SW The composite bar shown in Fig. P-273 is firmly attached to unyielding supports. An axial force P = 50 kips is applied at 60°F. Compute the stress in each material at 120°F. Assume a = 6.5 × 10- 6 in/(in-°F) for steel and 12.8 x 10-6 in/(in.°F) for aluminum.
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- 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 maxEQuiz No.3 584nkkz1 Cc7KLzOsdr9YUFgvcfFSqpsdlyljtewQ/formResponse Choose the correct answer What will be the value of load applied parallel to the diagonal of (100x100x100))mm cube since the splitting tensile strength=3 Mpa 57.8 KN O 46.7 KN O 47.1 KN O (28)=27.5 Mpa, then the fc(180)=-- for concrete produced using moistPabacboadcom Question Campletion Stutn MOD LE 4 A Movtng to another question will save this response Question 1 A hwo span beam subjected to shear and flexure only is reinforced as follows: @ MIDSPAN 2-016 mm 3-016 mm SECTION @ FACE OF SUPPORTS TOP BARS BOTTOM BARS 5-016 mm 2-016 mm Given: Stirrup diameter, dg - 10 mm Concrete fe=21 MPa Steel rebar fy 415 MPa Stirrup fy= 275 MPa Beam sizebxh=250 mm x 450 mm Assume all bars laid out in single layer. Calculate the following: Tensile steel ratio in positive bending at midspan = (in 5 decimal places) Design Moment strength of section at midspan for positive bending = kN-m (nearest whole number) kN m (nearest whole number) Nominal Moment strength of section at face of support for negative bending =
- Question Eight tensile bars with a diameter of 16mm, equally divided into Fwo layers, are developed, C=50mm Spacing between Stirrups = 150 mm center to Center and spacing be tween longitudinal bars is 80 mm if Atr=230 mm? [leastof Cs& cb], Fy = Fyt = 350 M Pa. use Fé = 25,35 and 45 MPa and assume Atr is Satis fied with ACI & BS, Find length of development vequired. > Subject: Advanced Concvete Design, %3DThe rigid plate ABC shown in figure 2, was positioned on top of two identical steel posts, and the aluminum post was 0.1mm shorter than the steel posts at temperature of 50°C, if the allowable compressive stress in steel is limited to 120 MPa and in aluminum to 150 MPa, what is the maximum possible load P that can be supported on the rigid plate at temperature of 10°C ?(2) PoL=280* 16' 34' PLL= 200k 18' Please select the ligutest W 12 A992 steel.
- For the column shown what is the nominal compressive strength (Pn) ? Pn=? W12x50 10 12X50 THT X combilevered WIA W1422 acts compositely with a 4-inch-thick floor slab whose effective width b is 90 inches. The beams are spaced at 7 feet 6 inches, and the span length is 30 feet. The superimposed loads are as follows: construction load = 20 psf, partition load = 10 psf, weight of ceiling and light fixtures = 5 psf, and live load = 60 psf, A992 steel is used, and fc=4 ksi. Determine whether the flexural strength is adequate. a. Use LRFD. b. Use ASD.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 Next
- A For the truss shown, find veyticat deflection bylunit lead Methed. EA 15 _atfd) Ioku for all members increase in temp is (50€) in (ad) short in (ad)is (3 EmfrThe 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. Use a501 plate and a502 gr2 rivet 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 rivets2. A column HP 14 x 102 of A572 Gr. 55 steel has a length of 15 ft is fixed at both ends. Compute the design compressive strength for LRFD and the allowable compressive strength for ASD. (Steel section properties are provided in the next page) ASTM Designation A572 Gr. 42 Gr. 50 Gr. 55 Gr. 60⁰ Gr. 65⁰ Yield Stress (ksi) 42 50 55 60 65 Fu Tensile Stressa (ksi) 60 65 70 75 80