2) Consider the truss shown below. The truss is subjected to a vertical load (P) at joint E. Determine a)the vertical displacement of point E (in the direction of force P), and b) the horizontal displacementof point E (perpendicular to the direction of force P).-|--PE

As per unit load method, the procedure is,i) introduce a virtual load (of 1 force unit) along the direction in which deflection is needed. Then we have three loading systems. Real system (given system of loading), virtual system for horizontal deflection (with unit load at point P along horizontal direction), and virtual system for vertical deflection (with unit load at point P along vertical direction)ii) evaluate internal forces in all members in all three systems (by applying method of joints for all the three loading systems with individual FBDs one for each system).iii) apply virtual work equations for real system and horizontal virtual system pair to evaluate horizontal deflection and for real system and vertical virtual system pair to evaluate vertical deflection.But for ease of calculation of internal forces, one combined FBD diagram is drawn for all three loadings is adopted here as shown. Then internal forces are evaluated for each system at later steps in the solutions as follows.Draw combined FBD diagram for all three loading systems.Here are x and y components of reaction force at pin support A, is reaction force at roller support B, P indicates real load, indicates applied virtual loads along horizontal and vertical directions.From isosceles right angle triangle DCE, length of side DE is . Similarly, the lengths of members .Assume all bars have same cross-sectional areas.At static equilibrium, i) equate algebraic sum of moment of forces about point A to zero.ii) equate algebraic sum of horizontal forces (along x-direction) to zero.iii) equate algebraic sum of vertical forces (along y-direction) to zero.Equations (i), (ii), and (iii) indicates reaction forces (in terms of real and virtual loads).Draw FBDs of joints.Joint - A:Equate algebraic sum of vertical forces to zero.Equate algebraic sum of horizontal forces to zero.Joint - B:Equate algebraic sum of vertical forces to zero.Equate algebraic sum of horizontal forces to zero.Joint - C:Equate algebraic sum of vertical forces to zero.Equate algebraic sum of horizontal forces to zero.Joint - E:Equate algebraic sum of vertical forces to zero.Note : Equations (1) to (7) are expressions for internal forces but not internal forces.In order to evaluate the internal forces for each individual loading system, we need to apply compatibility conditions to equations (1) to (7).Those compatibility conditions are i) in real loading system, .ii) in horizontal virtual loading system, .iii) in vertical virtual loading system, .Apply these compatibility conditions in equations (1) to (7) and determine the internal forces in each loading case.Tabulate the results as shown.MemberLengthReal SystemHorizontal virtual systemVertical virtual systemInternal force Internal force Internal force ACLP1PL1PLAD1.414L1.414P001.4142.827PLBC1.414L00000BDL-2P00-24PLCDL00000CELP1PL1PLDE1.414L-1.414P00-1.4142.827PLWrite expression for horizontal deflection.Here E indicates modulus of elasticity and A indicates cross-sectional area of members.Substitute all known values.Write expression for vertical deflection.Substitute all known values.Thus, the horizontal deflection of point E is and vertical deflection of point E is .

Engineering

Mechanical Engineering

2) Consider the truss shown below. The truss is subjected to a vertical load (P) at joint E. Determine a) of point E (in the direction of force P), and b) the horizontal displacement to the direction of force P). the vertical displacement of point E (perpendicular E

2) Consider the truss shown below. The truss is subjected to a vertical load (P) at joint E. Determine a) of point E (in the direction of force P), and b) the horizontal displacement to the direction of force P). the vertical displacement of point E (perpendicular E

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial load acts perpendicular to the geometric cross-section of the component. Based on the direction of the application of the load, an axial load can be tensile or compressive. The analysis of bodies under the action of axial loads is generally studied under the branch of mechanics, known as statics.

Question

2) Consider the truss shown below. The truss is subjected to a vertical load (P) at joint E. Determine a)
the vertical displacement of point E (in the direction of force P), and b) the horizontal displacement
of point E (perpendicular to the direction of force P).
-|--
P
E

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Step 1: Apply unit load method to evaluate deflections along horizontal and vertical directions.

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Step 2: Evaluate the reaction forces at A and B.

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Step 3: Evaluate expressions for internal forces using method of joints.

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Step 4: Evaluate internal forces and tabulate the results.

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Step 5: Evaluate the deflections along horizontal and vertical directions.

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