Sample Problemq. For the compound roofstructure shown, determine the2axial stress developed inD-Bmembers AE and EG. g = 20kN/m, AAE = AEG = 23 cm?.GJL=E9m? Solution4.37m4.37m1. Reaction forceql0=ZF, = FRA =FR177.4 kNqRB22. Axial force in EGD0 =ZMc=号(4.37+4.5)4- FR4 (4.37 +4.5) + FeG (1.2 +1)E=→ FeG = 356 kNFR4 =177.4kN16
Sample Problem • For the compound roof structure shown, determine the axial stress developed in members AE and EG. q = 20 kN/m, AAE = AEG = 23 cm2 . 1. Reaction force 0 177.4 kN 2 y RA RB ql F F F • Solution 2 0 4.37 4.5 2 4.37 4.5 1.2 1 356 kN C RA EG EG q M F F F 2. Axial force in EG 16
q3. Equilibrium at joint E2FeG = FAe COSα-4.37DLB=FAE4V4.372 +12= FAE = 366 kNGJL目E9m4.37m4.37m4. Axial stresses in membersqAE and EG356(10°)FeG=155MPaOEGDAEG23(10-4)A366(103)FAEC= 159 MPaEHA23(10-4)AEFR4=177.4kN17
3. Equilibrium at joint E 2 2 cos 4.37 4.37 1 366 kN EG AE AE AE F F F F 4. Axial stresses in members AE and EG 3 4 3 4 356 10 155 MPa 23 10 366 10 159 MPa 23 10 EG EG EG AE AE AE F A F A 17
Stresses Acting on Oblique SectionscosαFFkPαcosα=ocoSαAAPaacosα = cos'α0αpαk=asin 2αsinαTα =pαT-2a: defined as the angle measured from cross section normal tooblique cross section normal.18
2 cos cos cos cos cos sin sin 2 2 A A F F p A A p p α: defined as the angle measured from cross section normal to oblique cross section normal. Stresses Acting on Oblique Sections 18
: On cross section: maximum normal stress; zero shearing stress: On 45°oblique cross section: maximum shearing stressaa0450 = cos2(45°)= sin(90)Tmax22219
• On cross section: maximum normal stress; zero shearing stress. 0 0 0 2 0 max 45 45 sin(90 ) , cos (45 ) 2 2 2 • On 45°oblique cross section: maximum shearing stress. 19
: On 9oobligue cross section: zero normal and shearing stress20
• On 90°oblique cross section: zero normal and shearing stress. 20