This triangular loading has a, \begin{equation*} {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 It will also be equal to the slope of the bending moment curve. \newcommand{\mm}[1]{#1~\mathrm{mm}} To find the bending moments at sections of the arch subjected to concentrated loads, first determine the ordinates at these sections using the equation of the ordinate of a parabola, which is as follows: When considering the beam in Figure 6.6d, the bending moments at B and D can be determined as follows: Cables are flexible structures that support the applied transverse loads by the tensile resistance developed in its members. A fixed node will provide support in both directions down the length of the roof truss members, often called the X and Y-directions. 0000089505 00000 n The example in figure 9 is a common A type gable truss with a uniformly distributed load along the top and bottom chords. WebHA loads are uniformly distributed load on the bridge deck. WebThe only loading on the truss is the weight of each member. 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. \newcommand{\khat}{\vec{k}} So, the slope of the shear force diagram for uniformly distributed load is constant throughout the span of a beam. P)i^,b19jK5o"_~tj.0N,V{A. \[N_{\varphi}=-A_{y} \cos \varphi-A_{x} \sin \varphi=-V^{b} \cos \varphi-A_{x} \sin \varphi \label{6.5}\]. Cable with uniformly distributed load. 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). Formulas for GATE Civil Engineering - Fluid Mechanics, Formulas for GATE Civil Engineering - Environmental Engineering. %PDF-1.2 \), Relation between Vectors and Unit Vectors, Relations between Centroids and Center of gravity, Relation Between Loading, Shear and Moment, Moment of Inertia of a Differential Strip, Circles, Semicircles, and Quarter-circles, \((\inch{10}) (\lbperin{12}) = \lb{120}\). The equivalent load is the area under the triangular load intensity curve and it acts straight down at the centroid of the triangle. The rate of loading is expressed as w N/m run. (a) ( 10 points) Using basic mechanics concepts, calculate the theoretical solution of the Find the equivalent point force and its point of application for the distributed load shown. \newcommand{\inch}[1]{#1~\mathrm{in}} at the fixed end can be expressed as: R A = q L (3a) where . They are used for large-span structures. 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. \newcommand{\kgqm}[1]{#1~\mathrm{kg}/\mathrm{m}^3 } Support reactions. 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. 0000072621 00000 n \text{total weight} \amp = \frac{\text{weight}}{\text{length}} \times\ \text{length of shelf} Per IRC 2018 Table R301.5 minimum uniformly distributed live load for habitable attics and attics served with fixed stairs is 30 psf. The reactions of the cable are determined by applying the equations of equilibrium to the free-body diagram of the cable shown in Figure 6.8b, which is written as follows: Sag at B. They are used in different engineering applications, such as bridges and offshore platforms. 0000002965 00000 n \definecolor{fillinmathshade}{gray}{0.9} \newcommand{\aUS}[1]{#1~\mathrm{ft}/\mathrm{s}^2 } 0000001812 00000 n SkyCiv Engineering. fBFlYB,e@dqF| 7WX &nx,oJYu. WebAnswer: I Will just analyse this such that a Structural Engineer will grasp it in simple look. I have a 200amp service panel outside for my main home. \end{equation*}, The total weight is the area under the load intensity diagram, which in this case is a rectangle. WebThe only loading on the truss is the weight of each member. Based on their geometry, arches can be classified as semicircular, segmental, or pointed. A uniformly distributed load is a zero degrees loading curve, so the bending moment curve for such a load will be a two-degree or parabolic curve. In [9], the W \amp = w(x) \ell\\ The free-body diagram of the entire arch is shown in Figure 6.6b. \renewcommand{\vec}{\mathbf} A three-hinged arch is a geometrically stable and statically determinate structure. Questions of a Do It Yourself nature should be 0000003968 00000 n 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. Website operating 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. Similarly, for a triangular distributed load also called a. WebDistributed loads are forces which are spread out over a length, area, or volume. \Sigma M_A \amp = 0 \amp \amp \rightarrow \amp M_A \amp = (\N{16})(\m{4}) \\ A uniformly distributed load is a zero degrees loading curve, so a shear force diagram for such a load will have a one-degree or linear curve. \end{equation*}, Start by drawing a free-body diagram of the beam with the two distributed loads replaced with equivalent concentrated loads. A_y \amp = \N{16}\\ 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. WebCantilever Beam - Uniform Distributed Load. 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{\aSI}[1]{#1~\mathrm{m}/\mathrm{s}^2 } 0000017536 00000 n Thus, MQ = Ay(18) 0.6(18)(9) Ax(11.81). IRC (International Residential Code) defines Habitable Space as a space in a building for living, sleeping, eating, or cooking. Also draw the bending moment diagram for the arch. \Sigma F_x \amp = 0 \amp \amp \rightarrow \amp A_x \amp = 0\\ 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). If the cable has a central sag of 4 m, determine the horizontal reactions at the supports, the minimum and maximum tension in the cable, and the total length of the cable. First i have explained the general cantilever beam with udl by taking load as \"W/m\" and length as \"L\" and next i have solved in detail the numerical example of cantilever beam with udl.____________________________________________________IF THIS CHANNEL HAS HELPED YOU, SUPPORT THIS CHANNEL THROUGH GOOGLE PAY : +919731193970____________________________________________________Concept of shear force and bending moment : https://youtu.be/XR7xUSMDv1ICantilever beam with point load : https://youtu.be/m6d2xj-9ZmM#shearforceandbendingmoment #sfdbmdforudl #sfdbmdforcantileverbeam | Terms Of Use | Privacy Statement |, The Development of the Truss Plate, Part VIII: Patent Skirmishes, Building Your Own Home Part I: Becoming the GC, Reviewing 2021 IBC Changes for Cold-Formed Steel Light-Frame Design, The Development of the Truss Plate, Part VII: Contentious Competition. W \amp = \N{600} The horizontal thrusts significantly reduce the moments and shear forces at any section of the arch, which results in reduced member size and a more economical design compared to other structures. \end{equation*}, The line of action of this equivalent load passes through the centroid of the rectangular loading, so it acts at. w(x) \amp = \Nperm{100}\\ Analysis of steel truss under Uniform Load. \newcommand{\cm}[1]{#1~\mathrm{cm}} For additional information, or if you have questions, please refer to IRC 2018 or contact the MiTek Engineering department. Since youre calculating an area, you can divide the area up into any shapes you find convenient. Copyright 2023 by Component Advertiser To use a distributed load in an equilibrium problem, you must know the equivalent magnitude to sum the forces, and also know the position or line of action to sum the moments. 0000006097 00000 n by Dr Sen Carroll. Portion of the room with a sloping ceiling measuring less than 5 feet or a furred ceiling measuring less than 7 feet from the finished floor to the finished ceiling shall not be considered as contributing to the minimum required habitable area of that room. 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. *B*|SDZxEpm[az,ByV)vONSgf{|M'g/D'l0+xJ XtiX3#B!6`*JpBL4GZ8~zaN\&*6c7/"KCftl QC505%cV$|nv/o_^?_|7"u!>~Nk \end{align*}, This total load is simply the area under the curve, \begin{align*} 6.4 In Figure P6.4, a cable supports loads at point B and C. Determine the sag at point C and the maximum tension in the cable. WebThree-Hinged Arches - Continuous and Point Loads - Support reactions and bending moments. \newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}} A rolling node is assigned to provide support in only one direction, often the Y-direction of a truss member. \newcommand{\ihat}{\vec{i}} 6.2.2 Parabolic Cable Carrying Horizontal Distributed Loads, 1.7: Deflection of Beams- Geometric Methods, source@https://temple.manifoldapp.org/projects/structural-analysis, status page at https://status.libretexts.org. We know the vertical and horizontal coordinates of this centroid, but since the equivalent point forces line of action is vertical and we can slide a force along its line of action, the vertical coordinate of the centroid is not important in this context. For Example, the maximum bending moment for a simply supported beam and cantilever beam having a uniformly distributed load will differ. The formula for any stress functions also depends upon the type of support and members. The relationship between shear force and bending moment is independent of the type of load acting on the beam. suggestions. \begin{align*} Weight of Beams - Stress and Strain - So the uniformly distributed load bending moment and shear force at a particular beam section can be related as V = dM/dX. \begin{equation*} The sag at B is determined by summing the moment about B, as shown in the free-body diagram in Figure 6.9c, while the sag at D was computed by summing the moment about D, as shown in the free-body diagram in Figure 6.9d. Hb```a``~A@l( sC-5XY\|>&8>0aHeJf(xy;5J`,bxS!VubsdvH!B yg* endstream endobj 256 0 obj 166 endobj 213 0 obj << /Type /Page /Parent 207 0 R /Resources << /ColorSpace << /CS3 215 0 R /CS4 214 0 R /CS5 222 0 R >> /XObject << /Im9 239 0 R /Im10 238 0 R /Im11 237 0 R /Im12 249 0 R /Im13 250 0 R /Im14 251 0 R /Im15 252 0 R /Im16 253 0 R /Im17 254 0 R >> /ExtGState << /GS3 246 0 R /GS4 245 0 R >> /Font << /TT3 220 0 R /TT4 217 0 R /TT5 216 0 R >> /ProcSet [ /PDF /Text /ImageC /ImageI ] >> /Contents [ 224 0 R 226 0 R 228 0 R 230 0 R 232 0 R 234 0 R 236 0 R 241 0 R ] /MediaBox [ 0 0 595 842 ] /CropBox [ 0 0 595 842 ] /Rotate 0 /StructParents 0 >> endobj 214 0 obj [ /ICCBased 244 0 R ] endobj 215 0 obj [ /Indexed 214 0 R 143 248 0 R ] endobj 216 0 obj << /Type /Font /Subtype /TrueType /FirstChar 32 /LastChar 148 /Widths [ 278 0 0 0 0 0 0 0 0 0 0 0 0 333 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 722 722 722 0 0 0 778 0 0 0 0 0 0 722 0 0 0 722 667 611 0 0 0 0 0 0 0 0 0 0 0 0 556 611 556 611 556 333 611 611 278 0 0 278 889 611 611 611 0 389 556 333 611 0 778 0 556 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 500 500 ] /Encoding /WinAnsiEncoding /BaseFont /AIPMIP+Arial,BoldItalic /FontDescriptor 219 0 R >> endobj 217 0 obj << /Type /Font /Subtype /TrueType /FirstChar 32 /LastChar 146 /Widths [ 278 0 0 0 0 0 722 0 0 0 0 0 278 333 278 278 556 556 0 556 0 556 556 556 0 556 333 0 0 0 0 611 0 722 722 722 722 667 611 778 722 278 556 722 611 833 722 778 667 0 722 667 611 722 667 944 667 667 0 0 0 0 0 0 0 556 611 556 611 556 333 611 611 278 278 556 278 889 611 611 611 0 389 556 333 611 556 778 556 556 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 278 278 ] /Encoding /WinAnsiEncoding /BaseFont /AIEEHI+Arial,Bold /FontDescriptor 218 0 R >> endobj 218 0 obj << /Type /FontDescriptor /Ascent 905 /CapHeight 718 /Descent -211 /Flags 32 /FontBBox [ -628 -376 2034 1010 ] /FontName /AIEEHI+Arial,Bold /ItalicAngle 0 /StemV 144 /XHeight 515 /FontFile2 243 0 R >> endobj 219 0 obj << /Type /FontDescriptor /Ascent 905 /CapHeight 718 /Descent -211 /Flags 96 /FontBBox [ -560 -376 1157 1000 ] /FontName /AIPMIP+Arial,BoldItalic /ItalicAngle -15 /StemV 133 /FontFile2 247 0 R >> endobj 220 0 obj << /Type /Font /Subtype /TrueType /FirstChar 32 /LastChar 176 /Widths [ 278 0 355 0 0 889 667 0 333 333 0 0 278 333 278 278 556 556 556 556 556 556 556 556 556 556 278 278 0 584 0 0 0 667 667 722 722 667 611 778 722 278 500 0 556 833 722 778 667 778 722 667 611 722 667 944 0 0 611 0 0 0 0 0 0 556 556 500 556 556 278 556 556 222 222 500 222 833 556 556 556 556 333 500 278 556 500 722 500 500 500 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 222 222 333 333 0 556 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 737 0 400 ] /Encoding /WinAnsiEncoding /BaseFont /AIEEFH+Arial /FontDescriptor 221 0 R >> endobj 221 0 obj << /Type /FontDescriptor /Ascent 905 /CapHeight 718 /Descent -211 /Flags 32 /FontBBox [ -665 -325 2028 1006 ] /FontName /AIEEFH+Arial /ItalicAngle 0 /StemV 94 /XHeight 515 /FontFile2 242 0 R >> endobj 222 0 obj /DeviceGray endobj 223 0 obj 1116 endobj 224 0 obj << /Filter /FlateDecode /Length 223 0 R >> stream 6.6 A cable is subjected to the loading shown in Figure P6.6. Users however have the option to specify the start and end of the DL somewhere along the span. \newcommand{\inlb}[1]{#1~\mathrm{in}\!\cdot\!\mathrm{lb} } This equivalent replacement must be the. Variable depth profile offers economy. y = ordinate of any point along the central line of the arch. These loads are expressed in terms of the per unit length of the member. \newcommand{\psf}[1]{#1~\mathrm{lb}/\mathrm{ft}^2 } Determine the sag at B and D, as well as the tension in each segment of the cable. Taking the moment about point C of the free-body diagram suggests the following: Free-body diagram of segment AC. \newcommand{\ftlb}[1]{#1~\mathrm{ft}\!\cdot\!\mathrm{lb} } This chapter discusses the analysis of three-hinge arches only. Supplementing Roof trusses to accommodate attic loads. <> Most real-world loads are distributed, including the weight of building materials and the force 0000012379 00000 n For the least amount of deflection possible, this load is distributed over the entire length Cables: Cables are flexible structures in pure tension. 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. 0000004601 00000 n 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. 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*} WebThe Influence Line Diagram (ILD) for a force in a truss member is shown in the figure. The bending moment and shearing force at such section of an arch are comparatively smaller than those of a beam of the same span due to the presence of the horizontal thrusts. Consider a unit load of 1kN at a distance of x from A. This means that one is a fixed node 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. I am analysing a truss under UDL. R A = reaction force in A (N, lb) q = uniform distributed load (N/m, N/mm, lb/in) L = length of cantilever beam (m, mm, in) Maximum Moment. 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. W = w(x) \ell = (\Nperm{100})(\m{6}) = \N{600}\text{.} WebThe Mega-Truss Pick will suspend up to one ton of truss load, plus an additional one ton load suspended under the truss. 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. 0000009328 00000 n 0000010481 00000 n UDL Uniformly Distributed Load. They take different shapes, depending on the type of loading. \newcommand{\kg}[1]{#1~\mathrm{kg} } Arches are structures composed of curvilinear members resting on supports. Note the lengths of your roof truss members on your sketch, and mark where each node will be placed as well. 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. 0000155554 00000 n Trusses - Common types of trusses. \\ \newcommand{\unit}[1]{#1~\mathrm{unit} } to this site, and use it for non-commercial use subject to our terms of use. Since all loads on a truss must act at the joints, the distributed weight of each member must be split between the 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 . Trusses containing wide rooms with square (or almost square) corners, intended to be used as full second story space (minimum 7 tall and meeting the width criteria above), should be designed with the standard floor loading of 40 psf to reflect their use as more than just sleeping areas. Putting into three terms of the expansion in equation 6.13 suggests the following: Thus, equation 6.16 can be written as the following: A cable subjected to a uniform load of 240 N/m is suspended between two supports at the same level 20 m apart, as shown in Figure 6.12. 0000007236 00000 n Legal. The distributed load can be further classified as uniformly distributed and varying loads. \newcommand{\amp}{&} These spaces generally have a room profile that follows the top chord/rafter with a center section of uniform height under the collar tie (as shown in the drawing). 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. \newcommand{\gt}{>} From static equilibrium, the moment of the forces on the cable about support B and about the section at a distance x from the left support can be expressed as follows, respectively: MBP = the algebraic sum of the moment of the applied forces about support B. The derivation of the equations for the determination of these forces with respect to the angle are as follows: \[M_{\varphi}=A_{y} x-A_{x} y=M_{(x)}^{b}-A_{x} y \label{6.1}\]. They can be either uniform or non-uniform. The shear force equation for a beam has one more degree function as that of load and bending moment equation have two more degree functions. For those cases, it is possible to add a distributed load, which distribution is defined by a function in terms of the position along the member. M \amp = \Nm{64} 6.8 A cable supports a uniformly distributed load in Figure P6.8. 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. 0000006074 00000 n Determine the sag at B, the tension in the cable, and the length of the cable. One of the main distinguishing features of an arch is the development of horizontal thrusts at the supports as well as the vertical reactions, even in the absence of a horizontal load. 0000072700 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. Determine the horizontal reaction at the supports of the cable, the expression of the shape of the cable, and the length of the cable. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. You're reading an article from the March 2023 issue. It includes the dead weight of a structure, wind force, pressure force etc. g@Nf:qziBvQWSr[-FFk I/ 2]@^JJ$U8w4zt?t yc ;vHeZjkIg&CxKO;A;\e =dSB+klsJbPbW0/F:jK'VsXEef-o.8x$ /ocI"7 FFvP,Ad2 LKrexG(9v w(x) = \frac{\N{3}}{\cm{3}}= \Nperm{100}\text{.} 0000016751 00000 n They are used for large-span structures, such as airplane hangars and long-span bridges. \newcommand{\kNm}[1]{#1~\mathrm{kN}\!\cdot\!\mathrm{m} } These types of loads on bridges must be considered and it is an essential type of load that we must apply to the design. This is due to the transfer of the load of the tiles through the tile Note that while the resultant forces are, Find the reactions at the fixed connection at, \begin{align*} 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{\kgsm}[1]{#1~\mathrm{kg}/\mathrm{m}^2 } Three-pinned arches are determinate, while two-pinned arches and fixed arches, as shown in Figure 6.1, are indeterminate structures. DLs are applied to a member and by default will span the entire length of the member. The lengths of the segments can be obtained by the application of the Pythagoras theorem, as follows: \[L=\sqrt{(2.58)^{2}+(2)^{2}}+\sqrt{(10-2.58)^{2}+(8)^{2}}+\sqrt{(10)^{2}+(3)^{2}}=24.62 \mathrm{~m} \nonumber\]. \newcommand{\second}[1]{#1~\mathrm{s} } Determine the tensions at supports A and C at the lowest point B. WebConsider the mathematical model of a linear prismatic bar shown in part (a) of the figure. 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. Many parameters are considered for the design of structures that depend on the type of loads and support conditions. For example, the dead load of a beam etc. 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. In analysing a structural element, two consideration are taken. Maximum Reaction. DoItYourself.com, founded in 1995, is the leading independent Fairly simple truss but one peer said since the loads are not acting at the pinned joints, When applying the non-linear or equation defined DL, users need to specify values for: After correctly inputting all the required values, the non-linear or equation defined distributed load will be added to the selected members, if the results are not as expected it is always possible to undo the changes and try again. 0000103312 00000 n Arches can also be classified as determinate or indeterminate. QPL Quarter Point Load. \newcommand{\ft}[1]{#1~\mathrm{ft}} Therefore, \[A_{y}=B_{y}=\frac{w L}{2}=\frac{0.6(100)}{2}=30 \text { kips } \nonumber\]. 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. \newcommand{\psinch}[1]{#1~\mathrm{lb}/\mathrm{in}^2 } IRC (International Residential Code) defines Habitable Space as a space in a building for living, sleeping, eating, or cooking.
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