F1–19. If the54 eyebolt is made a yield F1–22. The pin is made of a material having a failure C H Aof P TaE material R 1 S T having RESS required shear stress of tfail = 100 MPa. Determine the minimum 1 stress of sY = 250 MPa, determine the minimum P diameter d of its shank. Apply a factor of safety F.S. = 1.5 required diameter of the pin to the nearest mm. Apply a F1–19. If the eyebolt is made of a material havingfactor a yield is made a material having a failure against yielding. of safetyF1–22. of F.S. The = 2.5pin against shearoffailure. shear stress of tfail = 100 MPa. Determine the minimum 1 stress of sY = 250 MPa, determine the minimum required F1–16 diameter d of its shank. Apply a factor of safety F.S. = 1.5 required diameter of the pin to the nearest mm. Apply a against yielding. factor of safety of F.S. = 2.5 against shear failure. F1–17. ÖDEV The strut NO is glued 3. to the horizontal member at surface AB. If the strut has a thickness of 25 mm and the glue d 80 kN Soruan average 1. Diyagonal AB yüzeyi boyunca, yatay duran elemana yapıştırılmıştır. can withstand shear stresseleman of 600 kPa, determine 30 kN the maximum force P that can be applied to the strut. Diyagonal elemanın kalınlığı 25mm ise ve yapıştırıcının taşıyabileceği ortalama kayma d gerilmesi 600 kPa ise diyagonal elemana uygulanabilecek en büyük P kuvvetini hesaplayınız. P n at ear, 80 kN kNSIMPLE CONNECTIONS DESIGN30OF 1.7 DESIGN OF SIMPLE55 CONNECTIONS F1–22 1.7 55 F1–19 PROBLEMS 50 mm PROBLEMS F1–23. If the bolt head and the supporting 1 bracket are F1–22 1 made of the same material having a failure shear stress of t , determine the maximum allowable force P = 120 MPa F1–19 fail F1–23. If the the bolt bolt head and the supporting bracket are F1–20. If the to bara assembly is made Member B is subjected compressive force of of atomaterial *1–76.having The lapbelt to The be subjected to aso force can*1–76. be isapplied to that it be does not pull •1–73. Member B is 3subjected of assembly lapbelt assembly is to subjected to a force 60! a compressive force that made of the same material having a failure shear stress of a yield stress of ksi, determine the minimum s = 50 3 A and B are both made of wood and are thick, in. of 800 N. Determine (a) the required thickness t of Y through of the800 plate. a factor safety of F.S.thickness = 2.5 8 N. Apply Determine (a)ofthe required t of 1800 lb. If A and B are both made of wood and are 8 in. thick, t , determine the maximum allowable force P = 120 MPa required dimensions 1>8 in. Apply a h 1 nearest ne54 to the nearest h of in. the belt if the allowable tensile stress for the material fail C H4determine A P T the E R 1smallest S T1Rand E dimension SA Sh2 to the B shear failure. to the nearest dimension h against of the belt if the allowable tensile stress for the material 4 in. the smallest F1–20. If1.5 the bar assembly isEach made of a material having that can be lap applied to the bolt so that it does not pull factorsoofthat safety F.S. = against yielding. bar has a zontal segment it does not fail in shear. The is (b) the required length (s ) = 10 MPa, d t allow l the horizontal segment that it does not fail in shear. The is (st)allow = 10 MPa, (b) the required lap length dl a yield stress of so determine thecanminimum s Y = 50 ksi, if through the plate.stress Applyofa factor of safety of F.S. = 2.5 thickness of 0.5 stress forF1–22. theaverage is tallow = 300 psi. the=aglue sustain ifanthe allowable shear # segment gshear a yield Thein. pin is made a materialishaving failure forof the tallow 300 psi.in. Apply glue sustainhaving an allowable 80 mm required dimensions and h2 atoyield the nearest 1>8 a h1segment F1–19. If the eyebolt is shear made stress of a F1–17 material having F1–22. The pin is the made of shear acan material a failure shear stress of against failure. and (c) required diameter (t ) = 0.75 MPa, dr of g equired shear stress of tfail = 100 MPa. Determine allow the minimum and (c) theminimum required diameter dr of (tallow )100 0.75.MPa, g = MPa factor of safety F.S. = 1.5 against yielding. Each bar has a stress of sY = 250 MPa, determine the minimum required shear stress of t Determine the = failshear stress for Soru 2. Şekildeki kayma gerilmesi # the olan themm. pinApply if the thebirpinmalzemeden is S. = 1.5 required diameter of the pinbulon to the göçme nearest a allowable pin if the allowable shear stressa forimal the pin is thickness 0.5 in.of safety F.S. = (t1.5 ) =required diameter d of its shank. Apply aoffactor diameter of the pin to the nearest mm. Apply 30 MPa. allow p factor of safety of F.S. = 2.5 against shear failure. edilmiştir. Bulon için gerekli en küçük çapı en yakın mm ölçüsünde hesaplayınız. Kayma 80 mm (t ) = 30 MPa. mm 75 allow p F1–18. Determine the maximum average shear stress against yielding. factor of safety of F.S. = 2.5 against shear failure. 15 kip h1 göçmesine karşı F.S.=2 kullanınız. developed in the 30-mm-diameter pin.emniyet 30katsayını kip B h2 800 lb B 15 kip 30 kN h2 C 13 12 30 kN ress 5 d C h A F1–20 AB 15 kip 13 12 h 5 B 15 kip 30 kN 800 lb h1 800 N 30 kip 80 kN A t 45 mm 45 mm t dl A 75 mm 30 mm 800 N 40 mm80 kN 30 mm dl F1–20 P F1–2340 mm F1–21. Determine the maximum force P that can be applied to the rod if it is made of material having a yield dr F1–22 dr # Y 1–73 F1–22 F1–24. Six nails are used to 800 hold at A against stress Prob. of s the possibility that failure = 250 MPa. ConsiderProb. 1–73 N the hanger P F1–23 F1–21. Determine the maximum force P that can be The lever isoccurs attached to the shaft A using a key that the column. Determine the minimum required diameter800 of N in1–74. the rodThe andlever at section a–a. Apply a shaft factorAofusing safety is attached to the a key that p F1–19 Soru 3. Şekildeki çubuğa uygulanabilecek en büyük P kuvvetini hesaplayınız. Çubuk akma applied to the rod if it is made of material having a yield F1–23. If the bolt head and the supporting bracket are dth d and of length 25 Ifyielding. the shaft is fixed and each to head the nearest if it is made are of material in. Prob. F.S. of =has 2 against Prob. 1–76 If thenail bolt and the1>16 supporting bracket amm. width d and length of 25 mm. If the shaft isF1–23. fixed and 1–76 Six nails are used to hold the hanger at A against stress of smaterial MPa. Consider the possibility failuret = F1–24. of the same aolan failuremalzemeden shear stress ofthathaving Y = 250having al force of 200made N ais applied perpendicular to the gerilmesi # yapılmıştır. a-a kesitinde göçme ihtimalini . Apply a factor of safety of 16 ksi fail made of the same material having a failure shear stress ofF.S. = 2 vertical force ofrod 200and N at is section applieda–a. perpendicular to the the column. Determine the minimum required diameter of occurs in the Apply a factor of safety t , determine the maximum allowable force P = 120 MPa 40 kN fail determine the dimension ddetermine if thebulundurunuz. allowable shear Akmaya against failure. tfailF.S.=2 ,shear determine the maximum allowable force P =shear 120 MPa gözönünde karşı emniyet katsayısını kullanınız. handle, the dimension d if the allowable having each nail totothethe nearest 1>16 in. if it is made of material of F.S. = made 2 against yielding. thatbar can be applied to the bolt so that it•1–77. does not pull The wood specimen is subjected pull of r the key isIftallow =stress 35assembly MPa. F1–20. the is of a material having that can be applied to the so specimen that it does not pull to the pull of •1–77. Thebolt wood is subjected fora the key isF1–18 tallow = 35 MPa. inimum having tfail . Apply a factor of safety of F.S. = 2 = 16 ksinormal through plate. a factor safety of F.S. = 2.5 testing machine. 10 kN in athrough tension If the allowable a yield stress of sYthe ksi,Apply determine theofminimum = 50 the plate. Apply a factor of safety of F.S. 2.5allowable normal 10 kN in a tension testing machine. If=the 300 lb/ft d Apply a shear failure. the d nearest 1>8 in. Apply against shear failure. stress foragainst the shear woodfailure. isstress )allow 12 MPa (stagainst required dimensions h2 to the a h1 and for = the woodand is (st)allow = 12 MPa and the P ar has a a allowable shear stress is tallowable determine the = 1.2 MPa, determine the MPa, stress factor of safety F.S. = a1.5 againstayielding. a allow = 1.2shear 40amm Each bar has is tallow A required dimensions b and t so that the specimen reaches 300 lb/ft thickness of 0.5 in. required dimensions b and t so that the specimen reaches 50 mm80 mm 20 mm 80 mm 20 mm these stresses simultaneously. Thestresses specimen has a width The of specimen has a width of P these simultaneously. 500 mm a 40 mm 25 mm. 500 mm B 25 mm. mm 75mm 60 mm 200 N 50 200 N ha1 30 kip Prob. 1–74 Section a-a Prob. 1–74 # h2 120 60 mm A mm 30 mm The joint is fastened together using bolts.together 1–75. The joint is two fastened using two bolts. kip F1–21 B 15 C ne the required diameter of the bolts if the failure Soru 4. Aşağıdaki birleşim iki bulon kullanılarak Determine the required diameter of the bolts if the failure Section a-a ess for the bolts is shear t Use a factor of = 350 MPa. fail 40 mm # olduğuna gerekli bulon çapını stress for the bolts is tfailgöre a factor of = 350 MPa. Use A 120 mm 15 kip F1–20 r shear of F.S. = 2.5. safety for shear of F.S.kullanınız. = 2.5. F1–21 emniyet katsayısı 30 mm 80 kN 75 mm 10 kN B 10 kN A 9 ft 30 mm F1–24 gerçekleştirilmiştir. Akma göçme gerilmesi t Kaymaya karşı F.S.=2.5 belirleyiniz. A t 40 mm A 9 ft 80 kN 30 mm P F1–23 can be b P F1–23 b F1–21. Determine the maximum force P that can be a yield 30 mm applied to the rod if it is made of material having a yield 30 mm F1–24. Six nails are used to hold the hanger at A against t failure F1–24. Six nails are used to hold the hanger at A against stress of sY = 250 MPa. Consider the possibility that failure the column. Determine the minimum required diameter of of safety the column. Determine the minimum required diameter of occurs in the rod and at section a–a. Apply a factor of safety each nail to the nearest 1>16 in. if it is made of material each nail to the nearest 1>16 in. if it is made of material of F.S. = 2 against yielding. having tfail = 16 ksi. Apply a factor of safety of F.S. = 2 having tfail = 16 ksi. Apply a factor of safety of F.S. = 2 against shear 40 kN failure. against shear failure. kN a # kN 30040 lb/ft Prob. 1–75 300 lb/ft 40 mm 10 kN 10 kN Prob. 1–77 P Prob. 1–75 Prob. 1–77 50 mm a A 120 mm B B A 60 mm #22 Section a-a F1–21 9 ft F1–24 9 ft F1–24 F1–24

Download
# Ödev3