Determine the design flexural strength in ft kips of a W12 x 35 of A242 (Fy = 50ksi) steel subject toa. Continuous lateral support. Blank 1b. An unbraced length of 10 ft with Cb = 1.0. Blank 2C. An unbraced length of 20 ft with Cb = 1.0. Blank 3W12x35...A =d =10.30in.^212.500in.tw =0.300in.bf =6.560in.tf =0.520in.T =10-1/8in.k =0.8200in.k1 =0.7500in.gage =3-1/2in.rt =1.740in.d/Af =3.66Ix =285.00in.^4Sx =45.60in.^35.250in.ly =|Sy =24.50in.^47.47in.13ry =1.540in.Zx =51.20in.43Zy =11,50in 43J =0.74in.4Cw =879in. ^6a =55.42in.Wno =19.60in.^2Sw =Of =16.80in.^49.75in.43in.43Qw =25.40Blank 1 Add your answerBlank 2 Add your answerBlank 3Add your answer

A W12x35 is given We need to find out Design flexural strength

Determine the design flexural strength in ft kips of a W12 x 35 of A242 (Fy = 50ksi) steel subject to a. Continuous lateral support. Blank 1 b. An unbraced length of 10 ft with Cb = 1.0. Blank 2 C. An unbraced length of 20 ft with Cb = 1.0. Blank 3 W12x35 ... A = d = 10.30 in.^2 12.500 in. tw = 0.300 in. bf = 6.560 in. tf = 0.520 in. T= 10-1/8 in. k = k1 = 0.8200 in. 0.7500 in. gage = 3-1/2 in. rt = 1.740 in. d/Af = 3.66 Ix = 285.00 in.^4 Sx = 45.60 in.^3 5.250 in. ly =| Sy = 24.50 in.^4 7.47 in.13 ry = 1.540 in. Zx = Zy = J = Cw = 51.20 in.43 11,50 in 43 0.74 in.4 879 in. ^6 a = 55.42 in. Wno = 19.60 in ^2 Sw = 16.80 in.^4 Qf = 9.75 in 43 Qw = 25.40 in.43

Determine the design flexural strength in ft kips of a W12 x 35 of A242 (Fy = 50ksi) steel subject to a. Continuous lateral support. Blank 1 b. An unbraced length of 10 ft with Cb = 1.0. Blank 2 C. An unbraced length of 20 ft with Cb = 1.0. Blank 3 W12x35 ... A = d = 10.30 in.^2 12.500 in. tw = 0.300 in. bf = 6.560 in. tf = 0.520 in. T= 10-1/8 in. k = k1 = 0.8200 in. 0.7500 in. gage = 3-1/2 in. rt = 1.740 in. d/Af = 3.66 Ix = 285.00 in.^4 Sx = 45.60 in.^3 5.250 in. ly =| Sy = 24.50 in.^4 7.47 in.13 ry = 1.540 in. Zx = Zy = J = Cw = 51.20 in.43 11,50 in 43 0.74 in.4 879 in. ^6 a = 55.42 in. Wno = 19.60 in ^2 Sw = 16.80 in.^4 Qf = 9.75 in 43 Qw = 25.40 in.43

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter10: Plate Girders

Section: Chapter Questions

Problem 10.7.8P

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Determine the design flexural strength in ft kips of a W12 x 35 of A242 (Fy = 50ksi) steel subject to a. Continuous lateral support. Blank 1 b. An unbraced length of 10 ft with Cb = 1.0. Blank 2 C. An unbraced length of 20 ft with Cb = 1.0. Blank 3 %3D W12x35 A = 10.30 in 42 d = 12.500 in. tw = 0.300 in. bf = 6.560 in. tf = 0.520 in. T= k = k1 = 10-1/8 in. 0.8200 in. 0.7500 in. 3-1/2 gage = rt = in. 1.740 in. d/Af = 3.66 Ix = 285.00 in.4 in.43 Sx = 45.60 5.250 in. ly = Sy = 24.50 in.4 7.47 in.43 ry = Zx = 1.540 in. 51.20 in 43 Round off you answer to two decimal places. Zy =| 11.50 in.^3 J= 0.74 in.^4 Cw = 879 in.^6 a = 55.42 in. Wno = 19.60 in.^2 Sw = 16.80 in. 14 Qf = Qw = 9.75 in.43 25.40 lin 43 Blank 1 Add your answer Blank 2 Add your answer Blank 3 Add your answer
Determine the design flexural strength in ft kips of a W12 x 35 of A242 (Fy = 50ksi) steel subject to a. Continuous lateral support. Blank 1 b. An unbraced length of 10 ft with Cb = 1.0. Blank 2 C. An unbraced length of 20 ft with Cb = 1.0. Blank 3 W12x35 ... 10.30 in2 d= 12.500 in. tw = 0.300 in. bf = 6.580 in. 0.520 in. T= 10-1/9 in. k= 0.8200 in. k1 = 0.7500 In. gage 3 1/2 In. it d'Af = 1.740 n. 3.88 205.00 n.4 Sx- 45.80 n.3 IX= 5.250 n. ly= Sy = 24.50 7.47 In.43 ry = 1.540 n. Zx = Zy = J= Cw = 51.20 n.3 n.3 11.50 0.74 in.4 879 n.6 a = 55.42 n. Wno = 19.60 m2 Sw= 18.80 n.4 9.75 in43 Qw =| 25.40 in.3
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Determine the design flexural strength in ft kips of a W12 x 35 of A242 (Fy = 50ksi) steel subject to a. Continuous lateral support. b. An unbraced length of 10 ft with Cb = 1.0 C. An unbraced length of 20 ft with Cb = 1.0. W12x35 10.30 ... A = d = in.^2 12.500 in. tw = 0.300 in. bf = 6.560 in. tf = 0.520 in. T= 10-1/8 in. k = 0.8200 in. k1 = 0.7500 in. gage = rt = 3-1/2 in. 1.740 in. d/Af = 3.66 Ix = Sx = 285.00 in.^4 45.60 in ^3 5.250 ly = 24.50 in ^4 Sy = 7.47 in.^3 1.540 ry = Zx = in. 51.20 in.^3 Zy = J= 11.50 in 43 0.74 in.^4 Cw = 879 in. ^6 a = 55.42 in. Wno = 19.60 in ^2 Sw = Qf = 16.80 in.^4 9.75 in 43 Qw = 25.40 in.^3 in.
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