1. A steel column has an unsupported height of 5.4 m is pin connected at both ends with sidesway is prevented. It carries an total axial load of 530 kN, Mx = 20 kNm, My = 3 kNm at the top and Mx = 13.5 kN, My = 1 kNm at the bottom which causes bending to both major and weak axis (single curvature for both axis). Except for bracings at the top and bottom there is no lateral bracing about either major and minor axis of the column. Fy= 250 MPa. USE ASD and check if the section is adequate

1. A steel column has an unsupported height of 5.4 m is pin connected at both ends with sidesway is prevented. It carries an total axial load of 530 kN, Mx = 20 kNm, My = 3 kNm at the top and Mx = 13.5 kN, My = 1 kNm at the bottom which causes bending to both major and weak axis (single curvature for both axis). Except for bracings at the top and bottom there is no lateral bracing about either major and minor axis of the column. Fy= 250 MPa. USE ASD and check if the section is adequate

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

A Mansard roof truss is loaded as shown. Determine the force in members GI, HI, and HJ. 1.2 kN 1.2 kN 1.2 kN 1.2 kN 1.2 kN B H OL El G K 4 m4m4m- 3 m A |C 2.25 m 4 m- 2.25 m
A Mansard roof truss is loaded as shown. Determine the force in members DF, DG, and EG. 3 m ✓ A 1.2 kN 1.2 kN 1.2 kN 1.2 kN 1.2 kN B D C E| k 2.25 m 4 m. F G 4 m- H -4 m I 4 K OL 2.25 m Show complete solutions and diagrams. Original work please I will upvote
If the maximum allowable force (tensile or compressive) in members BC in the truss shown known to be 72 kN. Determine the maximum load P that can applied to the truss. Given a = 2 m a a IC 2a H. O A.P = 90.00 KN O B. P = 54.0O KN O C.P= 36.00 KN O D.P= 72.00 KN
4.5 ft 5.0 ft 0.02 ft D [A B. 6 in 18 in The structural bracing system above is comprised of two solid round circular members AB and BC. There is a 0.02 ft gap between point C and the adjacent rigid wall, Member AB is made of steel with the following properties: EAB = 29000 ksi CAB = 6.6-10 6 joF VAB = 0.32 Member BC is made of reinforced nylon with the following properties: EBC = 6400 ksi CBC = 11.5-10 6 oF VBC = 0.45 Which of the following is closest to the change in temperature required to close the 0.02 ft gap from C to D? O +19.1 °F +2753 °F +229.4 °F +66.4 °F
Light-grade steel channel was used as a purlin of a truss. The top chord of the truss is inclined I V : 4 H and distance between trusses is equal to 6 m. The purlin has a weight of 79 N/m and spaced at 1.2 m. on centers. The dead load including the roof materials is 720 Pa, live load of 1000 Pa and wind load of 1.2 1.2 1440 Pa. Coefficient of Purlins pressure at leeward and windward are 0.6 and 0.2 respectively. Assume all loads passes through the centroid of the section. Properties of C 200 x 76 mm Sx = 6.19 x 104 mm Sy = 1.38 x 104 mm3 W = 79 N/m 1.2 Truss 1.2 12 12 1.2 Allowable bending stress Fbr = Fby = 207 MPa Truss %3D 6m O Calculate the bending stress, fox, for dead load and live load combination (D + L). Calculate the bending stress, foy, for dead load and live load combination (D + L). O Calculate the maximum ratio of actual to the allowable bending stress for load combination 0.75 (D + L + W) at the windward side. fbx = 151.14 MPa fby = 169.6 MPa Interaction = 1.25
Light-grade steel channel was used as a purlin of a truss. The top chord of the truss is inclined I V: 4 H and distance between trusses is equal to 6 m. The purlin has a weight of 79 N/m and spaced at 1.2 m. on centers. The dead load including the roof materials is 720 Pa, live load of 1000 Pa and wind load of 1.2 1.2 1440 Pa. Coefficient of Purlins pressure at leeward and windward are 0.6 and 0.2 respectively. Assume all loads passes through the centroid of the section. Truss Properties of C 200 x 76 mm Sx = 6.19 x 104 mm Sy = 1.38 x 104 mm W = 79 N/m 12 12 1.2 I Allowable bending stress Fbx= Fby = 207 MPa Truss %3D 6m O Calculate the bending stress, fox, for dead load and live load combination (D+ L). Calculate the bending stress, foy, for dead load and live load combination (D + L). O Calculate the maximum ratio of actual to the allowable bending stress for load combination 0.75 (D + L + W) at the windward side. fbx = 151.14 MPa fby = 169.6 MPa Interaction = 1.25
2. Determine the force in each member of the truss. IHG F D 2a L6@a²6a
An I-section girder has a web 600 mm x 12,5 mm and flanges 300 mm x 25mm. The girder must withstand a bending moment of 300 kN.m and a shearforce of 1 MN at a particular section. Determine (a) the maximum bendingstress induced in the section; (b) the maximum shear stress induced in thesection due to the shear force only; (c) the percentage of the shear force carriedby the web; (d) the percentage of the bending moment carried by the flanges.
10 kN 10 kN 10 kN 10 kN 10 kN 10 kN 10 kN Q2 (25%): For the simple truss shown, B, X VE A. There are five "zero-force members". Identify them without any calculations. B. Determine the forces in the member X 4 m X, 0. N M L. K marked with "X". 6 at 3 m = 18 m
Determine the force in membres DC, HC, and HI of the truss, and state if the members are in tension or compression. Use the method of sections. F -2 m 40 kN E G -2 m- 50 kN 14 H A D -2 m- C 1.5 m B -30 kN 1.5 m 40 kN 1.5 m
2: The truss is supported by a pin in point A and a roller at point E. Show your FBD 2 m F 5 kN. B 3 m A E 10 kN 5 m 5 m
Channel sections are used as a purlin. The top chords of the truss are sloped at 5H to 2V. Trusses are spaced 5m on centers and the purlins are spaced 1.4m on centers. Use the properties of the channel section BELOW. Loads: (Consider all loads pass thru the centroid of the section.) Dead load = 750 Pa Live load = 1,000 Pa Wind load = 1,400Pa Wind coefficients: Windward = 0.3 Leeward = -0.5 Use Fbx=Fby=250 MPa. Determine the maximum value of the interaction equation using the load combination of 0.75(D+L+W)