CHAPTER 1: FLUID PROPERTIES LEARNING OUTCOMES At the end of this topic‚ you should be able to: Define Fluid State differences between solid and fluid Calculate common fluid properties: i. Mass density ii. Specific weight iii. Relative density iv. Dynamic viscosity v. Kinematic viscosity INTRODUCTION Fluid Mechanics Gas Liquids Statics i F 0 F 0 i Laminar/ Turbulent Dynamics ‚ Flows Compressible/ Incompressible Air‚ He‚ Ar‚ N2‚ etc. Water‚ Oils‚ Alcohols‚
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same time. The wing design will be optimized. Before processing the wing design‚ it is necessary to understand the structure of the wing. In the figure-x ‚ it shown the elements and layout of wing structure of the Pilatus PC9/A which include spars‚ shear webs‚ ribs‚ skin and skin stingers. This section will be process the design of those elements base on the loads calculated in section 4. Note that the calculation is for one wing only as the load same for both wings as assumed. Figure xx: The wing
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experimental shear modulus (modulus of rigidity) with the tabulated values. 2. The maximum shear stress at the limit of proportionality or the proportional limit shear stress‚ 3. The general characteristics of the torque‚ angle of twist relationship. The proportional limit shear stress: the highest shear stress that the material can withstand and still return to its original geometry is at the limit of the proportional portion of the graph and is known as the proportional limit shear stress.
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MODULUS OF RIGIDITY Modulus of Rigidity Using the initial length‚ radius‚ measurements of torque and angle of twist‚ it is possible to calculate the Modulus of Rigidity for each specimen. The modulus of rigidity or shear modulus represents the ratio of the shear stress to the shear strain. It is denoted ‘G’. To calculate two separate values of G for each specimen‚ it is possible to use two methods. It is important to determine the polar moment of inertia to be used throughout the experiment:
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Halpin-Tsai model for transverse modulus has been introduced. E2 = Em (1+Vf) (1 - Vf) Unidirectional Ply- Shear Modulus: Shear modulus is defined as the ration of shear stress to the shear strain. The shear modulus is
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fy = 460 N/mm2‚ fy‚ links = 250 N/mm2. The beam is exposed to a mild condition. a) Determine whether the beam is singly or doubly reinforced beam. b) Determine the main reinforcement for the beam using a design chart. c) Determine the shear reinforcement for the beam. d) Check whether the deflection of the beam is acceptable. e) Sketch the detailed cross section of the beam indicating the reinforcements in (b) and (c). a) Determine whether the beam is singly or doubly reinforced
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ball bracket and find the tensile and shear loads applied to the two bolts that attach bracket to the channels shown in the Figure (b). [pic][pic] a) b) c) [pic] d) Problem 2 The figure below shows a cam and a cam-follower arm. If the load P = 1000 N‚ what spring force is needed at the right end to maintain a minimum load between cam and follower of 125 N? Find the maximum shear force and bending moment in the follower arm. Plot the shear and moment diagrams. Problem 3
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UNIVERSITI TUN HUSSEIN ONN MALAYSIA Faculty of Mechanical and Manufacturing Engineering __________________________________________________________________ COURSE INFORMATION COURSE TITLE: ENGINEERING LABORATORY III (BDA 27101) TOPIC 1: TENSILE TEST 1. INTRODUCTION The tensile experiment is the most common mechanical test that reveals several important mechanical properties‚ such as: modulus of elasticity‚ yield strength‚ ultimate tensile strength‚ ductility‚ and toughness. The material
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modulus of a material is 200kN/mm2 and its rigidity modulus is 80kN/mm2 . Determine its bulk modulus. 3. What is point of contraflexure? In which beam will it occur? 4. Derive the relation between the intensity of load and shear force‚ in bending theory. 5. Draw the shear stress distribution diagram for an Isection. 6. Derive an expression for the power transmitted by a circular shaft in S.I. units‚ which is subjected to a torque T in N m. 7. Distinguish between close coil and open coil helical
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Shear Stress Forces parallel to the area resisting the force cause shearing stress. It differs to tensile and compressive stresses‚ which are caused by forces perpendicular to the area on which they act. Shearing stress is also known as tangential stress. where V is the resultant shearing force which passes through the centroid of the area A being sheared. Problem 115 What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick? The shear strength is 350 MN/m2
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