1.1. Loads on Truss Girder

Transcription

1 Practice for Exam 1 Solution 1 / Loads on Truss Girder SDL 23 psf purlin_wt 10.0 plf 25 = 75 girder_wt k = 88.9 plf P D ext P D int SDL: (23 psf) (7.5 ft) (25.0 ft) = 4.31 k purlin wt: (10.0 plf) (25.0 ft) = 0.25 k girder wt: (88.9 plf) (7.5 ft ) = k P D ext 5.23 k SDL: = (23 psf) (15.0 ft) (25.0 ft) = 8.63 k purlin wt: (10.0 plf) (25.0 ft) = 0.25 k girder wt: (88.9 plf) (15.0 ft ) = k P D int k 15 = 135 Live Load Reduction: use one span of girder to compute tributary area, A If A T Then R 1 Plan View A T = (75 ft) (25 ft) = 1875 sf <= 200 sf 1.0 R >= 600 sf 0.6 L r = R 1 * 20 psf = 12.0 psf else = A T P Lr ext = (Lr) (7.5 ft) (25.0 ft) = 2.25 k P Lr int = (Lr) (15.0 ft) (25.0 ft) = 4.50 k P A 15 k Max. hord ompressive Force due DL Estimate location of max. compressive force in chords due to Dead Loads Deflected shape of analogous beam T T assume analogous beam is loaded with uniform load M Therefore, check top chords in middle of Span 2, and bottom chords near Supports

2 Practice for Exam 1 Solution 2 / 8 R k R k f k f k Therefore max chord compressive force due to DL occurs = k, in member 2 f 2 D = 25.8 k Max. diagonal Force due DL Estimate location of max. diagonal force due to dead loads V f k T f k T Therefore max diagonal force due to DL occurs = k T, in member 31 f 31 D = 32.7 k T max tensile force in Member 5 due to Lr Estimated deflected shape of truss due to elongation of Member 5 Shape of Influence Line

3 Practice for Exam 1 Solution 3 / 8 Loading to cause max tension force in M5 15 k 5 k Select Influence Line B from exam. Node I 5 Px f 5 Px Lr f 5 = k Lr f 5 = k T Max compressive force in Member 33 due to Lr Estimated deflected shape of truss due to elongation of Member 33 Shape of Influence Line 15 k 5 k 5 k 5 k 3 k

4 Practice for Exam 1 Solution 4 / 8 Select Influence Line from exam. Node I 33 Px Px f Lr f 33 = k Lr f 33 = k

5 E 331, Spring 2011 Practice for Exam 1 Solution 5 / 8 2. alculate the bar forces in the truss below k Write next to the chords in compression and T next to the chords in tension on the figure below. T T Write the bar force for each member on the figure below. Indicate tension forces with T and compression forces with. Use the joint and member numbers indicated in the figure above k T 12.2 k T 10.2 k T 30 k 45.2 k 52.0 k

6 E 331, Spring 2011 Practice for Exam 1 Solution 6 / 8 3. For the truss shown in Problem 2, calculate the horizontal deflection of Joint 1 due to the loading shown below. Indicate the direction of the deflection by drawing an arrow next to the deflection. All members are steel with cross sectional area of 2.0 in k 10 k E 29,000 ksi A 2.0 in 2 ft in^2 k in in Member L A f Load δ f U_1_H Δ 1_H Sum: Δ 1_Η = in Typical alcs for Member 1 in Table above L = Member Length = ft, from previous problem f = bar force in Member 1 due to load = k δ axial deformation due to actual loads = f x L / (A x E) = ( k) (7.62 ft x 12 in/ft) / ( 2.00 in2) (29,000 ksi) ) = in f U_1_H = bar force in Member 1 due to unit load at Joint 1 in the horizontal direction = -0.44, from table Δ 1_H = deflection at Joint 1 in horizontal direction due to Member 1 deformation = δ x f U_1_H = ( in) (-0.44) = in

7 E 331, Spring 2011 Practice for Exam 1 Solution 7 / 8 4. Explain why (draw a sketch if necessary): a. a cable is more structurally efficient than an arch. A cable has only one failure mode material failure, while an arch has two failure modes material failure and buckling). b. an arch is more structurally efficient than a beam An arch has a uniform stress distribution over its cross section and therefore all fibers are fully loaded at failure. A beam has a linearly varying stress distribution over it cross section and therefore only the outer fibers are fully loaded at failure. 5. For each loading shown below, sketch the beam profile (side view) that is most efficient. (a) (b) (c) Note: the most efficient profile is the profile resembling the shape of the moment diagram due to the applied loads. Assuming the beams fail in flexure. 6. Which of the two structure profiles would be more efficient for a 3 span continuous structure with uniform load? Explain. amel back Truss antilever Truss The cantilever truss would be more efficient since the truss is tallest at the supports, which is where the largest moment occurs in a continuous structure.

8 E 331, Spring 2011 Practice for Exam 1 Solution 8 / 8 7. State whether each truss below is: Unstable, Stable and Determinate or Indeterminate (a) ounting forces and equations: # Forces = 4 members + 3 reactions = 7 # Eqns = 4 joints x 2eqns/joint = 8 #Forces < Eqns, therefore structure is unstable Deflection shape: Member s joints can deflect without causing member deformation. Therefore, structure is unstable. Therefore, structure is unstable (b) ounting forces and equations: # Forces = 25 members + 3 reactions = 28 # Eqns = 14 joints x 2eqns/joint = 28 #Forces = Eqns, therefore structure is stable & determinate Deflection shape: Member s joints can deflect without causing member deformation. Therefore, structure is unstable. Therefore, structure is unstable (c) ounting forces and equations: # Forces = 4 members + 4 reactions = 8 # Eqns = 4 joints x 2eqns/joint = 8 #Forces = Eqns, therefore structure is stable & determinate Deflection shape: Member s joints cannot deflect without causing member deformation. Therefore, structure is stable. Therefore, structure is stable & determinate