Support reactions. Therefore, \[A_{y}=B_{y}=\frac{w L}{2}=\frac{0.6(100)}{2}=30 \text { kips } \nonumber\]. A three-hinged arch is subjected to two concentrated loads, as shown in Figure 6.3a. HWnH+8spxcd r@=$m'?ERf`|U]b+?mj]. WebThe Mega-Truss Pick will suspend up to one ton of truss load, plus an additional one ton load suspended under the truss. 0000010481 00000 n As per its nature, it can be classified as the point load and distributed load. 6.9 A cable subjected to a uniform load of 300 N/m is suspended between two supports at the same level 20 m apart, as shown in Figure P6.9. WebAttic truss with 7 feet room height should it be designed for 20 psf (pounds per square foot), 30 psf or 40 psf room live load? Another In. \newcommand{\lbm}[1]{#1~\mathrm{lbm} } 6.3 Determine the shear force, axial force, and bending moment at a point under the 80 kN load on the parabolic arch shown in Figure P6.3. We can see the force here is applied directly in the global Y (down). You may freely link Problem 11P: For the truss of Problem 8.51, determine the maximum tensile and compressive axial forces in member DI due to a concentrated live load of 40 k, a uniformly distributed live load of 4 k/ft, and a uniformly distributed dead load of 2 k/ft. A uniformly distributed load is a type of load which acts in constant intensity throughout the span of a structural member. \sum F_x \amp = 0 \rightarrow \amp A_x \amp = 0 Sometimes distributed loads (DLs) on the members of a structure follow a special distribution that cannot be idealized with a single constant one or even a nonuniform linear distributed load, and therefore non-linear distributed loads are needed. 0000155554 00000 n Users can also apply a DL to a member by first selecting a member, then right-clicking and selecting Add Distributed Load, which will bring you to the Distributed Load input screen with the member ID field already filled. submitted to our "DoItYourself.com Community Forums". Roof trusses can be loaded with a ceiling load for example. Since all loads on a truss must act at the joints, the distributed weight of each member must be split between the two joints. You may have a builder state that they will only use the room for storage, and they have no intention of using it as a living space. The value can be reduced in the case of structures with spans over 50 m by detailed statical investigation of rain, sand/dirt, fallen leaves loading, etc. \newcommand{\kg}[1]{#1~\mathrm{kg} } \end{align*}. \end{align*}, This total load is simply the area under the curve, \begin{align*} The snow load should be considered even in areas that are not usually subjected to snow loading, as a nominal uniformly distributed load of 0.3 kN/m 2 . { "1.01:_Introduction_to_Structural_Analysis" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.02:_Structural_Loads_and_Loading_System" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.03:_Equilibrium_Structures_Support_Reactions_Determinacy_and_Stability_of_Beams_and_Frames" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.04:_Internal_Forces_in_Beams_and_Frames" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.05:_Internal_Forces_in_Plane_Trusses" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.06:_Arches_and_Cables" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.07:_Deflection_of_Beams-_Geometric_Methods" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.08:_Deflections_of_Structures-_Work-Energy_Methods" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.09:_Influence_Lines_for_Statically_Determinate_Structures" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.10:_Force_Method_of_Analysis_of_Indeterminate_Structures" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.11:_Slope-Deflection_Method_of_Analysis_of_Indeterminate_Structures" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.12:_Moment_Distribution_Method_of_Analysis_of_Structures" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.13:_Influence_Lines_for_Statically_Indeterminate_Structures" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "01:_Chapters" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "license:ccbyncnd", "licenseversion:40", "authorname:fudoeyo", "source@https://temple.manifoldapp.org/projects/structural-analysis" ], https://eng.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Feng.libretexts.org%2FBookshelves%2FCivil_Engineering%2FBook%253A_Structural_Analysis_(Udoeyo)%2F01%253A_Chapters%2F1.06%253A_Arches_and_Cables, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), 6.1.2.1 Derivation of Equations for the Determination of Internal Forces in a Three-Hinged Arch. Cables are used in suspension bridges, tension leg offshore platforms, transmission lines, and several other engineering applications. When applying the DL, users need to specify values for: Heres an example where the distributed load has a -10kN/m Start Y magnitude and a -30kN/m end Y magnitude. If a Uniformly Distributed Load (UDL) of the intensity of 30 kN/m longer than the span traverses, then the maximum compression in the member is (Upper Triangular area is of Tension, Lower Triangle is of Compression) This question was previously asked in 8 0 obj \newcommand{\MN}[1]{#1~\mathrm{MN} } \newcommand{\aUS}[1]{#1~\mathrm{ft}/\mathrm{s}^2 } Minimum height of habitable space is 7 feet (IRC2018 Section R305). The free-body diagram of the entire arch is shown in Figure 6.6b. WebA uniform distributed load is a force that is applied evenly over the distance of a support. \newcommand{\ft}[1]{#1~\mathrm{ft}} DoItYourself.com, founded in 1995, is the leading independent \sum F_y\amp = 0\\ 0000004878 00000 n The examples below will illustrate how you can combine the computation of both the magnitude and location of the equivalent point force for a series of distributed loads. 0000003968 00000 n \newcommand{\kN}[1]{#1~\mathrm{kN} } f = rise of arch. 0000090027 00000 n A cable supports a uniformly distributed load, as shown Figure 6.11a. Once you convert distributed loads to the resultant point force, you can solve problem in the same manner that you have other problems in previous chapters of this book. {x&/~{?wfi_h[~vghK %qJ(K|{- P([Y~];hc0Fk r1 oy>fUZB[eB]Y^1)aHG?!9(/TSjM%1odo1 0GQ'%O\A/{j%LN?\|8`q8d31l.u.L)NJVK5Z/ VPYi00yt $Y1J"gOJUu|_|qbqx3.t!9FLB,!FQtt$VFrb@`}ILP}!@~8Rt>R2Mw00DJ{wovU6E R6Oq\(j!\2{0I9'a6jj5I,3D2kClw}InF`Mx|*"X>] R;XWmC mXTK*lqDqhpWi&('U}[q},"2`nazv}K2 }iwQbhtb Or`x\Tf$HBwU'VCv$M T9~H t 27r7bY`r;oyV{Ver{9;@A@OIIbT!{M-dYO=NKeM@ogZpIb#&U$M1Nu$fJ;2[UM0mMS4!xAp2Dw/wH 5"lJO,Sq:Xv^;>= WE/ _ endstream endobj 225 0 obj 1037 endobj 226 0 obj << /Filter /FlateDecode /Length 225 0 R >> stream In most real-world applications, uniformly distributed loads act over the structural member. \newcommand{\mm}[1]{#1~\mathrm{mm}} A uniformly distributed load is spread over a beam so that the rate of loading w is uniform along the length (i.e., each unit length is loaded at the same rate). \end{equation*}, Distributed loads may be any geometric shape or defined by a mathematical function. All information is provided "AS IS." w(x) = \frac{\N{3}}{\cm{3}}= \Nperm{100}\text{.} 0000003744 00000 n \end{align*}. \end{equation*}, \begin{align*} P)i^,b19jK5o"_~tj.0N,V{A. Point load force (P), line load (q). The programs will even notify you if needed numbers or elements are missing or do not meet the requirements for your structure. The following procedure can be used to evaluate the uniformly distributed load. Support reactions. It includes the dead weight of a structure, wind force, pressure force etc. The formula for any stress functions also depends upon the type of support and members. by Dr Sen Carroll. Determine the support reactions of the arch. Well walk through the process of analysing a simple truss structure. A uniformly distributed load is the load with the same intensity across the whole span of the beam. \newcommand{\Nperm}[1]{#1~\mathrm{N}/\mathrm{m} } The line of action of the equivalent force acts through the centroid of area under the load intensity curve. This page titled 1.6: Arches and Cables is shared under a CC BY-NC-ND 4.0 license and was authored, remixed, and/or curated by Felix Udoeyo via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. The three internal forces at the section are the axial force, NQ, the radial shear force, VQ, and the bending moment, MQ. Web48K views 3 years ago Shear Force and Bending Moment You can learn how to calculate shear force and bending moment of a cantilever beam with uniformly distributed load In order for a roof truss load to be stable, you need to assign two of your nodes on each truss to be used as support nodes. % WebDistributed loads are forces which are spread out over a length, area, or volume. Support reactions. The magnitude of the distributed load of the books is the total weight of the books divided by the length of the shelf, \begin{equation*} WebWhen a truss member carries compressive load, the possibility of buckling should be examined. Essentially, were finding the balance point so that the moment of the force to the left of the centroid is the same as the moment of the force to the right. kN/m or kip/ft). For example, the dead load of a beam etc. To determine the vertical distance between the lowest point of the cable (point B) and the arbitrary point C, rearrange and further integrate equation 6.13, as follows: Summing the moments about C in Figure 6.10b suggests the following: Applying Pythagorean theory to Figure 6.10c suggests the following: T and T0 are the maximum and minimum tensions in the cable, respectively. 6.5 A cable supports three concentrated loads at points B, C, and D in Figure P6.5. The free-body diagram of the entire arch is shown in Figure 6.4b, while that of its segment AC is shown in Figure 6.4c. Follow this short text tutorial or watch the Getting Started video below. A cable supports three concentrated loads at B, C, and D, as shown in Figure 6.9a. A_y = \lb{196.7}, A_x = \lb{0}, B_y = \lb{393.3} stream W = w(x) \ell = (\Nperm{100})(\m{6}) = \N{600}\text{.} \newcommand{\jhat}{\vec{j}} The rest of the trusses only have to carry the uniformly distributed load of the closed partition, and may be designed for this lighter load. \newcommand{\Pa}[1]{#1~\mathrm{Pa} } is the load with the same intensity across the whole span of the beam. Here is an example of where member 3 has a 100kN/m distributed load applied to itsGlobalaxis. Copyright 2023 by Component Advertiser home improvement and repair website. It also has a 20% start position and an 80% end position showing that it does not extend the entire span of the member, but rather it starts 20% from the start and end node (1 and 2 respectively). \newcommand{\aSI}[1]{#1~\mathrm{m}/\mathrm{s}^2 } The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Applying the equations of static equilibrium suggests the following: Solving equations 6.1 and 6.2 simultaneously yields the following: A parabolic arch with supports at the same level is subjected to the combined loading shown in Figure 6.4a. The highway load consists of a uniformly distributed load of 9.35 kN/m and a concentrated load of 116 kN. Determine the support reactions and the normal thrust and radial shear at a point just to the left of the 150 kN concentrated load. This is due to the transfer of the load of the tiles through the tile A roof truss is a triangular wood structure that is engineered to hold up much of the weight of the roof. You can learn how to calculate shear force and bending moment of a cantilever beam with uniformly distributed load (UDL) and also to draw shear force and bending moment diagrams. The free-body diagrams of the entire arch and its segment CE are shown in Figure 6.3b and Figure 6.3c, respectively. \Sigma F_x \amp = 0 \amp \amp \rightarrow \amp A_x \amp = 0\\ A_y \amp = \N{16}\\ The remaining portions of the joists or truss bottom chords shall be designed for a uniformly distributed concurrent live load of not less than 10 lb/ft 2 Note that, in footnote b, the uninhabitable attics without storage have a 10 psf live load that is non-concurrent with other IRC (International Residential Code) defines Habitable Space as a space in a building for living, sleeping, eating, or cooking. Arches are structures composed of curvilinear members resting on supports. \newcommand{\lbperin}[1]{#1~\mathrm{lb}/\mathrm{in} } 6.6 A cable is subjected to the loading shown in Figure P6.6. Use this truss load equation while constructing your roof. \newcommand{\inlb}[1]{#1~\mathrm{in}\!\cdot\!\mathrm{lb} } To determine the normal thrust and radial shear, find the angle between the horizontal and the arch just to the left of the 150 kN load. The bar has uniform cross-section A = 4 in 2, is made by aluminum (E = 10, 000 ksi), and is 96 in long.A uniformly distributed axial load q = I ki p / in is applied throughout the length. This step is recommended to give you a better idea of how all the pieces fit together for the type of truss structure you are building. This is a load that is spread evenly along the entire length of a span. Since youre calculating an area, you can divide the area up into any shapes you find convenient. problems contact webmaster@doityourself.com. 0000047129 00000 n The expression of the shape of the cable is found using the following equations: For any point P(x, y) on the cable, apply cable equation. They can be either uniform or non-uniform. +(B_y) (\inch{18}) - (\lbperin{12}) (\inch{10}) (\inch{29})\amp = 0 \rightarrow \amp B_y \amp= \lb{393.3}\\ W \amp = w(x) \ell\\ GATE CE syllabuscarries various topics based on this. 0000007236 00000 n Based on their geometry, arches can be classified as semicircular, segmental, or pointed. Distributed loads (DLs) are forces that act over a span and are measured in force per unit of length (e.g. These loads are expressed in terms of the per unit length of the member. x = horizontal distance from the support to the section being considered. For a rectangular loading, the centroid is in the center. 0000001392 00000 n %PDF-1.2 The horizontal thrust at both supports of the arch are the same, and they can be computed by considering the free body diagram in Figure 6.5b. The relationship between shear force and bending moment is independent of the type of load acting on the beam. Note that while the resultant forces are, Find the reactions at the fixed connection at, \begin{align*} A cantilever beam is a determinate beam mostly used to resist the hogging type bending moment. They are used for large-span structures, such as airplane hangars and long-span bridges. 0000011431 00000 n Three-pinned arches are determinate, while two-pinned arches and fixed arches, as shown in Figure 6.1, are indeterminate structures. In the literature on truss topology optimization, distributed loads are seldom treated. \newcommand{\slug}[1]{#1~\mathrm{slug}} Taking B as the origin and denoting the tensile horizontal force at this origin as T0 and denoting the tensile inclined force at C as T, as shown in Figure 6.10b, suggests the following: Equation 6.13 defines the slope of the curve of the cable with respect to x. This equivalent replacement must be the. I am analysing a truss under UDL. If we change the axes option toLocalwe can see that the distributed load has now been applied to the members local axis, where local Y is directly perpendicular to the member. The length of the cable is determined as the algebraic sum of the lengths of the segments. WebThree-Hinged Arches - Continuous and Point Loads - Support reactions and bending moments. \end{equation*}, The line of action of this equivalent load passes through the centroid of the rectangular loading, so it acts at. 0000016751 00000 n WebA 75 mm 150 mm beam carries a uniform load wo over the entire span of 1.2 m. Square notches 25 mm deep are provided at the bottom of the beam at the supports. WebAnswer: I Will just analyse this such that a Structural Engineer will grasp it in simple look. w(x) = \frac{\Sigma W_i}{\ell}\text{.} View our Privacy Policy here. 6.4 In Figure P6.4, a cable supports loads at point B and C. Determine the sag at point C and the maximum tension in the cable. These parameters include bending moment, shear force etc. 0000009351 00000 n 6.1 Determine the reactions at supports B and E of the three-hinged circular arch shown in Figure P6.1. In contrast, the uniformly varying load has zero intensity at one end and full load intensity at the other. H|VMo6W1R/@ " -^d/m+]I[Q7C^/a`^|y3;hv? A cantilever beam has a maximum bending moment at its fixed support when subjected to a uniformly distributed load and significant for theGATE exam. \newcommand{\kNm}[1]{#1~\mathrm{kN}\!\cdot\!\mathrm{m} } Applying the equations of static equilibrium to determine the archs support reactions suggests the following: Normal thrust and radial shear. To be equivalent, the point force must have a: Magnitude equal to the area or volume under the distributed load function. \end{equation*}, Start by drawing a free-body diagram of the beam with the two distributed loads replaced with equivalent concentrated loads. This means that one is a fixed node and the other is a rolling node. Assume the weight of each member is a vertical force, half of which is placed at each end of the member as shown in the diagram on the left. Users can also get to that menu by navigating the top bar to Edit > Loads > Non-linear distributed loads. (a) ( 10 points) Using basic mechanics concepts, calculate the theoretical solution of the \text{total weight} \amp = \frac{\text{weight}}{\text{length}} \times\ \text{length of shelf} *wr,. \newcommand{\cm}[1]{#1~\mathrm{cm}} If the number of members is labeled M and the number of nodes is labeled N, this can be written as M+3=2*N. Both sides of the equation should be equal in order to end up with a stable and secure roof structure. This confirms the general cable theorem. For Example, the maximum bending moment for a simply supported beam and cantilever beam having a uniformly distributed load will differ. Sometimes, a tie is provided at the support level or at an elevated position in the arch to increase the stability of the structure. 6.2 Determine the reactions at supports A and B of the parabolic arch shown in Figure P6.2. To apply a DL, go to the input menu on the left-hand side and click on the Distributed Load button. HA loads to be applied depends on the span of the bridge. Determine the total length of the cable and the length of each segment. I have a new build on-frame modular home. The concept of the load type will be clearer by solving a few questions. Questions of a Do It Yourself nature should be WebIn truss analysis, distributed loads are transformed into equivalent nodal loads, and the eects of bending are neglected. All rights reserved. A parabolic arch is subjected to a uniformly distributed load of 600 lb/ft throughout its span, as shown in Figure 6.5a. \end{align*}, \(\require{cancel}\let\vecarrow\vec Given a distributed load, how do we find the location of the equivalent concentrated force? Maximum Reaction. This is a quick start guide for our free online truss calculator. \(M_{(x)}^{b}\)= moment of a beam of the same span as the arch. Since all loads on a truss must act at the joints, the distributed weight of each member must be split between the \newcommand{\second}[1]{#1~\mathrm{s} } If those trusses originally acting as unhabitable attics turn into habitable attics down the road, and the homeowner doesnt check into it, then those trusses could be under designed. To develop the basic relationships for the analysis of parabolic cables, consider segment BC of the cable suspended from two points A and D, as shown in Figure 6.10a. This step can take some time and patience, but it is worth arriving at a stable roof truss structure in order to avoid integrity problems and costly repairs in the future. w(x) \amp = \Nperm{100}\\ Determine the total length of the cable and the tension at each support. 0000113517 00000 n You can add or remove nodes and members at any time in order to get the numbers to balance out, similar in concept to balancing both sides of a scale. The horizontal thrust at both supports of the arch are the same, and they can be computed by considering the free body diagram in Figure 6.5c. 0000006074 00000 n Under a uniform load, a cable takes the shape of a curve, while under a concentrated load, it takes the form of several linear segments between the loads points of application. Determine the support reactions and the bending moment at a section Q in the arch, which is at a distance of 18 ft from the left-hand support. at the fixed end can be expressed as Substituting Ay from equation 6.8 into equation 6.7 suggests the following: To obtain the expression for the moment at a section x from the right support, consider the beam in Figure 6.7b. Distributed loads (DLs) are forces that act over a span and are measured in force per unit of length (e.g. suggestions. 0000001790 00000 n This will help you keep track of them while installing each triangular truss and it can be a handy reference for which nodes you have assigned as load-bearing, fixed, and rolling. 0000006097 00000 n Per IRC 2018 section R304 habitable rooms shall have a floor area of not less than 70 square feet and not less than 7 feet in any horizontal dimension (except kitchens). Note the lengths of your roof truss members on your sketch, and mark where each node will be placed as well. trailer << /Size 257 /Info 208 0 R /Root 211 0 R /Prev 646755 /ID[<8e2a910c5d8f41a9473430b52156bc4b>] >> startxref 0 %%EOF 211 0 obj << /Type /Catalog /Pages 207 0 R /Metadata 209 0 R /StructTreeRoot 212 0 R >> endobj 212 0 obj << /Type /StructTreeRoot /K 65 0 R /ParentTree 189 0 R /ParentTreeNextKey 7 /RoleMap 190 0 R /ClassMap 191 0 R >> endobj 255 0 obj << /S 74 /C 183 /Filter /FlateDecode /Length 256 0 R >> stream 0000001531 00000 n 0000008311 00000 n \newcommand{\ftlb}[1]{#1~\mathrm{ft}\!\cdot\!\mathrm{lb} } WebThe Influence Line Diagram (ILD) for a force in a truss member is shown in the figure. \\ \end{equation*}, \begin{equation*} The equivalent load is the area under the triangular load intensity curve and it acts straight down at the centroid of the triangle. Uniformly distributed load acts uniformly throughout the span of the member. GATE Exam Eligibility 2024: Educational Qualification, Nationality, Age limit. \newcommand{\psf}[1]{#1~\mathrm{lb}/\mathrm{ft}^2 } 0000002965 00000 n Consider the section Q in the three-hinged arch shown in Figure 6.2a. 0000072414 00000 n The general cable theorem states that at any point on a cable that is supported at two ends and subjected to vertical transverse loads, the product of the horizontal component of the cable tension and the vertical distance from that point to the cable chord equals the moment which would occur at that section if the load carried by the cable were acting on a simply supported beam of the same span as that of the cable.

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