A plane truss is shown in figure. All the members of the planar truss have the same properties in terms of area of cross-section (A) and modulus of elasticity (E). The maximum axial force that any member can support is 8 kN in tension and 6 kN in compression. The maximum force P that can be supported at joint D would be 60⁰ m -4m- 60% 4 m- 60% _kN.

A plane truss is shown in figure. All the members of the planar truss have the same properties in terms of area of cross-section (A) and modulus of elasticity (E). The maximum axial force that any member can support is 8 kN in tension and 6 kN in compression. The maximum force P that can be supported at joint D would be 60⁰ m -4m- 60% 4 m- 60% _kN.

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

For the truss shown in the figure, the force in the member QR is po zero 2. 2 3. P 4. VP S P L
Calculate the forces in all members of the truss given in figure below: 34 KN 25 KN 60° 60° 60° 60° A 4.5 m 4.5 m 45 KN
3. For the truss shown in the figure for this question, members AB and CD shown in bold each have a weight of 2 N. 4 m IC Establish the force in members BC, E I 3. CG, DE, and GH. Ans. FBC = 8 N (tension) FCG = 1.67 N (compression) FDE = 0 FGH = 6.67 N (compression) %3D 5 N 2N 7 N %3D %3D
PLEASE HELP MEEE
Problem 1: Determine the forces in all members of the truss shown below. 2m B 45% 45 2m 45% с 4m 45% D r = 0.5m 30* 3 kN
Determine the force in all of the members of the truss shown in the figure. State whether the members are in tension or compression. 4 m H 2 m J K 2 m A E В C D 4 kN 2 kN 4 kN A, = 5 kN 2 m E. 5 kN 2 m 2 m 2 m
1) The truss (containing 12 pin joints, A, B, C, ... L) shown in the figure is simply supported at points H and L that are 4 meters apart. Angles A between different truss members are either 90° or 45° and the member lengths are either 1 m or m. H a) Compute the support reactions when P = 4 kN b) Draw a free body diagram from which internal forces in AB, EJ and IJ can be obtained. c) From the free body diagram of part b obtain forces in members AB, EJ and IJ (indicate if the internal forces are tension or compression) d) List all zero force members that carry no load. (a) (b) 3 (c) 3 (d) 2 RH = FAB = RL = FEJ = Fj =
The truss shown in the figure is applied with force P= 5kN %3D 1.50 m 1.50 m 150 m 2 m 2 m What is the force in member DE in kN? A.
Determine the force in each member of the truss as shown in the figure.
5. For the truss shown in the figure, calculate the stresses in members CE, DE, and DF. The cross sectional of each member is 1160 mm2.
1. A pin-connected truss composed of members AB and BC is subjected to a vertical force P = 40 kN at joint B (see figure). Each member is of constant cross- sectional area: AAB = 0.004 m² and ABC =0.002 m². The diameter d of all pins is 20 mm, clevis thickness t is 10 mm, and the thickness 1 of the bracket is 15 mm. Determine the normal stress acting in each member and the shearing and bearing stresses at joint C. Fc Clevised ends 2 m A 1.5 m P=40 kN
For the truss shown: a) Calculate support reactions. b) Draw the free body diagram for joint A. A 500 lb 500 lb F В 45 ° 45 5 ft E 5 ft D 10 ft 10 ft