= ======= == Q2- The architectural engineer proposed a width of 350 mm and a height of 750 mm for a cantilever beam shown below, as you a design engineer ,design this beam according to ACI-14 requirements and assume in your solution that: Mu 1300 kN.m fc 21 MPa and fy 420 MPa Bar diameter of 32mm for longitudinal reinforcement. Bar diameter of 12 mm for stirrups. Two layers of reinforcement for both tension and compression zone. ⚫Use bundle bars when it is possible. 350mm
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- 7 7a 7b 7c Alaterally supported beam was designed for flexure. The beam is safe for shear & deflection. The most economical section is structural tubing however the said section is not readily available at the time of the construction. If you are the engineer in charge of the construction what alternative section will be the best replacement? Why? The section is 8" x 8" x 7.94 mm thick: Use Fy=248 MPa: E=200,000 MPa AISC wall thickness Ix 106 S x 103 Jx 103 mm4 mm3 mm4 rx =ry Area Ag (mm2) mm Designation Weight/m 8x8 7.94 47.36 6,039 79.25 37.84 371.99 60.35 expla'n briefly your cho'ce. (transform your comparative analys's 'nto a narative form to support your cho'ce) 8x8 14.29 80.61 10,258 76.2 59.52 585.02 99.06 8x8 72.7 9,290 76.96 54.53 539.13 90.32 8x8 9.53 56.09 7,161 78.48 44.12 432.62 70.76 mm 12.7 Zx 103 mm3 437.53 714.48 650.57 512.92Please write the complete solutions and Legibly. Rate will be given! Do not just copy asnwers in CHEGG! Thank you :) A singly reinforced rectangular beam 270 mm x 430 is reinforced with 4-16-mm-dia. rebars with yield strength 420 MPa. If bar centroid is located 70 mm from the bottom edge of the beam and f'c = 21 MPa, calculate the design moment strength in kN-m. Write your answer with 2 decimal places only.Determine the Design Flexural Strength of W460x52 A992 Steel with the following cases: CIVIL ENGINEERING STEEL DESIGN With Continuous lateral bracing Unbraced length = 4m, Cb = 1.0 Unbraced length= 12m, Cb = 1.0 a.) b.) c.) Fy = 345 MPa Properties: d = 450mm tf = 10.8 mm 1x 212x10mm* Cw 306x10 mm" Sx** 944 x10¹mm¹ bf 142mm MY Ag - 6650mm2 Zx 1090x10'mm' Ho439mm ry = 31mm tw* 7.62mm J-211x10¹mm* ly = 6.37x10 mm rls - 38.4mm
- For the column shown what is the nominal compressive strength (Pn) ? Pn=? W12x50 10 12X50 THT X combilevered WI(2) PoL=280* 16' 34' PLL= 200k 18' Please select the ligutest W 12 A992 steel.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/m
- 6 6a 6b 6c Alaterally supported beam was designed for flexure. The beam is safe for shear & deflection. The most economical section is W 6 x 20 however the said section is not readily available at the time of the construction. If you are the engineer in charge of the construction what alternative section will be the best replacement? Why? Use: Fy=248 MPa: E=200,000 MPa Designation W8 x 24 W8 x 21 Weight Ag (mm2) 3787 36 31 width mm kg/m d (depth) 30 157.48 152.91 9.27 W6 x 20 expla'n briefly your cho'ce. (transform your comparative analys's 'nto a narative form to support your cho'ce) W8 x 28 42 204.72 165.99 11.81 5323 4568 201.42 164.97 10.16 3974 210.31 133.86 flange bf thickness tf Web 10.16 thickness tw 6.60 Elastic Properties mm 4 Ix x 106 7.24 6.22 6.35 17 41 34 31 mm 3 Sxx 103 220 398 342 298 mm rx 67.56 87.63 86.87 88.65 mm 4 lyx 106 6 9 8 4 mm 3 Syx 106 72 109 92 61 mm ry 38.10244 41.15 40.89 32 Plastic properties. mm 3 Zx x 103 244 446 380 334 mm 3 Zyx 103 110 166 140 93…Pabacboadcom 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 =(1) Given A992 steel Ag = 10 in %3D KLy- 80, ry what is $ Pa=? = 60 the' column streugth design
- 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 maxThe 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 ?O 88 130% v - + I Annotate T| Edit Trial expired Unlock Full Version ENGR 263 + A A A T O 4.10 Member AB is the beam under consideration. As shown in the illustration of the loading condition, member AB is an overhanging beam that supports a uniformly distributed roof load of 500 lb/ft. It also carries concentrated loads from a rooftop HVAC unit (4000 lb), an interior hanging display support (2000 lb), and a marquee sign (3000 lb). Marquee overnang Displau SLIPPort II Raof Kiots Ol I W = 500 Ib/Ft A B 4000 b 2000 000 I Steel beam !! I1 3 FE 5 FE - Ft RoOFtop HVAC unit 10 FE 4 Ft Marquee sign< R R2 Steel beam (negligible weight) Free-body diagram Dispiay Support Loading condition