Q1:If allowable stresses are o, 165MPa, o, 100MPa T 70MPa determine max. value of W. W 300mm Ti00mm Im 4m Im 300mm T 100mm
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Q: Q1:If allowable stresses are o, 165MPa.0, 100MPa = 70MPa determine max. value of W. M 300mm 100mm Im…
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- I 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/mEQuiz 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 moistWHERE: BOLT DIAMETER = 22mm Fy = 241Mpa Fu = 414Mpa
- Determine the bolt diameter if the shear stress is 150 MPa and the bearing stress is 337.50 MPa. p 100 mm 8881 B I 20 mm 国町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 max500 mm 62.5 mm 375 mm 62.5 mm P. Determine the total required number of 20mm bars to be used in two end faces only for the 300mm x 500mm rectangular column shown to support the following loads: Pu = 800 kN Mu = 120 kN-m Use f'c = 20.7 MPa and fy = 413.7 MPa Use 1 ksi = 6.893 MPa %3D 300 mm O O O O
- Determine the safe load of the column section shown, if it has a yield strength of 25 MPa. E = 200000 MPa. Use NSCP Specifications. Fyz248 mpa Properties of Channel Section d = 305 mm t₂ = 7.2 mm A = 3929 mm² t₁ = 12.7 mm Ix=53.7 x 10mm¹ x = 117 mm Properties of W 460 x 74 A = 9450 mm² b = 190 mm ly= 1.61 x 10 mm x = 17.7 mm tw = 9.0 mm rx = 188 mm ry = 41.9 mm d = 457 mm tr = 14.5 mm Ix = 333 x 10 mm Iy = 16.6 x 10mm* 7.21 When the height of column is 6 m. When the height of column is 10 m. Assume K= 1.0 457 CIVIL ENGINEERING- STEEL DESIGNA steel flat of rectangular section of size 70 x 6 mm is connected to a gusset plate by three bolts each having a shear capacity of 15kN in holes having diameter 11.5mm. If the allowable tensile stress in the flat is 150MPa, the maximum tension that can be applied to the flat is T4 35 15 1221 5885 20 20 15Designation Serial size 457x191 ■ Calculate the design tension resistance of the 457 × 191 × 67 UB. Given information from steel section table (assuming S275): Axis y-y cm³ 1296 67 Elastic modulus Wel Axis Z-Z cm³ 153 Mass per metre kg/m 67.1 Depth of section h mm 453.4 Width of section b mm 189.9 Plastic modulus Wpl Axis y-y cm³ 1471 Axis Z-Z cm³ 237 Thickness Thickness Root of web of flange radius tw tf mm r mm mm 8.5 12.7 Buckling parameter U 0.872 Torsional index X 37.9 10.2 Depth between fillets d mm 407.6 Ratios for local buckling Flange Web Cf/tf Cw/tw dm6 0.705 6.34 Warping Torsional constant lw constant IT cm4 37.1 48.0 Area of section cm² 85.5 Second moment of area Axis y-y cm4 29380 Axis Z-Z cm4 kNm 522 1452 Indicative values for S355 steel Mc.y.Rd Nb.z.Rd* for Lcr=3.5m kN 1600 Radius of gyration i Axis y-y cm 18.5 Axis Z-Z cm 4.12; Designation Serial size 67 457x191
- The center of gravity of the 800N highway sign is at point G. It has been determined that it can withstand a tensile force of 1200N. Other elements in the system are sufficiently durable. The sign is intended to be connected to the fixed joints D and C and AB rope as shown in the figure. Is such a design correct in terms of resistance? 6m B 2m 8m Al D 6m 2m 2mExercise 1 : Find the deflection at the point load and the slopes at the ends for the steel shaft which is simply supported at the bearing A and B as shown in Fig.1. Calculate the following: a. Deflection at the point load b. Slope at the ends - 1.25 x 10 mm 3000 N I2= 4x 10 mm B 150 mm 75 mm /- 125 mm Fig-() Take E-200 GPa. Given: Young's modulus, E = 200 GPa - 200 x 10 Pa (or) 200 x 10 N/m2 E - 200 x 10 N/m2 = 2x 10 N/mm2 Moment of inertin, Element i and 2, I, = 1.25 x 105 mm4 Element 3, 1, - 4 x 10 mm 3000 N my. : m2, 2 dyn. Fiy 150 mm da Fay d,3 Fay dye, Fay 125 mm t75 mm Fig. (i). For ciement 1, length L, = 150 mm For element 2, length L, - 75 mm For element 3, length L, = 125 mm Point load at node 2, F, =' 3000 N Fig.1OP N C What is the shear resistance of a universal beam (UB) with the following dimensions [EC3 Section 6.2 Resistance of cross sections]: Cross sectional area A = 105 cm² Height h = 584.6 mm Flange thickness t = 11.4 mm Flange width b = 213.2 mm y-y Web thickness tw = 7.8 mm Root radius r = 12.7 mm Yield strength fy = 355 N/mm² Calculate the shear resistance and give your answer to the nearest whole kN. Do not include units in your answer.