Method of Sections for Truss Analysis

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Doreen Montgomery
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1 RH 331 Note Set 5.2 F2013abn Method of Sections for Truss nalysis Notation: () = shorthand for compression = name for load or axial force vector (T) = shorthand for tension Joint onfigurations (special cases to recognize for faster solutions) ase 1) Two odies onnected or F has to be equal and opposite to F ase 2) Three odies onnected with Two odies in Line or or even F and F have to be equal, and F has to have zero force. ase 3) Three odies onnected and a Force 2 odies aligned & 1 ody and a Force are ligned Four odies onnected - 2 odies ligned and the Other 2 odies ligned F has to equal F, and [F has to equal ] or [F has to equal F] Method of Sections (relies on internal forces being in equilibrium with external forces on a section) 1. etermine support reaction forces. 2. ut a section in such a way that force action lines intersect. 3. Solve for equilibrium. Sum moments about an intersection of force lines of action 107

2 RH 331 Note Set 5.2 F2013abn dvantages: Quick when you only need one or two forces (only 3 equations needed) isadvantages: Not always easy to find a place to cut a section or see where force lines intersect F ompound Truss: truss assembled of simple trusses and additional links. It has b=2n-3, is statically determinate, rigid and completely constrained with a pin and roller. It can be identified by triangles with pins in the middle of some sides. Statically Indeterminate Trusses: Occur if there are more members than equations for all the joints OR if there are more reaction supports unknowns than 3 iagonal Tension ounters: rossed bracing of cables or slender members is commonly used in bridge trusses, buildings and towers. These trusses look indeterminate, but can be solved statically because the bracing cannot hold a compressive force. The members are excluded in the analysis. Method: 1. etermine support reaction forces. 2. ut a section in such a way that the tension counters are exposed. 3. Solve for force equilibrium in y with one counter. If the value is positive (in tension), this is the solution. 4 F G H 4. Solve for force equilibrium in y with the other counter. 108

3 RH 331 Note Set 5.2 F2013abn xample 1 (pg 99) (Support forces must be found as well). 109

4 RH 331 Note Set 5.2 F2013abn xample 2 Using the method of sections, determine member forces in F,,, and F. SOLUTION: section can t pass through 5 members, so there will have to be two sections. The first passes through F, and. F is shown assumed to be in compression, while the other forces are drawn assumed to be in tension. There can be only two intersections when two of the three forces are parallel at and : Σ M = 100 ( 6 ft ) ( 8 ft ) = 0 = 75 (T) Σ M = 100 ( 12 ft ) F( 8 ft ) = 0 F = 150 () ecause is the only unknown force with a y component, it is useful to sum forces in the y direction (although it also has the only remaining unknown x component): Σ F = ft y ( ) = ft (or Σ F = ( 6 ft ) = ft = 125 (T) x ) second section can be drawn through, F and F. There are three points of intersection of the unknown forces - at, F and. is not on the section, but we know where it is. Σ M = 300 ( 6 ft ) + F( 6 ft ) = 0 F = 300 () Σ M F = 200 ( 6 ft ) + y( 6 ft ) = 0 (sliding components to ) = y ( 100 ) = (T) or Σ M F = 200 ( 6 ft ) + x( 8 ft ) = 0 (sliding components to ) = x ( 100 ) = (T) Σ M = 200 ( 6 ft ) + F( 8 ft ) = 0 F = 150 () 110

5 RH 331 Note Set 5.2 F2013abn xample 3 (pg 90) etermine the member forces, and F

6 RH 331 Note Set 5.2 F2013abn xample 4 Using the method of sections, determine member forces in, and. SOLUTION: Find the support reactions from rigid body equilibrium, or in this case, from load tracing with symmetrical loads. raw a section line through the members of interest, cutting through no more than 3 members to separate the truss into two pieces. In this case, and can be cut through, while will need another section. raw one of the sections, exposing the member forces. rawing them out or away from the cut assumes tension. is drawn in compression. So is, but because it has a 45 degree angle, the components will have the same magnitude. Find a point to sum moments where two unknown forces intersect. This may be on a point of the section or off the section. X is such a location where the line of action of intersects that of. For every 15 ft to the left, the line slopes down 5 ft, so X is located (10 ft/ 5 ft)15 ft = 30 ft to the left of. Σ M = 450 ( 15 ft ) 300 ( 30 ft ) ( 30 ft ) = 0 X y = -75, so = y/sin45 = 106 tension y X (compression was assumed, but the answer was negative indicating our assumption wasn t verified). (Notice that x and y slid down to and then the lever arm for x was 0. The components can also slide to the other end point of the member to locate the lever arms) Summing at where and intersect means there will be no lever arms. Sliding the components of to means there will be no lever arm for y: Σ M = 450 ( 15 ft ) + x( 10 ft ) = 0 x = 675, so = x 10 = compression raw a section line that passes through and cuts through no more than three members. If we hadn t already found, we could sum moments at point X again to eliminate and from our equation, leaving. ut it is obvious that we have only one unknown y force, which is : ΣF y =. = 0 = 225 tension

Method of Sections for Truss Analysis

Method of Sections for Truss Analysis Notation: (C) = shorthand for compression P = name for load or axial force vector (T) = shorthand for tension Joint Configurations (special cases to recognize for