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The simply supported queen-post trussed beam shown in the photo is to be designed to support a uniform load of 2 kN/m. The dimensions of the structure are shown in Fig. 10-17a. Determine the force developed in member CE. Neglect the thickness of the beam and assume the truss members are pin connected to the beam. Also, neglect the effect of axial compression and shear in the beam. The cross-sectional area of each strut is 400 mm2, and for the beam 1 = 20(10) mm*. Take E = 200 GPa.

The simply supported queen-post trussed beam shown in the photo is to be designed to support a uniform load of 2 kN/m. The dimensions of the structure are shown in Fig. 10-17a. Determine the force developed in member CE. Neglect the thickness of the beam and assume the truss members are pin connected to the beam. Also, neglect the effect of axial compression and shear in the beam. The cross-sectional area of each strut is 400 mm2, and for the beam 1 = 20(10) mm*. Take E = 200 GPa.

Structural Analysis
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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10.7 COMPOSITE STRUCTURES 10.7 Composite Structures Composite structures are composed of some members subjected only to axial force, while other members are subjected to bending. If the structure is statically indeterminate, the force method can conveniently be used for its analysis. The following example illustrates the procedure. EXAMPLE 10.9 The simply supported queen-post trussed beam shown in the photo is to be designed to support a uniform load of 2 kN/m. The dimensions of the structure are shown in Fig. 10-17a. Determine the force developed in member CE. Neglect the thickness of the beam and assume the truss members are pin connected to the beam. Also, neglect the effect of axial compression and shear in the beam. The cross-sectional area of each strut is 400 mm2, and for the beam 1= 20(106) mm". Take E = 200 GPa. 2 kN/m 2 kN/m B 1 m 2 m ACE Primary structure 2 m 2 m Actual structure (a) + Fig. 10-17 SOLUTION Principle of Superposition. If the force in one of the truss members is known, then the force in all the other members, as well as in the beam can be determined by statics. Hence, the structure is indeterminate to the first degree. For solution the force in member CE is chosen as the redundant. This member is therefore sectioned to eliminate its capaci to sustain a force. The principle of superposition applied to the structure is shown in Fig. 10-176. A FCE FCESCECE Redundant FC applied (b) Compatibility Equation. With reference to the relative displacemen of the cut ends of member CE, Fig. 10-17b, we require 0 = ACE + FCEFCE CE (1)
Transcribed Image Text:10.7 COMPOSITE STRUCTURES 10.7 Composite Structures Composite structures are composed of some members subjected only to axial force, while other members are subjected to bending. If the structure is statically indeterminate, the force method can conveniently be used for its analysis. The following example illustrates the procedure. EXAMPLE 10.9 The simply supported queen-post trussed beam shown in the photo is to be designed to support a uniform load of 2 kN/m. The dimensions of the structure are shown in Fig. 10-17a. Determine the force developed in member CE. Neglect the thickness of the beam and assume the truss members are pin connected to the beam. Also, neglect the effect of axial compression and shear in the beam. The cross-sectional area of each strut is 400 mm2, and for the beam 1= 20(106) mm". Take E = 200 GPa. 2 kN/m 2 kN/m B 1 m 2 m ACE Primary structure 2 m 2 m Actual structure (a) + Fig. 10-17 SOLUTION Principle of Superposition. If the force in one of the truss members is known, then the force in all the other members, as well as in the beam can be determined by statics. Hence, the structure is indeterminate to the first degree. For solution the force in member CE is chosen as the redundant. This member is therefore sectioned to eliminate its capaci to sustain a force. The principle of superposition applied to the structure is shown in Fig. 10-176. A FCE FCESCECE Redundant FC applied (b) Compatibility Equation. With reference to the relative displacemen of the cut ends of member CE, Fig. 10-17b, we require 0 = ACE + FCEFCE CE (1)
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