CH. 5 TRUSSES BASIC PRINCIPLES Typical depth-to-span ratios range from 1:10 to 1:20 - Flat trusses require less overall depth than pitched trusses Spans: 40-200 Spacing: 10 to 40 on center - Residential & light commercial: trussed rafters of 2x4 or 2x6 @ 24 oc - Open web steel joists: 2 to 3 Roof loads transferred from decking to purlins attached to truss @ panel points - If concentrated loads between panel points or uniform loads applied to top chords, member must be designed for axial loading as well as for bending Like beams - Compression in top chord & tension in bottom chords - Forces in a parallel chord truss increase towards center - Bowstring truss, chord forces constant b/c depth varies from min @ supports to max @ center Truss members designed as columns if in compression If in tension the net area must be determined (actual area bolt holes) If concentrated loads or uniform loads on any chord member between panel points, member must resist bending stresses Regardless of material, trusses need to be designed so they are concentric: member is symmetric on both sides of centroid axis in the plane of the truss - Steel trusses with double angles back-to-back with 3/8 or 1/2 gusset plate with tee sections or wide flange With light loads, bars or rods can be used for tension members Wood trusses: web members between double top and bottom chords or with all members in same plane connected with gusset plate Centroidal axes of intersecting members must meet at a point to avoid eccentric loading - Steel composes of angles: standard to have gage lines rather than centroidal axes meet - Gage line: standard dim from corner edge of an angle to centerline of bolt holes & depends on size of angle TRUSS ANALYSIS First: determine loads in various members - General guidelines for truss analysis - Sum of vertical forces at any point equals zero - Sum of horizontal forces at any point equals zero - Sum of moments about any point equals zero - Forces acting upward or to the right are considered positive (+) - Forces acting downward or to the left are considered negative (-) - All forces should be indicated. If direction is unknown when beginning the analysis show the force in tension, acting away from the joint or cut section. If calculation of force is negative, this indicates direction is reversed.
- For analysis, trusses are assumed to have pivoting or rolling supports to avoid other stresses at these points - x component of a force (horizontal) is equal to the force times cosine of the angle the force vector makes with the x-axis F x = F cos a - y component of a force (vertical) is equal to the force times the cosine of the angle the force vector makes with the y-axis F y = F cos b - X-Y axis can be tilted to any convenient angle if required by the problem There are three methods to determine the forces in truss members - Method of joints: useful in determining the forces in all members or when only the forces in the members near the supports needs to be calculated - Method of section: when only necessary to find the forces in a few members, particularly ones that are not at or near the supports - Graphic method: complex trusses and avoids calculations inherent in analytic solutions but not as accurate Method of Joints Each joint is considered separately as a free body diagram which the equations of equilibrium are applied Starting from one support the force in each member is determined joint by joint until all calculated Example 5.1: find forces in the members using the method of joints - First: find reactions. Since loads are symmetric, - Note: if loads not symmetric, reactions could be found by taking moments about one reaction and setting the moments equal to zero R A = R F = 1/2(17000) = 8500lbf
- Second: start with joint A at reaction R A and draw joint as free body diagram - Direction of reaction is known (upward) and types of forces in the top and bottom chord can be assumed to be in compression and tension respectively show force F AB toward the joint and force F AD away from the joint - Calculate the angle between members AB and AD tan = 10/15 tan = 33.7 the compliment of this angle is 56.3 - Notice: the reaction force is positive b/c acting upward and y-component of F AB is negative since acting downward & force F AD has no vertical component - Solve for F AB : F AB = 8500/cos56.3 F AB = 15,320 lbf (compression) - Since positive, the assumption that force AB is in compression is correct - Solve for F Ad.the force in member AD is found in similar way knowing sum of horizontal forces equals zero - Third: consider joint B F Ad = F AB cos33.7 = 0 F Ad = 12,746 lbf (tension) - Since direction of force in member BD is not clear, assume tension (arrow away from joint) - Actually since the 5000 lbf load is acting down, there must be a force acting upward to counteract so member BD would have to be in compression. For purposes of illustration however assume it is in tension - Tilt X-Y axes so the X-axis aligns with top chord of truss - Angles between X-Y and forces can be determined with trigonometry - With axes tilted force F BC has no vertical component so force F BD can be found easily -5000 cos33.7 - F BD cos22.6 both forces are acting downward both negative values F AB = -4506 lbf
- The negative number indicates the assumption that BD was in tension is incorrect it is in compression - Find forces in BC knowing that the summation of forces in the X-axis equals zero 15,320 5,000 cos56.3 - F BC 4,506 cos67.4 = 0 F BC = 10,814 lbf (compression) - Finally: draw joint C as a free body diagram - Sum of the forces in the y-direction is zero so: 10,814 cos56.3-7000 + 10,814 cos56.3 - F CD = 0 F CD = 5000 lbf (tension) Method of Sections Portions of truss is cut through three members, one of which is the member under analysis Cut section is drawn as free body diagram and force in members is found by taking moments about various points knowing that M = 0 Example 5.2: determine the forces in members BD and BC using the method of sections - First: solve for reactions (which were determined to be 8500lbf at each reaction - Second: cut section thru two members under analysis. In this free body diagram there are five forces acting: Two known & Three unknown - To find force in member BD, take moments about point A. Selecting this point eliminates the unknowns of F BC and F AD b/c their lines of action pass thru the points so moment is zero. This leaves only F BD acting about A - Note: moments in a clockwise direction are positive and counterclockwise are negative
- Third: find dimension of the moment arm of BD: a line passing through A perpendicular to BD - With simple trigonometry the lengths of AB is found to be 9.01 ft and the angle between AB and the moment arm of BD is 22.6, then: cos22.6 = h/9.01 h = 8.32ft - Then the sum of moments about A equals zero or: (5000)(7.5) F BD (8.32) = 0 F BD = 4507lbf - Fourth: find value of FBC - To eliminate two unknowns take moments about point D. this is acceptable even though it is outside the free body diagram b/c equation of moment equilibrium holds at any point in the truss. - The moment arm from D perpendicular to BC must be found. It is the same dimension as previous moment arm calculated at 8.32ft, then the sum of moments about joint D is: (8500)(15) (5000)(7.5) F BC (8.32) = 0 F BC = 10,817 lbf - In both cases the answer was a positive number indicating that the original assumption of direction of force (compression) was correct