t 4#9 1) fé= 5000 psi r Grade 60 steel; 4 #9 Rebars, 5-9" t 4#9 NWC, 5-9" L = 25 ft & Mn = ? 4#9 14" $ 21" #3" Simple span 39
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- Problem. A bolted connection shown consists of two plates 300 mm x 12 mm connected by 4 - 22 mm diameter bolts. 12 mm bolts T12 mm te12 mm P 300 75 75 75 Edge distances = 75 mm %3D dnole for tensile and rupture = dp +3 mm %3D dnole for bearing strength for Lec = d + 1.5 mm Fy = 248 MPa Fu = 400 MPa %3D En 330 MPa %3D Use LRFD design method. Determine the design strength due to tensile rupture strength of plates. (kN)4. The tension member shown below is a C 310 x 30.8 of A572M gr 345 steel materials. Will it safely support a service dead load of 265 kN and a service live load of 555 kN. L 50 102.5 102.5 50 65 65 65 O O 22 veny diameter bolts C310 x 30.8 Iso A36M stool matorial and select a double anglo tonsion member to rosist(1) 21" 3" 4#9 bood 14" (1) Ec = ? (2) fr=? (3) Mer=? (5) #4 (4) When M = 70 & Mn = ? fé = 5 ksi Grade 60 Steel NWC. ft-k I 1R-2 Dr b=14", h=24" #4 tied stirrups. 4 #9 Rebars. fe=? Please write down units properly in the final answers. fs=?
- A76 x 76 x6 mm angular section shown is welded to an 8 mm thick gusset plate. The length L1 is 65 mm, L2 is 125 mm, and the cross sectional area of the angle is 929 sq.mm. Fy-248MPA and Fu -400MPA. Gusset Plate Alowable Sresse Alowable tensle stress igross ara)0.00 Alowable tereile stres net area)0SOF. Allowable shear stress (net ares)0.30F. Determine the value of P based on net area using a strength reduction coefficient of 85% O 138.24 KN O 140.14 kN O 157.93 kN O 304.00 N NextO cd O efgh O ij O klmn O qrst V What is the line of action of which the section will yield in tension? All holes are the same diameter. O aflrsmgb O op t P n j h C d10 – 7/8" Bolts Gusset Plate Plate thickness t = ½" 2" b. 2" Tension Member A36 Steel Plate 2" P 12" 2" g I 2" Calculate Tension Capacity 2" of line 1. abcde (A,„) 2. abfcgde (A„) 3. abfgde (A,), and 4. Yielding over Agross An A 3" 3"
- cd efgh ij O klmn u op V 0 0 0 t Р What is the line for which the section will rupture in tension? All holes are the same diameter. m n j h d3 10" 2" EI 4" -2" 2" 12" UT 8' A992 steel. &Mp= ?Material Strengths: Fy = 248 MPa, F = 400 MPa For the two lines of bolt holes, the pitch is the value that will give a net area DEFG equal to the one along ABC. The diameter of the holes is 22 mm. The thickness of the plate is 12 mm. Determine the LRFD design tensile strength based on yielding on gross area. D A to 50 1 ÓB 50 50 ic O 410 kN O 464 kN O 446 kN O 401 KN F IG Pitch Pitch KISS|Y
- sted 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| * A's O As,web As bw Steel areas: A's-2 + 14 = 308 mm² A₁=50 mm² (area of $8) Materials : C25/30 / B420C Longitudinal reinforcements A's=2 $ 14 As,web= 2 $ 12 As= 5 0 14 bw=300 mm d=670 mm d'=30 mm k₁= 0,85 k3 = 0,85 Pb=0.0205 (from table) Transverse Reinforcements $8/200 mm (2-leg stirrups) Material coefficients Ymc:1.5 (Concrete) Yms: 1,15 (Steel) Determine the reinforcement ratio and failure type of the beam As,web= 212= 226 mm² Asw=2x50=100 mm² (Total Area of two leg stirrup) A₁= 5 + 14=770 mm²Determine the maximum factored LRFD tensile force capacity in tension only (no block shear). The angle is ASTM A36 steel. X = 0.7 Y = 3.8 Z=1/2 Round your answer to 2 decimal places. Your Answer: Incorrect The answer is 77.15 ± 1%. 1/2" X" bolts 1½" Gusset plate -L3 X 3 X Z Tu I