Civil 3130: Structural Analysis (Trusses) Lecture Notes PDF

CIVL 3130: Structural Analysis Trusses  Objectives Study the characteristics and behavior of trusses. Analyze determinate trusses by method of joints and method of sections to determine bar forces. Also learn to visually identify bars with zero...

CIVL 3130: Structural Analysis Trusses  Objectives Study the characteristics and behavior of trusses. Analyze determinate trusses by method of joints and method of sections to determine bar forces. Also learn to visually identify bars with zero force. Classify determinate and indeterminate truss structures and determine the degree of indeterminacy. Determine if a truss structure is stable or unstable.  Collapse of the I-35W Bridge in Minneapolis, Minnesota (1967 – Aug. 1st, 2007) https://www.mprnews.org/story/2009/03/25/attorney-ntsb-is- wrong-about-cause-of-bridge-collapse The bridge had a catastrophic failure during rush hour in which 13 people were killed. The failure was due to overload of an under-designed gusset connection. The bridge was statically determinate, hence causing complete collapse of the structure, which highlights the importance of providing a redundant design with indeterminate structures.  Characteristics and Behaviour of Trusses A truss is a structural element composed of a stable arrangement of slender interconnected bars (Figure a). Although joints, typically formed by welding or bolting truss bars to gusset plates, are rigid (Figure b), the designer normally assumes that members are connected at joints by frictionless pins (Figure c) Since no moment can be transferred through a frictionless pin joint, truss members are assumed to carry only axial force—either tension or compression.  Characteristics and Behaviour of Trusses In addition to varying the area of truss members, the designer can vary the truss depth to reduce its weight (Figure a). In regions where the bending moment is large—at the center of a simply supported structure or at the supports in a continuous structure—the truss can be deepened (Figure b)  (a) Characteristics and Behaviour of Trusses The diagonals of a truss typically slope upward at an angle that ranges from 45° to 60°. In a long-span truss the distance between panel points should not exceed 15 to 20 ft (5 to 7 m) to limit the unsupported length of the compression chords, which must be designed as columns (buckling may govern). The slenderness of tension members must be limited also to reduce vibrations produced by wind and live load. Tacoma bridge: Spaces Images/Mint Images Limited/Alamy Stock Photo If a truss carries equal or nearly equal loads at all panel points (Figure a), the direction in which the diagonals slope will determine if they carry tension or compression forces. Figure b shows the difference in forces set up in the diagonals of two trusses that are identical in all respects. (b) Characteristics and Behaviour of Trusses Bracings (transverse and diagonal members within horizontal planes of the top and bottom chords) are used to stiffen or stabilize trusses in the out-of-plane direction. Another function of the bracings (including the top and bottom chord) is the transmission of wind (or lateral) forces to the end supports.  Types of Trusses The triangular form is geometrically stable and will not collapse under load. A pin-connected rectangular element will collapse under the smallest lateral load. Joint D can be formed by extending bars from joints B and C. Similarly, Joint E Simple truss is formed by extending bars from joints C and D. Trusses formed in this manner are called simple trusses. If two or more simple trusses are connected by a pin or a pin and a tie, or three nonparallel nonconcurrent bars, the resulting truss is termed a compound truss compound truss. If a truss-usually one with an unusual shape-is neither a simple nor a compound truss, it is termed a complex truss. complex truss  Analysis of Trusses The entire structure is considered as a rigid body and, the equations of static equilibrium together with any condition equations that may exist are applied. The analysis used to evaluate the bar forces is based on the following three assumptions: 1.Bars are straight and carry only axial load (i.e., bar forces are directed along the longitudinal axis of truss members). This assumption also implies that we have neglected the deadweight of the bar. If the weight of the bar is significant, we can approximate its effect by applying one-half of the bar weight as a concentrated load to the joints at each end of the bar. 2.Members are connected to joints by frictionless pins. No moments can be transferred between the end of a bar and the joint to which it connects. (If joints are rigid and members stiff, the structure should be analyzed as a rigid frame.) 3.Loads are applied only at joints.  Analysis of Trusses As a sign convention, tensile force is positive, and a compression force is negative. A tension bar will result in a force that acts away from the free body of each joint. If a bar is in compression, the axial forces at the ends of free body of the bar act inward and tend to compress the bar. Bar forces, or internal axial forces, may be analyzed by considering the equilibrium of a joint—the method of joints—or by considering the equilibrium of a section of a truss—the method of sections.  Method of Joints To determine bar forces by the method of joints, we analyze free-body diagrams of joints. The free-body diagram is established by imagining that we cut the bars by an imaginary section through the bars next to the joint, but not through the joint. For example, to determine the bar forces in members AB and BC, we use the isolated free body of joint B. Since the bars carry axial force, the line of action of each bar force is directed along the longitudinal axis of the bar. Because all forces acting at a joint pass through the pin, they constitute a concurrent force system. For this type of force system, only two equations of statics (that is, ΣFx = 0 and ΣFy = 0) are available to evaluate unknown bar forces. Hence, we can only analyze joints that contain a maximum of two unknown bar forces.  Example 1 Analyze the truss in the figure below by the method of joints. → +  Fx = 0 -𝐴𝑥 + 22 kips = 0; 𝐴𝑥 = 22 kips +   Fy = 0 -𝐴𝑦 + 𝐷𝑦 = 0; 𝐴𝑥 𝐴𝑦 = 𝐷𝑦 𝐷𝑦 ⟳+ ∑𝑀𝐴 = 0 𝐴𝑦 -𝐷𝑦 x 11 + 22 x 12 kips. ft = 0; 𝐷𝑦 = 22 x 12 kips. ft /11 = 24 kips 𝐴𝑦 = 𝐷𝑦 = 24 kips  Example 1 +   Fy = 0 0 = −24 + YAB and YAB = 24 kips To find 𝑭𝑨𝑩 , isolate Joint A YAB X AB FAB and replace 𝑭𝑨𝑩 by its = = 3 4 5 rectangular components 𝑿𝑨𝑩 4 4 and 𝒀𝑨𝑩 X AB = YAB = (24) = 32 kips 3 3 5 5 FAB = YAB = (24) = 40 kips 3 3  Example 1 → +  Fx = 0 0 = −22 + X AB + FAD FAD = −32 + 22 = −10 kips (compression) To find 𝐹𝐴𝐷  Example 1 +   Fy = 0 To find 𝐹𝐵𝐷 , YBD = 0 isolate Joint B Therefore, 𝐹𝐵𝐷 = 0 To find 𝐹𝐵𝐶 , → +  Fx = 0 0 = FBC − 40 𝐹𝐵𝐶 = 40 kips tension To find 𝐹𝐷𝐶 , → +  Fx = 0 0 = −(−10) + X DC and X DC = −10 kips +   Fy = 0 0 = 24 + YDC and YDC = −24 kips  Section Method To analyze a stable truss by the method of sections, we imagine that the truss is divided into two free bodies by passing an imaginary cutting plane through the structure. Although there is no restriction on the number of bars that can be cut, we often use sections that cut three bars since three equations of static equilibrium are available to analyze a free body. Section 2-2  Example 2 Determine the forces in bars HG and HC of the truss in the figure by the method of sections.  Example 2 Solution To simplify the computations, we select a moment center (point a that lies at the intersection of the lines of action of forces F1 and F3). Next, Force F2 is extended along its line of action to point C and replaced by its rectangular components X2 and Y2. The distance x between a and the left support is established by proportion using similar triangles, that is, aHB and the slope (1:4) of force F1. 1 4 = 18 x + 24 x = 48 ft Sum moments of the forces about point a and solve for Y2.  Example 2 ⟳+ ∑𝑀𝑎 = 0 0 = −60(48) + 30(72) + Y2 (96) 𝑌2 = 7.5 kips (tension) Based on the slope of bar HC, establish X2 by proportion. Y2 X 2 = 3 4 4 X 2 = Y2 = 10 kips 3 Now compute the force F1 in bar HG. Select a moment center at the intersection of the lines of action of forces F2 and F3, that is, at point C (Figure c). Extend force F1 to point G and break into rectangular components. Sum moments about point C.  Example 2 ⟳+  M c = 0 0 = 60(48) − 30(24) + X 1 (24) X 1 = −90 kips (compression) Establish Y1 by proportion. X 1 Y1 = 4 1 X1 Y1 = = −22.5 kips 4  Zero Bars Case 1. If no external load is applied to a joint that consists of two bars, the force in both bars must be zero Case 2. If no external load acts at a Joint composed of three bars—two of which are collinear—the Force in the Bar That Is Not Collinear Is Zero Zero Bars - Example  Determinacy and Stability Since there are two equilibrium equations for each joint in a truss, ෍ 𝐹𝑥 = 0 𝑎𝑛𝑑 ෍ 𝐹𝑦 = 0 the total number of equilibrium equations available to solve for the unknown bar forces b and reactions r equals 2n (where n represents the total number of joints). Therefore, it must follow that if a truss is stable and determinate, the relationship between bars, reactions, and joints must satisfy the following criteria: 𝑟 + 𝑏 = 2𝑛 (determinate if satisfied) Note: the restraints exerted by the reactions must not constitute either a parallel or a concurrent force system. The degree of indeterminacy D equals D = 𝑟 + 𝑏 − 2𝑛 If 𝑟 + 𝑏 > 2𝑛, then the number of unknown forces exceed the available equations of statics, and the truss is indeterminate. If 𝑟 + 𝑏 < 2𝑛, there are insufficient bar forces and reactions to satisfy the equations of equilibrium, and the structure is unstable. Determinacy and Stability unstable 𝑓  Example 3 Verify that the truss in the figure is stable and determinate by demonstrating that it can be completely analyzed by the equations of statics for a force of 4 kips at joint F.  Example 3 Solution I. Since the structure has four reactions, we must analyze it by the method of joints. We first determine the zero bars. II. Since joints E and I are connected to only two bars and no external load acts on the joints, the forces in these bars are zero (see Case 1 of zero bars). III. With the remaining two bars connecting to joint D, applying the same argument would indicate that these two members are also zero bars. IV. Applying Case 2 of zero bars to joint G would indicate that bar CG is a zero bar. V. Next we analyze in sequence joints F, C, G, H, A, and B. Since all bar forces and reactions can be determined by the equations of statics, we conclude that the truss is stable and determinate.