Lab S2 Influence Lines

Most structures are designed for loads that stay where they are: the weight of a floor, the water in a tank. A bridge, a crane girder or a gantry rail carries loads that move, and the designer must answer a question that a shear-force diagram cannot: where must the vehicle stand to produce the largest reaction at a pier, the largest shear at a section, or the largest moment at midspan? The influence line answers it directly.

Kebanyakan struktur direka untuk load yang kekal di tempatnya: berat lantai, air dalam tangki. Jambatan, crane girder atau gantry rail pula membawa load yang bergerak, dan pereka perlu menjawab soalan yang tidak dapat dijawab oleh shear-force diagram: di mana kenderaan perlu berada untuk menghasilkan reaction terbesar pada pier, shear terbesar pada sesuatu section, atau moment terbesar di midspan? Influence line menjawabnya secara terus.

For a linear elastic structure the response is proportional to the load and the effects of several loads add (superposition). That is what makes the influence line useful: once \(\eta_R(x)\) is known for a unit load, the effect of any pattern of concentrated or distributed loads follows by multiplication and addition (Section 5).

Bagi struktur linear elastic, response berkadar dengan load dan kesan beberapa load boleh dijumlahkan (superposition). Inilah yang menjadikan influence line berguna: sebaik sahaja \(\eta_R(x)\) diketahui untuk unit load, kesan sebarang susunan concentrated load atau distributed load diperoleh melalui pendaraban dan penambahan (Section 5).

The model is a cantilever bridge of the type patented by Heinrich Gerber in 1866 and used, at a very different scale, in the Forth Bridge: two anchor spans with cantilever arms, and a suspended span that rests on the tips of the arms through hinges.

Model ini ialah cantilever bridge jenis yang dipatenkan oleh Heinrich Gerber pada 1866 dan digunakan, pada skala yang jauh berbeza, pada Forth Bridge: dua anchor span dengan cantilever arm, dan suspended span yang terletak di hujung arm melalui hinge.

Members. Beam ABC is simply supported at A and B with a cantilever BC of 190 mm. Beam DEF is its mirror image. The suspended span CD (250 mm) rests on the cantilever tips: the hinges transmit a vertical force but no bending moment.

Anggota. Beam ABC disokong mudah di A dan B dengan cantilever BC sepanjang 190 mm. Beam DEF ialah imej cerminnya. Suspended span CD (250 mm) terletak di hujung cantilever: hinge memindahkan daya menegak tetapi tidak memindahkan bending moment.

Supports and hingesSupport dan hinge

A support or a joint is described by what it stops: every movement it prevents produces a force (or moment) in that direction, and every movement it allows produces none.

Sesuatu support atau sambungan diterangkan berdasarkan apa yang ia halang: setiap pergerakan yang dihalang menghasilkan daya (atau moment) pada arah itu, dan setiap pergerakan yang dibenarkan tidak menghasilkan apa-apa.

Because a hinge carries no moment, each hinge adds one equation, \(M = 0\) at that point, and this is what makes the model statically determinate (next paragraph). In the model the supports at A and F can also pull downwards: the gauges there register uplift as a negative reading.

Oleh kerana hinge tidak membawa moment, setiap hinge menambah satu persamaan, \(M = 0\) pada titik itu, dan inilah yang menjadikan model ini statically determinate (perenggan seterusnya). Dalam model ini, support di A dan F juga boleh menarik ke bawah: gauge di situ merekodkan uplift sebagai bacaan negatif.

Determinacy. No horizontal load is applied, so there are four unknown vertical reactions \(R_A, R_B, R_E, R_F\). Statics gives two useful equations for the whole structure (\(\sum F_y = 0\), \(\sum M = 0\)) and each internal hinge one more (\(M = 0\) at C and at D): four equations for four unknowns. The structure is statically determinate, every influence line follows from equilibrium alone with no need for \(EI\), and every influence line is made of straight segments.

Determinacy. Tiada load mendatar dikenakan, jadi terdapat empat reaction menegak yang tidak diketahui, \(R_A, R_B, R_E, R_F\). Statik memberi dua persamaan berguna bagi keseluruhan struktur (\(\sum F_y = 0\), \(\sum M = 0\)) dan setiap hinge dalaman memberi satu lagi (\(M = 0\) di C dan di D): empat persamaan untuk empat anu. Struktur ini statically determinate; setiap influence line diperoleh daripada equilibrium sahaja tanpa memerlukan \(EI\), dan setiap influence line terdiri daripada segmen lurus.

Sign convention. Support reactions are positive upwards. The hinge forces \(R_C\) and \(R_D\) are the forces the suspended span presses on the cantilever tips (positive downwards on the tip); they are what the gauges at C and D register. Positions are measured from A, \(0 \le x \le 1130\) mm.

Sign convention. Support reaction positif ke atas. Hinge force \(R_C\) dan \(R_D\) ialah daya yang dikenakan oleh suspended span ke atas hujung cantilever (positif ke bawah pada hujung); inilah yang direkod oleh gauge di C dan D. Kedudukan diukur dari A, \(0 \le x \le 1130\) mm.

The derivation uses one idea three times: take moments about a support. The order matters: the suspended span is analysed first, because it is the only member whose loads reach the others.

Derivation ini menggunakan satu idea sebanyak tiga kali: ambil moment terhadap support. Urutan penting: suspended span dianalisis dahulu, kerana hanya anggota ini yang load-nya sampai kepada anggota lain.

For \(R_A\): remove the support at A and lift that end by one unit. Beam ABC is rigid between its support and the hinge, so it rotates about B: A rises by 1 and the tip C, 190 mm on the other side of B, drops by \(190/250 = 0.76\). The suspended span follows C and stays straight, dropping linearly to zero at D; beam DEF does not move. The result is exactly the influence line of Section 3, signs included.

Bagi \(R_A\): buang support di A dan angkat hujung itu sebanyak satu unit. Beam ABC adalah rigid antara support dan hinge, jadi ia berputar terhadap B: A naik sebanyak 1 dan hujung C, 190 mm di sebelah lain B, turun sebanyak \(190/250 = 0.76\). Suspended span mengikut C dan kekal lurus, turun secara linear ke sifar di D; beam DEF tidak bergerak. Hasilnya tepat sama dengan influence line dalam Section 3, termasuk tanda.

Two remarks. The principle gives exact ordinates only for statically determinate structures, where the released structure is a mechanism of straight rigid links; for an indeterminate beam the released structure bends and the principle gives the shape only. And the kinks of the deflected shape lie at the hinges and at the released restraint, never inside a rigid member: a quick visual check of any influence line you compute.

Dua catatan. Prinsip ini memberi ordinate yang tepat hanya bagi struktur statically determinate, di mana released structure ialah mechanism yang terdiri daripada rigid link yang lurus; bagi beam indeterminate, released structure melentur dan prinsip ini hanya memberi bentuk. Selain itu, bengkokan pada deflected shape terletak di hinge dan di restraint yang dilepaskan, tidak sekali-kali di dalam anggota rigid: semakan visual yang pantas bagi mana-mana influence line yang anda kira.

By superposition, for a single load \(P\) at \(x_1\), a series of loads \(P_i\) at \(x_i\), and a uniformly distributed load \(w\) between \(a\) and \(b\):

Melalui superposition, bagi satu load \(P\) di \(x_1\), satu siri load \(P_i\) di \(x_i\), dan uniformly distributed load \(w\) antara \(a\) dan \(b\):

\[ R = P\,\eta_R(x_1),\qquad R = \sum_i P_i\,\eta_R(x_i),\qquad R = \int_a^b w\,\eta_R(x)\,\mathrm{d}x = w\times(\text{area under the IL from } a \text{ to } b). \]

With masses in kg the reactions come out in kgf, which is what the laboratory gauges read.

Dengan jisim dalam kg, reaction yang diperoleh dalam kgf, iaitu unit bacaan gauge makmal.

5.1 Critical position of a series of loads5.1 Critical position bagi satu siri load

Every influence line of the model is piecewise linear, so \(\sum_i P_i\,\eta_R(x_i)\) is a piecewise linear function of the position of the load group, and a piecewise linear function reaches its extremes at a vertex. The largest (or most negative) value therefore occurs when one of the loads stands on a vertex of the influence line; only a few positions need to be tried.

Setiap influence line model ini adalah piecewise linear, jadi \(\sum_i P_i\,\eta_R(x_i)\) ialah fungsi piecewise linear terhadap kedudukan kumpulan load, dan fungsi piecewise linear mencapai nilai ekstrem pada vertex. Nilai terbesar (atau paling negatif) oleh itu berlaku apabila salah satu load berada pada vertex influence line; hanya beberapa kedudukan perlu dicuba.

5.2 Distributed load: area under the influence line5.2 Distributed load: area di bawah influence line

The areas under \(\eta_A\) are, by the triangle rule, \(+\tfrac12(250)(1.0) = +125\) mm over AB, \(\tfrac12(190)(-0.76) = -72.2\) mm over BC and \(\tfrac12(250)(-0.76) = -95.0\) mm over CD, and zero over DEF.

Area di bawah \(\eta_A\), menggunakan petua segi tiga, ialah \(+\tfrac12(250)(1.0) = +125\) mm pada AB, \(\tfrac12(190)(-0.76) = -72.2\) mm pada BC dan \(\tfrac12(250)(-0.76) = -95.0\) mm pada CD, dan sifar pada DEF.

What is measured. A 2.5 kg mass is placed at each of the seventeen halts in turn (Test 1), then the 5 kg + 2.5 kg pair (Test 2). At each halt the six dial gauges are read. One division corresponds to 0.05 kgf, so the experimental ordinate per unit load is

Apa yang diukur. Jisim 2.5 kg diletakkan pada setiap tujuh belas halt secara bergilir (Test 1), kemudian pasangan 5 kg + 2.5 kg (Test 2). Pada setiap halt, enam dial gauge dibaca. Satu division bersamaan 0.05 kgf, jadi ordinate eksperimen per unit load ialah

\[ \eta_{\text{test}} = \frac{(\text{reading in divisions})\times 0.05}{m},\qquad m = 2.5\ \text{kg (Test 1)}. \]

For Test 2, reading × 0.05 is the reaction in kgf, comparable with the compound-load table.

Bagi Test 2, bacaan × 0.05 ialah reaction dalam kgf, boleh dibandingkan dengan jadual compound load.

How to compare. Report the difference \(\eta_{\text{test}}-\eta_{\text{theory}}\) at every halt. A percentage difference is meaningful only where the theoretical ordinate is not small: at halt 2, where \(\eta_A = 0.2\), a gauge error of one division (0.02 per unit load) is already 10 %, and where the theoretical value is zero the percentage is undefined. Judge the agreement by the absolute difference and by the shape of the line.

Cara membandingkan. Laporkan perbezaan \(\eta_{\text{test}}-\eta_{\text{theory}}\) pada setiap halt. Perbezaan peratusan hanya bermakna apabila ordinate teori tidak kecil: di halt 2, di mana \(\eta_A = 0.2\), error gauge satu division (0.02 per unit load) sudah 10 %, dan apabila nilai teori sifar, peratusan tidak tertakrif. Nilai persetujuan berdasarkan perbezaan mutlak dan bentuk garis.

Equilibrium check. With the mass at any halt the four support gauges should add up to the applied mass: \((a+b+e+f)\times0.05 \approx 2.5\) kgf. Check it before leaving the laboratory, while the readings can still be repeated. The Lab tab does this for you.

Equilibrium check. Dengan jisim di mana-mana halt, jumlah bacaan empat support gauge sepatutnya sama dengan jisim yang dikenakan: \((a+b+e+f)\times0.05 \approx 2.5\) kgf. Semak sebelum keluar makmal, ketika bacaan masih boleh diulang. Tab Makmal melakukannya untuk anda.

Combining the first two in quadrature gives about ±0.03 per unit load, the same at every halt. Against an ordinate of 1.0 at A this is 3 %; against 0.2 at halt 2 it is 15 %. That is the quantitative reason for comparing by absolute difference.

Menggabungkan dua yang pertama secara quadrature memberi lebih kurang ±0.03 per unit load, sama pada setiap halt. Berbanding ordinate 1.0 di A, ini 3 %; berbanding 0.2 di halt 2, ia 15 %. Itulah sebab kuantitatif untuk membandingkan dengan perbezaan mutlak.

Bridge decks are designed for standard vehicle and lane loads placed, by means of influence lines or influence surfaces, where they produce the largest effects at each section. For continuous (indeterminate) girders the influence lines are curved and need an elastic analysis, but the Müller–Breslau sketch still tells the engineer which spans to load. The cantilever bridge, with its determinate hinges, made the long steel spans of the late nineteenth century possible because its forces could be found by statics alone; the price is end supports that must be designed for uplift, as Worked example 2 shows.

Deck jambatan direka untuk standard vehicle load dan lane load yang diletakkan, menggunakan influence line atau influence surface, di kedudukan yang menghasilkan kesan terbesar pada setiap section. Bagi continuous girder (indeterminate), influence line melengkung dan memerlukan analisis elastik, tetapi lakaran Müller–Breslau masih memberitahu jurutera span mana yang perlu dibebani. Cantilever bridge, dengan hinge yang determinate, membolehkan span keluli yang panjang pada akhir abad kesembilan belas kerana dayanya boleh dicari dengan statik sahaja; harganya ialah end support yang mesti direka untuk uplift, seperti yang ditunjukkan dalam Worked example 2.