Mechanics of Solids. Mechanics Of Solids. Suraj kr. Ray Department of Civil Engineering

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1 Mechanics Of Solids Suraj kr. Ray Department of Civil Engineering 1

Compressive force: When two equal magnitude force occurred on a body in such a way that their senses (Directions) are opposite and toward the CG of the body, is known as compressive force.

2 Mechanics of Solids is a branch of applied mechanics that deals with the behaviour of solid bodies subjected to various types of loading. Other names for this field of study are strength of materials and mechanics of deformable bodies. Introduction:- 2. Compressive force: When two equal magnitude force occurred on a body in such a way that their senses (Directions) are opposite and toward the CG of the body, is known as compressive force. Compression is a force which tends to squeeze or crush a body. A force exerted on a body can cause a change in either the shape or the motion of the body. The unit of force is the Newton, N. No solid body is perfectly rigid and when forces are applied to it, changes in dimensions occur. Such changes are not always perceptible to the human eye since they are so small. For example, the span of a bridge will sag under the weight of a vehicle and a spanner will bend slightly when tightening a nut. It is important for engineers and designers to appreciate the effects of forces on materials, together with their mechanical properties. The three main types of mechanical force that can act on a body are: (i) tensile, (ii) compressive, and (iii) Shear. 1. Tensile force: When two equal magnitude force occurred on a body in such a way that their senses (Directions) are opposite and outward from the CG of the body, is known as tensile force. Tension is a force which tends to stretch a body. Examples: A pillar or column supporting bridge or slab is in compression. The sole of a shoe is in compression. 3. Shear force: When two equal magnitude forces occurred on a body in such a way that their senses (Directions) are opposite and line of actions are also different, is known as shear force. Shear is a force that tends to slide one face of the material over an adjacent face. Example: A rivet holding two plates together is in shear stress if tensile force is applied between the plates. A horizontal beam is subjected to shear force. Examples: A rope or Cable of crane carrying load is in tension. When nut is tightened, bolts are in tension. 2

Problem: In Shown fig. state the types of S.I. unit of stress: So, Pascal is the S.I. unit of stress. Solution: A is supporting member, is in Compression. B is Horizontal beam, is in Shear.

A =cross-sectional area = 75 mm 2 =75 x 10-6 m 2 F = 15 KN = 15 x 10 3 N Stress in bar, Try Yourself: Problem: A circular wire has a tensile force of 60 N applied to it and this force produces a

3 Problem: In Shown fig. state the types of S.I. unit of stress: So, Pascal is the S.I. unit of stress. Solution: A is supporting member, is in Compression. B is Horizontal beam, is in Shear. C is holding load, is in Tension. Problem: A rectangular bar having a cross-sectional area of 75 mm 2 has a tensile force of 15 KN applied to it. Determine the stress in the bar. A =cross-sectional area = 75 mm 2 =75 x 10-6 m 2 F = 15 KN = 15 x 10 3 N Stress in bar, Try Yourself: Problem: A circular wire has a tensile force of 60 N applied to it and this force produces a stress of 3.06 MPa in the wire. Determine the diameter of the wire. F = 60 N σ = 3.06 MPa = 3.06 x 10 6 Pa d =? (say d is dia. of bar) Stress: Stress is nothing but only a ratio of applied force (say F) to the cross-sectional area (say A). Types of stress Tensile stress Compressive stress Shear stress Symbol used σ(greek word sigma) σ τ (Greek word tau) i.e. diameter of wire, d = 5.0 mm Sign convention for stress: Tensile stress or tension is taken as or with positive (+ve) sign. Compressive stress or compression is taken as or, with (-ve) sign. From definition: 3

Let s try it: 1. In given fig. if, Exerted force = 900 KN, dia. of rope =10 mm. find stress in rope. Strain: Strain is a ratio of change in length (say δ) to the original length (say L).

A hollow circular column supporting a building is attached to a metal plate and bolted into the concrete foundation as shown in fig.

4 Let s try it: 1. In given fig. if, Exerted force = 900 KN, dia. of rope =10 mm. find stress in rope. Strain: Strain is a ratio of change in length (say δ) to the original length (say L). Types of strain Tensile strain Compressive strain Shear strain Symbol used ε (Greek word epsilon) ε γ(greek word gamma) 2. A hollow circular column supporting a building is attached to a metal plate and bolted into the concrete foundation as shown in fig. The column outside diameter is 100 mm and an inside diameter is 75 mm. the metal plate dimensions are 200 mm x 200 mm x 10 mm. the load P is estimated at 800 KN. Determine the compressive stress in the column. % strain, S.I. unit of strain: Problem: A bar 1.60 m long contracts axially by 0.1 mm when a compressive load is applied to it. Determine the strain and the percentage strain. L = 1.60 m = δ = 0.1 mm 3. A square sectioned support of side 12 mm is loaded with a compressive force of 10 KN. Determine the compressive stress in the support. [69.44MPa] 4. A tube of outside diameter 60 mm and inside diameter 40 mm is subjected to a load of 60 KN. Determine the stress in the tube. [38.2 MPa] 5. A rectangular bar having a crosssectional area of 80 mm 2 has a tensile force of 20 KN applied to it. Determine the stress in the bar. [250 MPa] % strain, Problem: A wire of length 2.50 m has a percentage strain of 0.012% when loaded with a tensile force. Determine the extension of the wire. L = 2.50 m = 2.50 x 10 3 mm % = 0.012%, =? 4

Problem: If in shown fig. inner & outer dia. of a hollow circular tube are respectively d 1 =10 mm, d 2 = 10.5 mm, and the shortening due to the load is 0.012 mm.

5 Problem: If in shown fig. inner & outer dia. of a hollow circular tube are respectively d 1 =10 mm, d 2 = 10.5 mm, and the shortening due to the load is mm. determine the compressive stress and strain. Mechanical properties of materials: 1. Elasticity: It is the property of material by virtue of which body regain its original shape and size after removal of external forces. 2. Plasticity: It is the property of material by virtue of which a body doesn t return to the original shape and size after removal of forces too. L = 40 mm P = 100 N d 1 = 10 mm d 2 = 10.5 mm δ = mm Cross-sectional area of the tube: A = = 2 Hooke s law: Within the limit of proportionality, the extension of a material is proportional to the applied force. In other words, Within the limit of proportionality of a material, the strain produced is directly proportional to the stress producing it. Mathematically it can be written as, Compressive stress in the tube: Compressive strain in the tube: Let s try it: -4 2 Young s modulus of elasticity: Within the limit of proportionality, Hence, This constant of proportionality is called young s modulus of elasticity and is given the symbol E. So, 1. A pipe has an outside diameter of 25 mm, an inside diameter of 15 mm and length 0.40 m and it supported a compressive load of 40 KN. The pipe shortens by 0.5 mm when the load is applied. Determine (a) the compressive stress, (b) the compressive strain in the pipe when supporting the load. [127.3 MPa, ] 2. A circular hole of diameter 50 mm is to be punched out of a 2 mm thick metal plate. The shear stress needed to cause fracture is 500 MPa. Determine (a) the minimum force to be applied to the punch, and (b) the compressive stress in the punch at this value. [157.1 KN, MPa] Its unit is same as unit of stress, i.e. Pascal. Stiffness: it is the ratio of Force (F) to the change in length or extension (δ). Value of stiffness is said to be high if have a large value of Young s modulus. From definition, 5

6 Then, i.e. Problem A bar of thickness 15 mm and having a rectangular cross-section carries a load of 120 kn. Determine the minimum width of the bar to limit the maximum stress to 200 MPa. The bar, which is 1.0 m long, extends by 2.5 mm when carrying a load of 120 kn. Determine the modulus of elasticity of the material of the bar. Problem: A wire is stretched 2 mm by a force of 250 N. Determine the force that would stretch the wire 5 mm, assuming that the limit of proportionality is not exceeded. When δ = 2 mm, F = 250 N From Hook s law, 2 = k x 250 From which constant Again when, When δ = 5 mm, then 5=kF i.e. From which, Force F = 5 (125) = 625 N Hence, to stretch the wire 5 mm a force of 625 N is required. Problem: A copper rod of diameter 20 mm and length 2.0 m has a tensile force of 5 kn applied to it. Determine (a) the stress in the rod, (b) by how much the rod extends when the load is applied. Take the modulus of elasticity for copper as 96 GPa. d = 20 mm, L = 2.0 m, F = 5 KN, E = 96 GPa. t = 15 mm, F = 120 kn, b =?, σ = 200 MPa L = 1.0 m δ = 2.5 mm E =? Again since, Modulus of elasticity, thickness of bar. load. width of the bar. stress length of the bar. extension. modulus of elasticity. (a) Stress, (b) Extension, i.e. extension of rod is mm. 1. A brass tube has an internal diameter of 120 mm and an outside diameter of 150 mm and is used to support a load of 5 kn. The tube is 500 mm long before the load is applied. Determine by how much the tube contracts when loaded, taking the modulus of elasticity for brass as 90 GPa. [4.37µm] 2. In an experiment to determine the modulus of elasticity of a sample of mild steel, a wire is loaded and the corresponding extension noted. The Load (N) Extension (mm) results of the experiment are as shown. Draw the load/extension graph. The mean diameter of the wire is 1.3 mm and its length is 8.0 m. Determine the modulus of elasticity E of the sample, and the stress at the limit of proportionality. 6

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CHAPTER 3 THE EFFECTS OF FORCES ON MATERIALS

CHAPTER 3 THE EFFECTS OF FORCES ON MATERIALS CHAPTER THE EFFECTS OF FORCES ON MATERIALS EXERCISE 1, Page 50 1. A rectangular bar having a cross-sectional area of 80 mm has a tensile force of 0 kn applied to it. Determine the stress in the bar. Stress