Theory of Structures II Lecture Notes PDF

Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 1 Deflection of Structures Deflections of structures can occur from various...

Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 1 Deflection of Structures Deflections of structures can occur from various sources, such as loads, temperature, fabrication errors, or settlement. For good design, however, deflections must be limited in order to provide integrity and stability of the structure and prevent cracking of attached brittle materials such as concrete, plaster, or glass. Furthermore, a structure must not vibrate or deflect severely. Besides being able to calculate deflections for these purposes, we must also be able to calculate deflections at specified points in a structure to analyze statically indeterminate structures. In these lectures, the deflections to be considered apply only to structures having a linear elastic material response, and as a result, a structure subjected to a load will return to its original undeformed position after the load is removed. 1  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 1 The methods used to determine the deflection can be broadly classified as: 1. Geometrical Methods 2. Energy Method In the geometrical methods, the basic equations of equilibrium, compatibility, boundary conditions, and the material stress-strain relations are used to generate the governing differential equations. These equations are then solved by analytic, graphic or combined method. Some of the geometrical methods are: 1. The double integration method, 2. Moment area method, and 3. The conjugate beam method. In the energy method, the energy principles are used to formulate and solve the problem of deflection behavior. Deflection Diagrams (Elastic Curve): Before the slope or displacement of any point on a beam or frame is determined, it is often helpful to sketch the deflected shape of the structure when it is loaded. This deflection diagram represents the elastic curve or centerline deflection of the members. For most problems the elastic curve can be sketched without much difficulty. When doing so, however, it is necessary to account for the restrictions at the supports and connections. In a general sense, supports that resist a force, such as a pin, restrict displacement; and those that resist moment, such as a fixed wall, restrict rotation. 2 Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 1 Table (1) In Table 1, a roller, pin, and fixed support prevent displacement, and furthermore a fixed support prevents rotation. Also, the deflection of frame members that are fixed connected (4) causes the joint to rotate the connected members by the same amount 𝜽. On the other hand, if a pin connection is used at the joint, the members will each have a different slope or rotation at the pin, since the pin cannot support a moment (5). 3  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 1 Deflection Diagram of Beams. To illustrate the elastic curve of the beams, the moment diagram for this beam or frame be drawn first. Depending on the sign convention for moments it will be easy to construct the elastic curve and vice versa. Example 1: 4  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 1 Example 2: Other Examples: 5  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 1 Benefits of Elastic Curve: 6  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 1 Energy Methods The most energy methods are based on the conservation of energy, which states that the work 𝑼𝒆 done by all the external forces acting on a structure is transformed into internal work or strain energy 𝑼𝒊 , which occurs when the structure deforms. If the material’s elastic limit is not exceeded, this internal elastic strain energy will return the structure to its undeformed state when the loads are removed. The conservation of energy principle can be stated mathematically as 𝑼𝒆 = 𝑼𝒊 External Work-Force: When a force F undergoes a displacement dx in the same direction as the force, the work done is 𝒅𝑼𝒆 = 𝑭. 𝒅𝒙. If the total displacement is x, the work becomes 𝑥 𝑼𝒆 = ∫ 𝐹 𝑑𝑥 0 Consider now the effect caused by an axial force applied to the end of a bar as shown in Fig. below. 𝟏 𝑼𝒆 = 𝑷∆ 𝟐 7  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 1 External Work-Moment: The work of a moment is defined by the product of the magnitude of the moment 𝑴 and the angle 𝜽 through which it rotates, that is, 𝒅𝑼𝒆 = 𝑴 𝒅𝜽, Fig. If the total angle of rotation is u radians, the work becomes 𝜃 𝑼 𝒆 = ∫ 𝑀 𝑑𝜃 0 As in the case of a force, if the moment is applied gradually to a structure having linear elastic response from zero to M, the work is then 𝟏 𝑼𝒆 = 𝑴𝜽 𝟐 Examples: u P 1 𝑼𝒆 = 𝑃. 𝑢 2 Q - +𝜽 + 8  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 1 1 𝑼𝒆 = 𝑄. 𝑦 2 𝑴𝒐 𝜽 𝜽 1 𝑼𝒆 = 𝑴𝒐. 𝜽 2 Forms of Internal work (strain energy) When structure is subjected to external loading and hence external work is done, internal work or strain energy is generated and stored in the in divided member of structure. The internal work U strain energy due to various force effects are: 𝒍 𝑭𝒙 𝟐. 𝒅𝒙 𝑼𝒊 = ∫ 𝟐𝑬𝑨 𝟎 For prismatic member of length L and subjected to constant axial force 𝐹𝑥 : 𝑭𝒙 𝟐. 𝑳 𝑼𝒊 = 𝟐𝑬𝑨 Bending moment: 𝒍 𝑴𝒙 𝟐. 𝒅𝒙 𝑼=∫ 𝟐𝑬. 𝑰 𝟎 9  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 1 Torsion moment: 𝒍 𝑻𝒙 𝟐. 𝒅𝒙 𝑼𝒊 = ∫ 𝟐𝑮𝑱 𝟎 Shearing force: 𝒍 𝑽𝒙 𝟐. 𝒅𝒙 𝑼𝒊 = ∫ 𝟐𝑮𝑨𝒔 𝟎 Where: A : Cross-Section Area. L : Length of the Member. I : Moment of Inertia. E : Modulus of elasticity. G : Modulus of rigidity. J : Polar moment of inertia. In case of a rigid plane frame loaded in its own plane, while the twisting moments are absence, axial forces, shear in forces and bending moment act in plane of the frame. Consequently, the strain energy is the same of strain energy of these loads. 𝑭𝒙 𝟐. 𝒅𝒙 𝑽𝒙 𝟐. 𝒅𝒙 𝑴𝒙 𝟐. 𝒅𝒙 𝑼𝒊 = ∫ +∫ +∫ 𝟐𝑬. 𝑨 𝟐𝑮. 𝑨𝒔 𝟐𝑬. 𝑰 The strain energy due to axial forces and shearing forces is generally small compared to strain energy due to bending moment. Hence, it is common practice in the analysis of structure to ignore the strain energy due to the axial force and shearing forces. 10  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 1 In trusses only, axial forces are present. Hence, the total strain energy of truss may be expressed as: 𝑭𝒙 𝟐. 𝑳 𝑼𝒊 = ∑ 𝟐𝑬. 𝑨 The summation is conducted for all members of the truss. 11  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 1 Virtual Work Method (Unit Load Method) The principle of virtual work was developed by John Bernoulli in 1717 and is sometimes referred to as the unit-load method. Virtual work is the work done by a real force acting through a virtual displacement or a virtual force acting through a real displacement. The term “virtual” is used to describe the load, since it is imaginary and does not actually exist as part of the real loading. The virtual work equation for the force is The virtual work equation for the moment is 12  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 1 The internal virtual work or strain energy (𝑈𝑖 ) due to various force effects are: Bending: 𝑳 𝒎𝒙. 𝑴𝒙 𝟏. 𝜃 = ∫ 𝒅𝒙 𝑬𝑰 𝟎 Axial Force: 𝑳 𝒇𝒙. 𝑭 𝒙 1. ∆= ∫ 𝒅𝒙 𝑨𝑬 𝟎 The strain energy due to axial force in a TRUSS becomes as below: 𝒏 𝒇𝒙. 𝑭 𝒙. 𝑳 1. ∆= ∑ 𝑨𝑬 𝒊 where; 𝑚𝑥 is virtual bending moment; 𝑀𝑥 is real bending moment; 𝑓𝑥 is virtual axial force; 𝐹𝑥 is real axial force. 13  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 1 Example: Determine the rotation at point A for the beam shown in Fig. below. Assume EI=constant. Mo A B L Solution: For Rotation at A, To find virtual bending moment 𝒎𝒙 , we must apply a unit virtual moment at same point, 𝜹𝑸𝑨 = 𝟏 −1 ∑𝑀𝐵 = 0 → 𝑅𝑎. 𝐿 + 𝛿𝑄𝐴 = 0 → 𝑅𝑎 = 𝐿 −𝑥 𝑥 𝑚𝑥 = 𝑅𝑎. 𝑥 + 𝛿𝑄𝐴 = +1=1− 𝐿 𝐿 Then, to find the real bending moment 𝑴𝒙 we will apply the real load as below 𝑀𝑜 x Mo ∑𝑀𝐵 = 0 → 𝑅𝑎. 𝐿 − 𝑀𝑜 = 0 → 𝑅𝑎 = 𝐿 𝜃𝐴 𝑀𝑜 𝑀𝑥 = 𝑅𝑎. 𝑥 = 𝑥 Ra 𝐿 Rb Using principle of virtual work: 𝐿 𝑀𝑥. 𝑚𝑥 1. Ө𝐴 = ∫ 𝑑𝑥 𝐸𝐼 0 𝐿 1 𝑀𝑜 𝑥 1 ∗ Ө𝐴 = ∫ ( 𝑥) (1 − ) 𝑑𝑥 𝐸𝐼 𝐿 𝐿 0 𝑀𝑜. 𝐿 Ө𝐴 = 6𝐸𝐼 14