Normal Stress Problem 104 A hollow steel tube with an inside diameter of 100 mm must carry a tensile loadof 400 kN. Determine the outside diameter of the tube if the stress is limited to 120 MN/m2. Solution 104 P= A P= A where: P=400kN=400000N =120MPa A=41 D2–41 (1002) A=41 (D2–10000) thus, 400000=120[41 (D2−10000)] 400000=30 D2−300000 D2=30 400000+300000 D=119 35mm answer Problem 105 page 12 Given: Weight of bar = 800 kg Maximum allowable stress for bronze = 90 MPa Maximum allowable stress for steel = 120 MPa Required: Smallest area of bronze and steel cables Solution 105 By symmetry: Pbr=Pst=21(7848) Pbr=3924N Pst=3924N For bronze cable: Pbr= brAbr 3924=90Abr Abr=43 6mm2 For steel cable: answer Pst= stAst 3924=120Ast Ast=32 7mm2 answer . Problem 106 page 12 Given: Diameter of cable = 0.6 inch Weight of bar = 6000 lb Required: Stress in the cable Solution 106 MC=0 5T+10 3 34T =5(6000) T=2957 13lb T= A 2957 13= 41 (0 62) =10458 72psi answer . aluminum.5 in2 Required: Stress in steel.Problem 107 page 12 Given: Axial load P = 3000 lb Cross-sectional area of the rod = 0. and bronze sections Solution 107 For steel: stAst=Pst st(0 5)=12 answer For aluminum: alAal=Pal al(0 5)=12 al=24ksi st=24ksi answer For bronze: brAbr=Pbr br(0 5)=9 br=18ksi answer . use P=10000N=10kN answer .Problem 108 page 12 Given: Maximum allowable stress for steel = 140 MPa Maximum allowable stress for aluminum = 90 MPa Maximum allowable stress for bronze = 100 MPa Required: Maximum safe value of axial load P Solution 108 For bronze: brAbr=2P 100(200)=2P P=10000N For aluminum: alAal=P 90(400)=P P=36000N For Steel: stAst=5P P=14000N For safe P. 4 in2 Cross-sectional area of wire AC = 0.Problem 109 page 13 Given: Maximum allowable stress of the wire = 30 ksi Cross-sectional area of wire AB = 0.5 in2 Required: Largest weight W Solution 109 For wire AB: By sine law (from the force polygon): TABsin40 =Wsin80 TAB=0 6527W ABAAB=0 6527W 30(0 4)=0 6527W W=18 4kips For wire AC: TACsin60 =Wsin80 TAC=0 8794W TAC= ACAAC 0 8794W=30(0 5) W=17 1kips Safe load W=17 1kips answer . Problem 110 page 13 Given: Size of steel bearing plate = 12-inches square Size of concrete footing = 12-inches square Size of wooden post = 8-inches diameter Maximum allowable stress for wood = 1800 psi Maximum allowable stress for concrete = 650 psi Required: Maximum safe value of load P Solution 110 For wood: Pw= wAw Pw=1800[41 (82)] Pw=90477 9lb From FBD of Wood: P=Pw=90477 9lb . For concrete: P c= cA c Pc=650(122) Pc=93600lb From FBD of Concrete: P=Pc=93600lb Safe load P=90478lb answer . . 8 in2 Required: Stresses in members CE.Problem 111 page 14 Given: Cross-sectional area of each member = 1. DE. and DF Solution 111 From the FBD of the truss: MA=0 24RF=16(30) RF=20k At joint F: FV=0 53DF=20 DF=3331k(Compression) At joint D: (by symmetry) . BD=DF=3331k(Compression) ΣFV=0 DE=53BD+53DF DE=53(3331)+53(3331) DE=40k(Tension) At joint E: FV=0 53CE+30=40 CE=1632k(Tension) Stresses: Stress = Force/Area CE=1 81632=9 26ksi (Tension) answer DE=401 8=22 22ksi (Tension) answer DF=1 83331=18 52ksi (Compression) answer . and CF Solution 113 For member BD: (See FBD 01) MC=0 3(54BD)=3(60) BD=75kN Tension BD= BDA 75(1000)= BD(1600) BD=46 875MPa (Tension) For member CF: (See FBD 01) answer MD=0 4(1 2CF)=4(90)+7(60) CF=275 77kN Compression CF= CFA 275 77(1000)= CF(1600) CF=172 357MPa (Compression) For member BC: (See FBD 02) answer . BD. Required: Stresses in members BC.Problem 113 page 15 Given: Cross sectional area of each member = 1600 mm2. MD=0 4BC=7(60) BC=105kN Compression BC= BCA 105(1000)= BC(1600) BC=65 625MPa (Compression) answer . Problem 114 page 15 Given: Maximum allowable stress in each cable = 100 MPa Area of cable AB = 250 mm2 Area of cable at C = 300 mm2 Required: Mass of the heaviest bar that can be supported Solution 114 FH=0 TABcos30 =RDsin50 RD=1 1305TAB FV=0 TABsin30 +TAB+TC+RDcos50 =W TABsin30 +TAB+TC+(1 1305TAB)cos50 =W 2 2267TAB+TC=W TC=W−2 2267TAB MD=0 6(TABsin30 )+4TAB+2TC=3W 7TAB+2(W−2 2267TAB)=3W 2 5466TAB=W TAB=0 3927W TC=W−2 2267TAB TC=W−2 2267(0 3927W) TC=0 1256W Based on cable AB: TAB= ABAAB 0 3927W=100(250) W=63661 83N Based on cable at C: T2= CAC . 0 1256W=100(300) W=238853 50N Safe weight W=63669 92N W=mg 63669 92=m(9 81) m=6490kg m=6 49Mg answer .