Design For Static Loading
Design For Static Loading
Design For Static Loading
Instructional Objectives
At the end of this lesson, the students should be able to understand Types of loading on machine elements and allowable stresses. Concept of yielding and fracture. Different theories of failure. Construction of yield surfaces for failure theories. Optimize a design comparing different failure theories
3.1.1Introduction
Machine parts fail when the stresses induced by external forces exceed their strength. The external loads cause internal stresses in the elements and the component size depends on the stresses developed. Stresses developed in a link subjected to uniaxial loading is shown in figure-3.1.1.1. Loading may be due to: a) The energy transmitted by a machine element. b) Dead weight. c) Inertial forces. d) Thermal loading. e) Frictional forces.
3.1.1.1A- Stresses developed in a link subjected to uniaxial loading In another way, load may be classified as: a) Static load- Load does not change in magnitude and direction and normally increases gradually to a steady value. b) Dynamic loadLoad may change in magnitude for example, traffic of varying weight passing a bridge.Load may change in direction, for example, load on piston rod of a double acting cylinder. Vibration and shock are types of dynamic loading. Figure-3.1.1.2 shows load vs time characteristics for both static and dynamic loading of machine elements.
Load
Time
Static Loading
Load
Load
Load
Time
Time
Time
Dynamic Loading
3.1.2
Determination of stresses in structural or machine components would be meaningless unless they are compared with the material strength. If the induced stress is less than or equal to the limiting material strength then the designed component may be considered to be safe and an indication about the size of the component is obtained. The strength of various materials for engineering applications is determined in the laboratory with standard specimens. For example, for tension and compression tests a round rod of specified dimension is used in a tensile test machine where load is applied until fracture occurs. This test is usually carried out in a Universal testing machine of the type shown in clipping- 3.1.2.1 . The load at which the specimen finally ruptures is known as Ultimate load and the ratio of load to original cross-sectional area is the Ultimate stress.
3.1.2.1V
Similar tests are carried out for bending, shear and torsion and the results for different materials are available in handbooks. For design purpose an allowable stress is used in place of the critical stress to take into account the uncertainties including the following: 1) Uncertainty in loading. 2) Inhomogeneity of materials. 3) Various material behaviors. e.g. corrosion, plastic flow, creep. 4) Residual stresses due to different manufacturing process.
5) Fluctuating load (fatigue loading): Experimental results and plot- ultimate strength depends on number of cycles. 6) Safety and reliability. For ductile materials, the yield strength and for brittle materials the ultimate strength are taken as the critical stress. An allowable stress is set considerably lower than the ultimate strength. The ratio of ultimate to allowable load or stress is known as factor of safety i.e.
material is ductile. However, there are many instances when a ductile material may fail by fracture. This may occur if a material is subjected to (a) (b) (c) (d) (e) Cyclic loading. Long term static loading at elevated temperature. Impact loading. Work hardening. Severe quenching.
Yielding and fracture can be visualized in a typical tensile test as shown in the clipping- Typical engineering stress-strain relationship from simple tension tests for same engineering materials are shown in figure- 3.1.3.1.
(True)
Stress
U
f (Engineering)
y P
Elastic range
Plastic range
Strain
3.1.3.1F- (a) Stress-strain diagram for a ductile material e.g. low carbon steel.
Stress
U Y
(True) f (Engineering)
0.2 % offset
Stress
Strain
f (Ultimate fracture)
Strain
Strain
For a typical ductile material as shown in figure-3.1.3.1 (a) there is a definite yield point where material begins to yield more rapidly without any change in stress level. Corresponding stress is y . Close to yield point is the proportional limit which marks the transition from elastic to plastic range. Beyond elastic limit for an elastic- perfectly plastic material yielding would continue without further rise in stress i.e. stress-strain diagram would be parallel to parallel to strain axis beyond the yield point. However, for most ductile materials, such as, low-carbon steel beyond yield point the stress in the specimens rises upto a peak value known as ultimate tensile stress o . Beyond this point the specimen starts to neck-down i.e. the reduction in cross-sectional area. However, the stress-strain curve falls till a point where fracture occurs. The drop in stress is apparent since original crosssectional area is used to calculate the stress. If instantaneous cross-sectional area is used the curve would rise as shown in figure- 3.1.3.1 (a) . For a material with low ductility there is no definite yield point and usually off-set yield points are defined for convenience. This is shown in figure-3.1.3.1. For a brittle material stress increases linearly with strain till fracture occurs. These are demonstrated in the clipping- 3.1.3.2 .
3.1.3.2V
3.1.4.1 Maximum
According to this, if one of the principal stresses 1 (maximum principal stress), 2 (minimum principal stress) or 3 exceeds the yield stress, yielding would occur. In a two dimensional loading situation for a ductile material where tensile and compressive yield stress are nearly of same magnitude 1 = y 2 = y Using this, a yield surface may be drawn, as shown in figure- 3.1.4.1.1. Yielding occurs when the state of stress is at the boundary of the rectangle. Consider, for example, the state of stress of a thin walled pressure vessel. Here 1= 22, 1 being the circumferential or hoop stress and 2 the axial stress. As the pressure in the vessel increases the stress follows the dotted line. At a point (say) a, the stresses are still within the elastic limit but at b, 1 reaches y although 2 is still less than y. Yielding will then begin at point b. This theory of yielding has very poor agreement with experiment. However, the theory has been used 2 successfully for brittle materials.
+y a -y
..
b 1
+y -y
3.1.4.1.1Ftheory
3.1.4.2 Maximum
According to this theory, yielding will occur when the maximum principal strain just exceeds the strain at the tensile yield point in either simple tension or compression. If 1 and 2 are maximum and minimum principal strains corresponding to 1 and 2, in the limiting case
1 = 2 = 1 ( 1 2 ) E 1 ( 2 1 ) E 1 2 2 1
This gives, E1 = 1 2 = 0 E 2 = 2 1 = 0
The boundary of a yield surface in this case is thus given as shown in figure-
3.1.4.2.1
2 +y -y +y 1
2=0+1
-y
1=0+2
3.1.4.3 Maximum
According to this theory, yielding would occur when the maximum shear stress just exceeds the shear stress at the tensile yield point. At the tensile yield point 2= 3 = 0 and thus maximum shear stress is y/2. This gives us six conditions for a three-dimensional stress situation:
1 2 = y 2 3 = y 3 1 = y
2 +y +y 1
-y
-y
3.1.4.3.1F- Yield surface corresponding to maximum shear stress theory In a biaxial stress situation ( figure-3.1.4.3.1) case, 3 = 0 and this gives
1 2 = y 1 2 = y 2 = y 1 = y 1 = y 2 = y if 1 > 0, 2 < 0 if 1 < 0, 2 > 0 if 2 > 1 > 0 if 1 < 2 < 0 if 1 > 2 > 0 if 2 < 1 < 0
This criterion agrees well with experiment. In the case of pure shear, 1 = - 2 = k (say), 3 = 0 and this gives 1- 2 = 2k= y This indicates that yield stress in pure shear is half the tensile yield stress and this is also seen in the Mohrs circle ( figure- 3.1.4.3.2) for pure shear.
2 1
3.1.4.4 Maximum
According to this theory failure would occur when the total strain energy absorbed at a point per unit volume exceeds the strain energy absorbed per unit volume at the tensile yield point. This 1 y y by 2 1 y y 2
1 ( 11 + 2 2 + 33 ) = may be given 2 1 ( 11 + 2 2 + 33 ) = 2
1 2 + 2 1 22 = 1 y y y
This is the equation of an ellipse and the yield surface is shown in figure3.1.4.4.1 .
2 y
y E(1+ ) -y y E(1 )
y -y
It has been shown earlier that only distortion energy can cause yielding but in the above expression at sufficiently high hydrostatic pressure 1 = 2 = 3 = (say), yielding may also occur. From the above we may write 2 ( 3 2 ) = 2 y and if ~ 0.3, at stress level
lower than yield stress, yielding would occur. This is in contrast to the experimental as well as analytical conclusion and the theory is not appropriate.
3.1.4.5 Distortion
According to this theory yielding would occur when total distortion energy absorbed per unit volume due to applied loads exceeds the distortion energy absorbed per unit volume at the tensile yield point. Total strain energy ET and strain energy for volume change EV can be given as
ET = 1 3 ( 11 + 2 2 + 33 ) and E V = av av 2 2
Ed = ET- EV =
2(1 + ) 2 1 + 2 2 + 32 1 2 2 3 31 6E
E dy =
2(1 + ) 2 y 6E
The failure criterion is thus obtained by equating Ed and Edy , which gives
( 1 2 ) + ( 2 3 )
2
+ ( 3 1 ) = 2 2 y
2
12 + 2 2 1 2 = y 2 i.e. 1 + 2 1 2 = 1 y y y y
This is an equation of ellipse and the yield surface is shown in figure-3.1.4.5.1 . This theory agrees very well with experimental results and is widely used for ductile materials.
2 2
2 45o
y
y
-y -y
0.577 y
3.1.5
A comparison among the different failure theories can be made by superposing the yield surfaces as shown in figure- 3.1.5.1.
2 y -y
It is clear that an immediate assessment of failure probability can be made just by plotting any experimental in the combined yield surface. Failure of ductile materials is most accurately governed by the distortion energy theory where as the maximum principal strain theory is used for brittle materials.
A shaft is loaded by a torque of 5 KN-m. The material has a yield point of 350 MPa. Find the required diameter using (a) Maximum shear stress theory (b) Maximum distortion energy theory Take a factor of safety of 2.5.
A.1:
max
x y 2 = + 2
Y 350 x106 = 2xF.S. 2x 2.5
Since x = y = 0, max=25.46x103/d3 =
This gives d=71.3 mm. (b) Maximum distortion energy theory In this case 1 = 25.46x103/d3 2 = -25.46x103/d3 According to this theory,
2 2 2 ( 1 2 ) + ( 2 3 ) + ( 1 3 ) = 2 ( Y
F.S
Find the factor of safety using (a) Maximum shear stress theory (b) Maximum distortion energy theory. Take the tensile yield strength of the material as 400 MPa.
x=40 MPa
=20 MPa
y=125 MPa
3.1.6.1F
A.2:
From the Mohrs circle, shown in figure-3.1.6.2 1 = 42.38 MPa 2 = -127.38 MPa (a) Maximum shear stress theory
1 2 = Y 2 2xF.S
F.S
2
x=20 MPa
1
y=120 MPa
=-20 MPa
80 MPa
44.72 MPa
2
x=20 MPa
1
y=120 MPa
=-20 MPa
80 MPa
44.72 MPa
3.1.6.2F