Mechanics of Materials
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1 Mechanics of Materials 2. Introduction Dr. Rami Zakaria References: 1. Engineering Mechanics: Statics, R.C. Hibbeler, 12 th ed, Pearson 2. Mechanics of Materials: R.C. Hibbeler, 9 th ed, Pearson 3. Mechanics of Materials: J.M. Gere & B.J. Goodno, 8 th ed, Cengage Learning
2 Terminology Term Definition Notes Stress Deformation Load/Loading Resultant Force Strain Member Compression Tension Translation Rotation the intensity of the internal force acting on an area Changing of the shape of a subject Forces/force curve Overall force The intensity of deformation at a point in an object A structure unit (such as a bar, a beam, etc.) Applying two forces towards each others Applying two forces away from each others Moving along a straight line Moving in a circular way 2
3 What is Mechanics of Materials? Mechanics of materials is a branch of applied mechanics that studies the behavior of solid bodies (such as bars and beams) subjected to external loading. This involves finding the internal effects (stress and strain ) in a solid body, which is important for the safe design of structures (in bridges, buildings, airplanes, etc ). Stress is associated with the strength of the material from which the body is made, while strain is a measure of the deformation of the body. 3
4 This steel framework is subjected to stress. How does this affect the design of the framework and its connections? We will review some of the important principles of statics and how to determine the internal resultant loadings in a body. Afterwards we will learn the concepts of normal stress (σ) and shear stress (τ) and study members subjected to an axial load or direct shear. 4
5 A SHORT REVIEW OF STATICS 5
6 External Loads Surface forces: caused by the direct contact of one body with another. They are distributed over the area of contact between the bodies. -If it is applied on a small area (in comparison with the total surface area of the body), then the surface force can be idealized as a single concentrated force (at one point on the body). Body forces: They are developed when one body exerts a force on another, without direct physical contact between them. For examples: the effects caused by the earth s gravitation (weight!), or its electromagnetic field. Although body forces affect all the particles of the body, we usually represent them by a single concentrated force (for example centre of Gravity). -If it is applied along a narrow strip of area, the loading can be idealized as a distributed line of forces. Active Forces Reactive forces 6 Support reactions
7 Equilibrium of a Rigid Body Why do we need statics here?! Because in Mechanics of Materials we need to know all forces acting on the body. When in equilibrium, the net force and net moment acting on a body are zero. Equations of equilibrium : F 0 M 0 F x M 0, x 0, F y M y 0, 0, F z M 0 z 0 7 F x M 0, o 0 F y 0,
8 Steps for Solving Free Body Equilibrium problems: 1. If not given, establish a suitable coordinate system (x,y or x,y,z). 2. Draw a free body diagram (FBD) of the object. 3. Apply the three equations of equilibrium (E-of-E) to solve for the unknowns. The FBD will show us all the external loads (known & unknown forces & moments) and the support reactions. 8 How can I find the support reactions? As a general rule, if a support prevents translation of a body in a given direction, then a force is developed on the body in the opposite direction. Similarly, if a support prevents rotation, then a couple moment is exerted on the body in the opposite direction.
9 Prevents translation in all directions, and prevents rotation Support Reaction Prevents translation in one direction, and prevents rotation Prevents translation in one direction Prevents translation in all directions
10 Distributed Loads q is a line load (N/m or lb/ft) The resultant force F R of q, acts through the centroid C (or centre of area) and it is equivalent to the area under the distributed loading curve. Separate Parts FBD Overall FBD Ref [3]
11 Notice the internal forces and moments: 11 Ref [3]
12 In case of 3D systems p is a surface pressure (N/m 2 or lb/ft 2 ) w is a body force (N/m 3 or lb/m 3 ) Remember: When we have a simple structure, members can be subjected to either compression or tension forces. We can calculate these forces using the concepts of statics. 12 When we know the forces at the ends of each of the members, we can determine the internal forces in these members, by looking at a cross-section at an arbitrary point.
13 Distributed Loading Review We will analyze the most common case of a distributed pressure loading. This is a uniform load along one axis of a flat rectangular body. In such cases, w(x) is the load intensity distribution function and it has units of force per length. df w( x) dx The net force on the beam is given by + F df w( x) dx R L L Here A is the area under the loading curve w(x). A 13 Ref [1]
14 Location of the Resultant Force The force df will produce a moment of (x)(df) about point O. The total moment about point O is given as M xw( x) dx Ro xdf L L Assuming that F R acts at x, it will produce the moment about point O as M ( x) F ( x) w( x) dx Ro R L 14 Ref [1]
15 Comparing the last two equations, we get F R acts through a point C, which is called the geometric center or centroid of the area under the loading curve w(x). 15 Ref [1]
16 Example: Determine the magnitude and location of the equivalent resultent force (F R ) acting on the shaft Ref [1]
17 Note: for simples shapes we can find the shift easily using the geometric centroid Area C x C y b.h b/2 h/2 b.h/2 (a +b)/3 h/3 π.a 2 a a 17 π.a 2 /2 a 4.a /3.π
18 Examples: The rectangular load: F R x ft 4000lb The triangular loading: F R x m 1800N 18
19 Example about support reaction calculations Given: The 4kN load at B of the beam is supported by pins at A and C. Find: The support reactions at A and C. FBD of the beam: Solution: M A = (F C sin 45 ) (1.5) (4) (3) = 0 F c = kn A X A Y A 1.5 m 45 F C C 1.5 m B 4 kn + F X = A X cos 45 = 0; + F Y = A Y sin 45 4 = 0; A X = 8.00 kn A Y = 4.00 kn Ref [1]
20 Simple Truss 20 Ref [1]
21 The solution has finished until here! But let s see the reaction components at joint (A) 21 All reaction forces together Ref [1]
22 Internal Resultant Loadings Statics is used to determine the resultant loadings that act within a body. Consider a body in equilibrium by four external forces (neglect the weight for now). In order to obtain the internal loadings acting on a specific region, let s pass an imaginary section (cut) through the region of interest. Separate the two parts and draw the freebody diagram of one of them. There is a distribution of internal forces acting on the area of the cross section. We can represent these forces by a resultant force F R and moment M Ro, at any point O (usually the centroid). Here, we are interested in the normal and perpendicular components of F R and M Ro. 22
23 Therefore, we can define 4 types of loading: Normal force, N: This force acts perpendicular to the area. It is developed when the external loads tend to push or pull on the two segments of the body. Shear force, V: The shear force lies in the plane of the area. It is developed when the external loads tend to cause the two segments to slide over one another. Bending moment, M: It is developed when the external loads tend to bend the body about an axis lying on the area. Torsional moment (torque), T: It is developed when the external loads tend to twist one segment of the body about an axis perpendicular to the area. This exists only in a 3D system 23 Ref [2]
24 Important Points: Mechanics of materials is a study of the relationship between the external loads applied to a body and the stress and strain caused by the internal loads within the body. A support produces a force in a particular direction on its attached member if it prevents translation of the member in that direction, and it produces a couple moment on the member if it prevents rotation. In general the internal resultant loadings acting on a body are :a normal force, shear force, torsional moment, and bending moment. We can determine them by making a cross-section through the body. 24
25 Procedure for Analysis: (Calculating The resultant internal loadings at a point located on the section of a body ) 1. Determine the Support Reactions. 2. Draw the Free-Body Diagram. only in a 3D system ( indicate the unknown resultants N, V, M, and T ) 3. Write and solve the Equations of Equilibrium. Note: If the solution is negative, the assumed directional sense of the resultant is opposite to that shown on the free-body diagram 25
26 Examples: 26 Ref [2]
27 27 Ref [2]
28 EXAMPLE Ref [2]
29 29 Ref [2]
30 30 Ref [2]
31 0 F BD 5 EXAMPLE 1.3 (CONTINUED) F F x y F BD 4 BA 0 F F BA 7750lb 4650lb 31 Ref [2]
32 32 Ref [2]
33 33 Ref [2]
34 Problems: Student s solution: 34
35 35
36 36
37 Student s solution: 37
38 38
39 Student s solution: 39
40 40
41
42 42
43 Student s solution:
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