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- FOR THE GIVEN PROPERTIES DESIGN d': 15 mm THE REINFORCEMENT OF : 400 mm b d = $80 mm fc = 33 MPa fy = 320 MPa Mu: 1 000 000 000 N mm A. USE B.USE OF NSCP 2001 NSCP 2015 BEAM THE BEAM 132 13 Gra mm of BELOW. #H 1606 210135 03 M 12437 P0122399/100mA 02338941 242The given plate below with width of 200 mm andthickness of 16 mm is to be connected to two plates ofthe same width with half the thickness by 20 mmdiameter rivets as shown. The rivet hole is 2 mm greaterthan the rivet diameter. Allowable tensile stress on netarea is 0.6Fy. Allowable bearing stress is 1.35Fy. plate: fy = 241 mpa fu = 4141 mpa rivet: fv = 152 mpa a. Determine maximum load P without exceeding allowable tensile stress on plate b. Determine maximum load P without exceeding allowable shear stress on rivets c. Determine maximum load P without exceeding allowable bearing stress between plates and rivetsDate: Course: Time: Pind the vertical movement os is 120 KN Stsel Aluminum 3 m Im B P=120 kat Aluminum A = 900 mm² E = 70 GN m² steel A= 600 mm E 200 GNM² %3D
- Ro blem 2 20" го Given Base Rate A36 (Fye36 Ksi) WI2X5 20" B= N= 20 in t= 17/2" W 12 x 50 (de12.2", bf: 8.08 in) fc: 5 ksi A= A2 Determine the max Tu that Can be applied to this base Plate ? biet: ) determine u banaal on A, (2) Chorse the Smaller af tw valusSubject/ Plates and shells thery Square Plate Caxa) simplysupported Subjected to Con centrated force (4) over a central on all Face edges is Find q mn ? area (a *a). Use Book Theory Dear, Please help me ... of Plates and Shells for S. TimoshenkoI need a clear answer by hand, not by keyboard and fast answer within 20 minutes. Thank you | dybala For the beam loaded as shown, find the maximum tensile and compressive bending stresses and maximum shearing stress.. IN.A=4*10mm* 80mm 100mm 60mm 60mm 1.5KN 2kN/m 1mal A |-----|- | Im ++ Im->> 3m 3kN/m
- 7 30 Enercise:- diameter of 12mm A cylindrical rod 100 mm long with is subjected to a 30000 N. Assuming that the tensile load of deformation is entirely elastic and allowable elevation in pligmeter is allowable reduction in dligmeter is 0.015mm. choose amory only meeting requirent U (Poiss ons vahu) 0.33 below Material Aluminium alloy Bronze Titanium allow manimum 0.3 mm while manimum E (GPa) 70 100 110 0.34 3Use R as 899sted ingid 250MM B k steel bar bolt AUR rigid bar 12 mm thick 8 mm Brass 350mm Jointd 16 mene thick 350mm JointB 250 mm Xg = 20x10% Eg = 90 G Pa -6 A, = 12x10/0 Es = 200 G Pa thickness> 16 mm 8 mm bolt %3D 12 mmi thick shear strength f bolf bearing streng th of holt= 100 la 50 MPa %3D Assume no failure will take place in steel or brass. Temperature chauge brass steel on Temperature change on Determine thai can be applied to system - the max
- The rubber band given below is subjected to the following tensile loading. Calculate the minimum thickness of the rubber (tr) and the minimum steel pin diameter (Dpin) so that the structure does not fail. Consider: Allowable tensile strength of the rubber= 20MPa Allowable shear strength of the steel = 200MPa Reflection: 1) How would you solve this problem if a Factor of Safety was given? 2) Are there any other dimensions worth calculating for the rubber belt?The thin uniform plate shown has a mass of 100kg and is subjected to a force and couple momentalong its edges. If it is supported in the horizontalplane by a roller at A, a ball-and-socket joint at B,and a cord at C, determine the components ofreaction at these supports to maintain equilibrium. Please provide all written work and show FBD. Thank you!A0AC ** od of added a fost o pointed to the P = 1400 b P-1400 b