab-cd-ej-fh site: BCT Engineering --- # Truss ILDs for AB, CD, EJ, FH Theory of Structures I · Chapter 5: Influence Lines for Simple Structures - Asked in: 2069 Bhadra, 12 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/ild-truss-ab-cd-ej-fh - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5 ## Question Using influence line diagram, obtain member force in AB, CD, EJ and FH for the following loaded pin-jointed truss as shown in figure below. [Figure: pin-jointed truss with top joints A to G and bottom joints N, M, L, K, J, I, H, 6 panels of 3 m = 18 m, height 4 m; N hinge, I roller; verticals and diagonals as drawn.] ## Short answer Using section methods, the influence ordinates per unit load on the bottom chord are: AB: 0, −0.6, −0.45, −0.3, −0.15, 0, 0.15; CD: 0, −0.3, −0.6, −0.9, −0.45, 0, 0.45; EJ: 0, 0.2, 0.4, 0.6, 0.8, 0, 0.2; FH: 0, 0, 0, 0, 0, 0, 1.25. Member forces are computed as sum of actual loads multiplied by corresponding ordinates or UDL areas. ## Answer **Assumed truss (figure not available):** top joints $A, B, C, D, E, F, G$ (left to right) directly above bottom joints $N, M, L, K, J, I, H$, 6 panels of 3 m (18 m overall), height 4 m. Verticals at all joints; diagonals $AM$, $BL$, $CK$, $DJ$, $EI$, $FH$ (all sloping down to the right). Supports as given: $N$ hinge and $I$ roller ($x = 15$ m), so the last panel $IH$ (members $FG$, $GH$, $FH$, $IH$) overhangs 3 m beyond the roller. The loads are not given in the text, so the ILDs are drawn for a unit load on the bottom chord at $x$ from $N$: $$ R_N = \frac{15-x}{15}, \qquad R_I = \frac{x}{15} \qquad (0\le x\le 18) $$ For a load at $H$: $R_N = -0.2$, $R_I = 1.2$. Diagonal length $\sqrt{3^2+4^2} = 5$ m, $\sin\theta = 0.8$. #### Method - **AB (top chord, panel $NM$):** section through $AB$, $AM$, $NM$; moments about $M$ ($x=3$, where $AM$ and $NM$ meet), lever arm 4 m: for a load at or right of $M$, $F_{AB} = -\dfrac{3R_N}{4} = -0.75R_N$; for a load at $N$, 0. - **CD (top chord, panel $LK$):** section through $CD$, $CK$, $LK$; moments about $K$ ($x=9$): load at or right of $K$: $F_{CD} = -\dfrac{9R_N}{4}$; load at or left of $L$ (right part, only $R_I$ at 6 m from $K$): $F_{CD} = -\dfrac{6R_I}{4} = -1.5R_I$. - **EJ (vertical):** at the unloaded top joint $E$, $F_{EJ} = -0.8F_{EI}$, and $F_{EI} = V/0.8$ from the shear $V$ in panel $JI$, so $F_{EJ} = -V$. For a load at or left of $J$: $V = R_N - 1$, so $F_{EJ} = 1 - R_N = \dfrac{x}{15}$; for a load at or right of $I$: $F_{EJ} = -R_N$ (0 at $I$, $+0.2$ at $H$). - **FH (diagonal, overhang panel $IH$):** joint $G$ is unloaded and has only the two non-collinear members $FG$ and $GH$, so $F_{FG} = F_{GH} = 0$. At joint $H$ the remaining members are $IH$ and $FH$. For a load left of $I$, $H$ is unloaded and $F_{FH} = 0$. For the load at $H$: $0.8\,F_{FH} = 1$, so $F_{FH} = +1.25$ (tension). #### Ordinates (per unit load on the bottom chord; + tension, − compression) | Member | N | M | L | K | J | I | H | |---|---|---|---|---|---|---|---| | AB | 0.000 | -0.600 | -0.450 | -0.300 | -0.150 | 0.000 | +0.150 | | CD | 0.000 | -0.300 | -0.600 | -0.900 | -0.450 | 0.000 | +0.450 | | EJ | 0.000 | 0.200 | 0.400 | 0.600 | 0.800 | 0.000 | 0.200 | | FH | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 1.250 | #### Shapes - **AB:** 0 at $N$, compression peak $-0.6$ at $M$, decreasing linearly to 0 at the roller $I$, then tension on the overhang ($+0.15$ at $H$). - **CD:** compression, a triangle with peak $-0.9$ under $K$, zero at $I$, then tension on the overhang ($+0.45$ at $H$). - **EJ:** tension, rising linearly to $+0.8$ at $J$, falling to 0 at $I$, then rising again to $+0.2$ at $H$. - **FH:** zero from $N$ to $I$, then rising linearly to $+1.25$ (tension) at $H$. The force in any member for the actual loads is $F = \sum P\,y$ (concentrated loads) or $w\times$ area (UDL), using these ordinates. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For top chord AB in panel NM, the moment centre is M (x = 3 m) with lever arm 4 m. Why M? A. Because M is the hinge support B. Because AM and NM, the other two cut members, meet at M C. Because the load is at M D. Because M is the mid-point of AB Answer: B. The section cuts AB, AM and NM; AM and NM intersect at M, so taking moments about M leaves only AB. (Key point: Moment centre for AB) 2. What is the peak ordinate of the CD influence line (moment centre K at x = 9 m, lever arm 4 m, span 18 m)? A. -1.125 B. -0.750 C. -2.250 D. -0.375 Answer: A. At K, R_N = 0.5, so M_K = 0.5 x 9 = 4.5 per unit load and F_CD = -4.5/4 = -1.125 (compression). (Key point: CD ordinate) 3. How is the force in the vertical EJ obtained? A. From moments about N B. F_EJ = -0.8 F_EI at the unloaded joint E, so it equals minus the shear of panel JI C. It is always zero D. It equals the reaction R_H Answer: B. At unloaded joint E the vertical balances the diagonal's vertical component, and the diagonal carries panel shear, so F_EJ = -V of panel JI. (Key point: Vertical EJ from panel shear) 4. For diagonal FH, F_FH = -1.25 R_H for load left of I. Why the factor 1.25? A. Because the diagonal is 1.25 m long B. Because 1/sin(theta) = 1/0.8 converts shear into diagonal force C. Because of a 25 percent impact allowance D. Because R_H is multiplied by the height 4 m / 3.2 m Answer: B. The diagonal (3-4-5 geometry) has sin(theta) = 0.8, so F = V/0.8 = 1.25 V. 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