Design of Mechanical Failure Prevention 1: Static Failure Prevention · 2018. 3. 11. · 2.19 STATIC FAILURE • Static failure is a failure due to the action of extreme load upon

Chapter 2: Design of

Mechanical Failure Prevention 1: Static Failure Prevention

DR. AMIR PUTRA BIN MD SAAD

C24-322

[email protected] | [email protected]

mech.utm.my/amirputra

2.1 INTRODUCTION

Sometimes the service or operating loads can be readily determined, as are those

on some engines, compressors, and electric generators that operate at known

torques and speeds. Often the loads are difficult to determine, as are those on

automotive chassis components (which depend on road surfaces and driving

practices) or on the structure of an airplane (which depends on air turbulence and

pilot decisions). Sometimes experimental methods are used to obtain a statistical

definition of applied loads. In other instances engineers use records of service

failures together with analyses of strength in order to infer reasonable estimates of

loads encountered in service. The determination of appropriate loads is often a

difficult and challenging initial step in the design of a machine or structural

component.

2.2 EQUILIBRIUM EQUATIONS AND FREE-BODY DIAGRAM

A system with zero acceleration is said to be in equilibrium, if that system is

motionless or, at most, has constant velocity. For a non-accelerating body, these

equations can be simply expressed as

𝑀 = 0𝐹 = 0

Engineers need to know how the physical characteristics of the materials of which

components are fabricated relate to one another.

For an accelerating body they are

𝑀 = 𝐼𝛼𝐹 = 𝑚𝑎

2.2 FREE-BODY DIAGRAM cont.

The analysis of a very complex structure or machine can greatly simplify by

successively isolating each element and studying and analysing it by the use of

free-body diagrams. When all the members have been treated in this manner, the

knowledge obtained can be assembled to yield information concerning the

behaviour of the total system. Thus, free-body diagramming is essentially a means

of breaking a complicated problem into manageable segments, analysing these

simple problems, and then, usually, putting the information together again.

2.2 FREE-BODY DIAGRAM cont.

Using free-body diagrams for force analysis serves the following important purposes:• The diagram establishes the directions of reference axes, provides a place to

record the dimensions of the subsystem and the magnitudes and directions of the known forces, and helps in assuming the directions of unknown forces.

• The diagram simplifies your thinking because it provides a place to store one thought while proceeding to the next.

• The diagram provides a means of communicating your thoughts clearly and unambiguously to other people.

• Careful and complete construction of the diagram clarifies fuzzy thinking by bringing out various points that are not always apparent in the statement or in the geometry of the total problem. Thus, the diagram aids in understanding all facets of the problem.

• The diagram helps in the planning of a logical attack on the problem and in setting up the mathematical relations.

• The diagram helps in recording progress in the solution and in illustrating the methods used.

• The diagram allows others to follow your reasoning, showing all forces.

2.2 EQUILIBRIUM EQUATIONS AND

FREE-BODY DIAGRAM cont.

These equations apply with respect to each of any three mutually

perpendicular axes (commonly designated X, Y, and Z), although in many

problems forces and moments are present with respect to only one or two of

these axes. Remember the right hand principle.

Free-body diagrams help simplifying the analysis of a very complex structure or

machine by successively isolating each element and then studying and analyzing it.

2.2 FREE-BODY DIAGRAM

My suggestion:

Buy several pen colors to draw free body diagram, shear force diagram, bending moment diagram and torsion diagram and etc..

2.2 FREE-BODY DIAGRAM: 2D

Example 2.1:

Determine all the reaction forces at the supported point.

(a) Simple Support

(b) Built-in Support


Example 2.2:

Draw the free-body diagram.

(a) Ball-and-socket Joint

(b) Fixed Support


Solution 2.2:

(a) Ball-and-socket Joint

(b) Fixed Support


Example 2.3:


(a) (b) (c)

(d) (e) (f)


Example 2.4:


(a) (b)

(c) (d)

(e) (f)


Example 2.5:


(a) (b) (b)

2.3 FORCE FLOW CONCEPT

Example 2.6:

Draw the force flow for all the members.

2.3 FORCE FLOW CONCEPT

Solution 2.6:

Force flow lines and critical sections in yoke connection.

2.4 TWO FORCE MEMBER

(a) (b)

Internal ForceTwo Force Member

2.5 THREE FORCE MEMBER

Example 2.7:

Draw the free body diagram for ABC component.

2.5 THREE FORCE MEMBER

Solution 2.7:

Free Body Diagrams

• BD is a two-force member

• Lever ABC is a three-force member

Equations of Equilibrium

Solving,

kNF

kNFA

32.1

07.1

045sin3.60sin ;0

040045cos3.60cos ;0

3.604.0

7.0tan 1

FFF

NFFF

Ay

Ax

2.6 NORMAL STRESS

The normal stress 𝜎 is said to be uniformly distributed with

𝜎 =𝐹

𝐴

2.7 DIRECT SHEAR STRESS

Direct shear is usually assumed to be uniform across the cross section, and is given by

𝜏 =𝑉

𝐴

where 𝑍 =𝐼

𝑐is called the section modulus

2.8 BENDING STRESS

𝜎𝑚𝑎𝑥 =𝑀𝑐

𝐼𝜎𝑚𝑎𝑥 =

𝑀

𝑍𝜎 =

𝑀𝑦

𝐼

2.8 BENDING STRESS

• A beam having a T section with the dimension shown in Fig. 2.8 is subjected to a bending moment of 1600 N.m that causes tension at the top surface. Locate the neutral axis and find the maximum tensile and compressive bending stresses.

• The area of the composite section is A=1956 mm2

2.9 TRANSVERSE SHEAR

STRESS IN BEAM

Transverse Shear Stress Distribution:

𝜏 =𝑉

𝐼𝑏න𝑦=𝑦0

𝑦=𝑐

𝑦𝑑𝐴


STRESS IN BEAM

Effect of beam length on bending and shear loading. The principle behind thisgeneralization is illustrated in above figure, where the same loads are shown appliedto a long and short beam.


STRESS IN BEAM

Solid Rectangle Section:

𝜏 =𝑉𝑄

𝐼𝑡=

6𝑉

𝑏ℎ3ℎ2

4− 𝑦2

The result indicates that the shear stress distribution over the cross section isparabolic, as plotted in above figure. The shear force intensity varies fromzero at the top and bottom, 𝑦 = ±ℎ/2, to a maximum value at the neutralaxis at 𝑦 = 0.

𝜏𝑚𝑎𝑥 =3

2

𝑉

𝐴


STRESS IN BEAM

𝜏𝑚𝑎𝑥 =4

3

𝑉

𝐴

Solid Round Section:

𝜏𝑚𝑎𝑥 = 2𝑉

𝐴

Hollow Round Section (thin-wall tubing):

T = torque ,

l = length,

G = modulus of rigidity

J = polar second moment of area

2.10 TORSION STRESS

𝜏𝑚𝑎𝑥 =𝑇𝑟

𝐽

𝜃 =𝑇𝑙

𝐺𝐽

2.11 SHEAR, BENDING MOMENT AND TORSION DIAGRAMS

2.12 POWER

2.13 STRESS ELEMENT AND

COMPLEX STRESS

Determine all the element stresses.

Example 2.8:

(a) (b)

(c) (d)

(e) (f)

2.14 MOHR’S CIRCLE: PLANE STRESS

2.14 MOHR’S CIRCLE: PLANE STRESS

𝜎 =𝜎𝑥 + 𝜎𝑦

2+𝜎𝑥 − 𝜎𝑦

2cos 2𝜙 + 𝜏𝑥𝑦 sin 2𝜙

𝜏 = −𝜎𝑥 − 𝜎𝑦

2sin 2𝜙 + 𝜏𝑥𝑦 cos 2𝜙

Analytical Approach:

2.15 STRESS CONCENTRATION FACTOR, Kt

• Any discontinuity in a machine part alters the stress distribution in theneighborhood of the discontinuity so that the elementary stress equations nolonger describe the state of stress in the part at these locations.

• Stress concentrations can arise from some irregularity not inherent in themember, such as tool marks, holes, notches, grooves, or threads.

𝐾𝑡 =𝜎𝑚𝑎𝑥𝜎𝑛𝑜𝑚

𝐾𝑡𝑠 =𝜏𝑚𝑎𝑥𝜏𝑛𝑜𝑚

2.15 STRESS CONCENTRATION FACTOR, Kt

From Text Book. Refer Figure 4.35 and Figure 4.36.

r/d: 2/40 = 0.05D/d: 60/40 = 1.5Kt = 2.38

2.16 EXAMPLE

EXAMPLE 2.16:

2.17 CONVERSION TABLE

2.18 MECHANICAL PROPERTIES

EXERCISES 1.1:

Sketch a free body diagram of each element in the figure. Compute themagnitude and direction of each force using an algebraic or vector method,as specified.

A pin in a knuckle joint carrying a tensile load F deflects somewhat on account of this loading,making the distribution of reaction and load as shown in part (b) of the figure. A commonsimplification is to assume uniform load distributions, as shown in part (c). To further simplify,designers may consider replacing the distributed loads with point loads, such as in the modelsshown in parts d and e. If a = 12 mm, b = 18 mm, d =12 mm and F = 4 kN, estimate themaximum bending stress and the maximum shear stress due to V for the three simplifiedmodels. Compare the three models from a designers’s perspective in terms of accuracy, safetyand modeling time.

EXERCISES 1.2:

(i) Draw shear force diagram and bending moment diagram.

(ii) Determine the critical point.

(iii) Determine the principal stress and maximum shear stress

EXERCISES 1.3:

If Fy = 800 N (i) Determine the precise location of the critical stress element. (ii)Sketch the critical stress element and determine magnitudes and directions for allstresses acting on it. (iii) For the critical stress element, determine the principalstress and maximum shear stress.

EXERCISES 1.4:

(a) (b)

2.19 STATIC FAILURE

• Static failure is a failure due tothe action of extreme load uponto the design.

• The extreme loads can be formsof such as water, river, ocean,rain, snow, storm, thunder etc.

• Unexpected natural disaster isnot an engineer concerns.

2.20 GUARDING STATIC FAILURE CRITERIA

• There is no universal theory of failure for the general case of material propertiesand stress state. Instead, over the years several hypotheses have been formulatedand tested, leading to today’s accepted practices most designers do.

• The generally accepted theories are:

Ductile materials (yield criteria)» Maximum Shear Stress (MSS)*

» Distortion Energy (DE)*

» Ductile Coulomb-Mohr (DCM)

Brittle materials (fracture criteria)

» Maximum Normal Stress (MNS)*

» Brittle Coulomb-Mohr (BCM)

» Modified Mohr (MM)

2.20 GUARDING STATIC FAILURE CRITERIA

2.21 SAFETY FACTORS

2.22 MAXIMUM SHEAR STRESS THEORY

The maximum-shear-stress theory predicts thatyielding begins whenever the maximum shearstress in any element equals or exceeds themaximum shear stress in a tension testspecimen of the same material when thatspecimen begins to yield.

𝑛 =𝑆𝑦

2𝜏𝑚𝑎𝑥

* Knowledge of Complex Stress or Mohr Circle is compulsory.

2.23 DISTORTION ENERGY THEORY

𝜎′ = 𝜎𝑥2 − 𝜎𝑥𝜎𝑦 + 𝜎𝑦

2 + 3𝜏𝑥𝑦2

𝜎′ = Von Mises Stress

𝑛 =𝑆𝑦

𝜎′

The distortion-energy theory predicts thatyielding occurs when the distortion strainenergy per unit volume reaches or exceeds thedistortion strain energy per unit volume foryield in simple tension or compression of thesame material.

* Knowledge of Complex Stress or Mohr Circle is NOT compulsory.

2.24 MAXIMUM NORMAL STRESS THEORY

The maximum-normal-stress (MNS) theorystates that failure occurs whenever one of thethree principal stresses equals or exceeds thestrength.

𝑆𝑢𝑡 and 𝑆𝑢𝑐 are the ultimate tensile andcompressive strengths, respectively, given aspositive quantities.

𝑛 =𝑆𝑢𝑡𝜎1

𝑛 =𝑆𝑢𝑐𝜎2

* Knowledge of Complex Stress or Mohr Circle is compulsory.

2.25 EXAMPLE

Figure 2.0: Schematic drawing of the design.

The horizontal shaft ABCD is mounted in bearings at B and D as shown. A belt passes

around the 250 mm diameter pulley fixed to the shaft at A, and a gear pinion of 150 mm

pitch diameter is mounted on the shaft at C. Shaft diameters and axial disposition of the

components are as sketched. The belt strand tensions are horizontal and in the ratio

F1/F2 = 4, while the vertical reaction on the pinion, P, acts tangentially to the pinion's

pitch circle. Ascertain the shaft's safety factor when transferring 20 kW from belt to

pinion at a steady 7.5 Hz, taking the yield strength of the ductile shaft material to be 500

MPa and ultimate tensile strength is 700 MPa. Stress concentration are neglected here.

Example:

Documents

Design of Mechanical Failure Prevention 1: Static Failure Prevention · 2018. 3. 11. · 2.19 STATIC FAILURE • Static failure is a failure due to the action of extreme load upon