Q.3 The steel column shown Fig. 3 is pinned at both ends and unbraced about both x-x andbraced at mid-height y-y axes. Calculate the axial capacity of a W14 x 99 section A992 (Fy= 50 ksi) to assess its the adequacy of a for the given conditions.PD = 250 kPL = 500 kL=32'ŽFIG. 3TL=16'

The axial capacity needed for the steel column is 1100KExplanation:Step 1: Pu=1.2(250)+1.5(500) this…

Answered: Q.3 The steel column shown Fig. 3 is…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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I the Design the reinforcements of the given T beam below. bf=900mm bw=420mm tf=120mm d=550mm d'=80mm fc'=34MPA fy=415MPA USE NSCP 2015 mm2 A.Mu = 1400KN-m, As = mm2 B.Mu = 1700kN-m, As = mm2, As' = CS Scanned with CamScanner
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AW14 x 43 column in a braced frame has a yield stress of 50 ksi and a height of 10 ft. The column is pinned at the top and bottom and has no intermediate bracing. Determine the available strength of the column considering the critical stress for flexural buckling. From AISC Manual Table 1-1, the properties of a W14 x 43 are: A = 12.6 in^2 tw = 0.305 in bf/2tf = 7.54 h/tw = 37.4 rx = 5.82 in ry = 1.89 in Ix = 428 in4 ly = 45.2 in4 G= 11,200 ksi J = 1.05 in^4 Cw=1950 in^6 What is the nominal axial strength considering flexural buckling stress of the column section (in kips). O 310 O 468 O 447 O 375 O 420 O 436 O 517 O 280
The W36 x 135 beam shown carries service loads shown (excluding weight). It has been decided to add a 10-meter steel column at point B to eliminate excessive deflection. If the column being added is fixed at the bottom and pinned at the top, design the column that can withstand as support (Use a recommended value for K. Evaluate using LRFD and ASD). a. Use WT18 section b. Use W36 section c. Using the section at (b), determine the dimension of the base plate if the foundation dimension will be 1500mm x 1500mm. Pp = 88 kN PL = 163 kN Wp = 21 kN/m WL = 58 kN/m C B 6 m 12 m
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Calculate the forces in members BC, BE, and EF. Solve for each force from an equilibrium equation which contains that force as the only unknown. The forces are positive if in tension, negative if in compression. 4.0 m po Answers: BC= A 4.0 m BE = EF= i i i B 4.0 m 4.0 m 2.5 m E 12.5 KN 4.0 m kN 2 2 2 kN kN D
4-12 Compute M, for the beam shown in Fig. P4-12. Use f. = 4500 psi and fy = 60,000 psi. Does the compression steel yield in this beam at nominal Video Solution Strength? 10 in. 5 in. 5 in. 5 in. T 2.5 in. 2 No. 7 bars 20 in. 2.5 in. 6 No. 8 bars
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Given the truss shown, solve the following. Use A36 steel and assume there are two lines of three 19 mm bolts in each flange. Hole diameter is assumed to be three mm more than the bolt diameter. Neglect the weight of the section. a.What is the capacity of member CG? b. Design member CG using a W section. c. What is the tensile capacity of the section for CG based on gross area? The following W Section were considered for the design: W10x33 W14x38 W8x40 W16x36 W10x39 W8x13 189 KN 1.50 HL B 145 KN 1.50 E + 137 KN 121 KN -3.00 + 12.00 3.00 154 KN + W4x13 W6x9 W10x12 200 KN -1.50 1.50 1 030 T 1.20 +2.40 0.60 L i 0.30
Determine the design strength and the allowable strength for the compression member shown. Use NSCP 2015 and a steel with Fy =50ksi. L3X2.5X1/4 kL = 8ft PL 3/8 X 8 L3X2.5X1/4

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Q.3 The steel column shown Fig. 3 is pinned at both ends and unbraced about both x-x and braced at mid-height y-y axes. Calculate the axial capacity of a W14 x 99 section A992 (Fy = 50 ksi) to assess its the adequacy of a for the given conditions. PD = 250 k PL = 500 k L=32' Ž FIG. 3 T L=16'