Design against static loading, Lecture notes of Machine Design

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Prepared by: Dr. Gagandeep Bhardwaj, Assistant Professor, MED Email: gagandeep.med@thapar.edu Contact No. 8954388548

DESIGN AGAINST STATIC LOADING

05/02/2019 Dr. Gagandeep Bhardwaj, AP MED, TIET 2

MODES OF FAILURE

A static load is defined as a force, which is gradually applied to a mechanical component and which does not change its magnitude or direction with respect to time.

A mechanical component may fail, that is, may be unable to perform its function satisfactorily, as a result of any one of the following three modes of failure:

• Failure by elastic deflection (lateral or torsional rigidity)
• Failure by general yielding (Ductile materials)  General yielding : Considerable portion of the component is subjected to plastic deformation, called general yielding.  Localized yielding : The localized yielding in the region of stress concentration is restricted to a very small portion of the component and is not considered significant.
• Failure by fracture ( brittle material ): the ultimate tensile strength of the material is an important property to determine the dimensions of these components.

FACTOR OF SAFETY

The magnitude of factor of safety depends upon the following factors :  Effect of failure:  Type of Load:  Degree of Accuracy in Force Analysis:  Material of Component:  Reliability of Component:  Cost of Component:  Testing of Machine Element:  Service Conditions:  Quality of Manufacture:

FACTOR OF SAFETY

• For cast iron components, ultimate tensile strength is considered to be the failure criterion. F.o.s = 3 to 5
• For ductile materials, yield strength is considered to be the criterion of failure. F.o.s = 1.5 to 2
• For components made of ductile materials and those subjected to external fluctuating forces. F.o.s = 1.3 to 1.
• The design of certain components such as cams and followers, gears, rolling contact bearings or rail and wheel is based on the calculation of contact stresses by the Hertz’ theory (local stresses). F.o.s = 1.8 to 2. surface endurance limit.
• Certain components, such as piston rods, power screws or studs, are designed on the basis of the buckling consideration. F.o.s = 3 to 6 A higher factor of safety increases the reliability of the component. However, it increases the dimensions, the volume of material and consequently the cost of the machine component.

05/02/2019 Dr. Gagandeep Bhardwaj, AP MED, TIET 7

SHEAR LOAD

Riveted Joint

Shear Deformation

Shear Stress

Elements loaded in pure shear

Shear Strain

Bolted joint, showing three areas of direct shear.

Direct shear loading (showing failure in double shear)

05/02/2019 Dr. Gagandeep Bhardwaj, AP MED, TIET 8

BENDING LOAD

(a) Distribution of Bending stress; (b) Stress

DESIGN OF SIMPLE MACHINE PARTS

• The dimensions of simple machine parts are determined on the basis of pure tensile stress, pure compressive stress, direct shear stress, bending stress or torsional shear stress.
• The analysis is simple but approximate, because a number of factors such as principal stresses, stress concentration, and reversal of stresses is neglected.
• It is incorrect to assume allowable stress as data for design. The allowable stress is to be obtained from published values of ultimate tensile strength and yield strength for a given material.

PRINCIPAL STRESSES

Stresses on an inclined plane:

Principal Plane

Principal Stresses

Plane of max. shear

Max. shear stress Mohr’s^ circle

2-D state of stress

REPRESENTING STRESSES ON STRESS ELEMENT

Problem 1: Two plates, subjected to a tensile force of 50 kN, are fixed together by means of three rivets as shown in Figure. The plates and rivets are made of plain carbon steel 10C4 with a tensile yield strength of 250 N/mm^2. The yield strength in shear is 50% of the tensile yield strength, and the factor of safety is 2.5. Neglecting stress concentration, determine (i) the diameter of the rivets; and (ii) the thickness of the plates.

NUMERICAL PROBLEM

NUMERICAL PROBLEM

Problem 2: A beam of uniform rectangular cross-section is fixed at one end and carries an electric motor weighing 400 N at a distance of 300 mm from the fixed end. The maximum bending stress in the beam is 40 MPa. Find the width and depth of the beam, if depth is twice that of width.

Solution: W = 400 N; L = 300 mm; σ b = 40 MPa = 40 N/mm^2 ; h = 2 b

Section modulus:

Maximum bending moment (at the fixed end):

NUMERICAL PROBLEM

Problem 3: A hollow shaft of 40 mm outer diameter and 25 mm inner diameter is subjected to a twisting moment of 120 N-m, simultaneously, it is subjected to an axial thrust of 10 kN and a bending moment of 80 N-m. Calculate the maximum compressive and shear stresses. Solution: do = 40 mm ; di = 25 mm ; T = 120 N-m = 120 × 103 N-mm ; P = 10 kN = 10 × 103 N ; M = 80 N-m = 80 × 103 N-mm We know that cross-sectional area of the shaft,

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