ASM Metals HandBook Vol. 8 - Mechanical Testing and Evaluation

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(Eq 22)

Because of equilibrium considerations, the radial stress must be continuous at r = R

. Applying this condition to

Eq 19 and 20, the location of the neutral axis can be found:

(Eq 23)

Figure 9 is a schematic of the distributions of σ

and σ

. In the figure, σ

is continuous and compressive

throughout the plate thickness, while σ

changes from tension to compression at the neutral axis. The bending

moment, which according to this solution is independent of R

, becomes:

(Eq 24)

where h is half thickness, and b is the plate width.

Fig. 9 Schematic of circumferential, σ

, and radial, σ

, stresses in a plate during bending.

Source: Ref 7

The maximum radial stress occurs at the neutral axis. Its magnitude from Eq 19 and 23 is:

(Eq 25)

The ratio between (σ

)

max

and the tangential stress at r = R

for four plates of different thicknesses, all bent to an

inside radius of R

= 25 mm (1 in.), is given in Table 1. This table shows that for plastic bending to (R

/2h) >

10, the magnitude of (σ

)

max

becomes very small. In such cases, the effect of σ

can be neglected in the analysis

and the elastic-plastic bending solution based on the simple-beam theory can be used.

Table 1 Ratio between maximum radial stress and tangential stress for plate of various

thicknesses

Thickness

in.

mm in.

/ 2h

max /

(σ

at r = R

)

1 1.59

0.0625

16.49 0.030

1 3.17

0.125 8.48 0.059

1 6.35

0.25 4.47 0.112

1 12.7

0.5 2.45 0.203

See text for explanation of symbols

More elaborate analyses of plastic bending (Ref 6 and 8) commonly involve complicated numerical

computations. Hence, no equations can be given. A comparison between the results of the analysis in Ref 7 and

a solution for rigid work-hardening material behavior (Ref 6) is shown in Fig. 10. In this case, the result from

Ref 6 is for a model material with a high rate of strain hardening.

= 70 + 300

0.5

MPa

(Eq 26)

In using the analysis from Ref 6, = 120 MPa (17.4 ksi), which is the approximate average flow stress for

bending to = 0.11, and = 169.5 MPa (24.6 ksi), as shown in Fig. 10(b), have been assumed. As expected,

some differences in the predicted stress distributions are observed.

Fig. 10 Comparison of results for determining plastic bending in a plate. (a) Distribution

of tangential and radial stresses for a 25 mm (1 in.) thick plate bent to R

= 100 mm (4 in.).

(b) Stress-strain diagrams used in the analyses for (a)

However, by using = 169.5 MPa (24.6 ksi) in the solution from Ref 6, a close agreement (percent different <

6) between the estimates for the fiber stresses at r = R

and r = R

is obtained. Because the prediction of

maximum fiber stress and strain is of special interest, the following procedure based on the solution in Ref 6 is

suggested. First, find the maximum fiber strain:

(Eq 27)

where ε

and ε

represent the maximum fiber strain at the outer and inner radii, respectively. Using the stress-

strain equation or stress-strain curve for the material, determine as the flow stress at ε

. The maximum fiber

stress is 2 / .

References cited in this section

6. P. Dadras and S.A. Majlessi, Plastic Bending of Work Hardening Materials, ASME Trans. J. Eng. Ind.,

Vol 104, 1982, p 224–230

7. R. Hill, The Mathematical Theory of Plasticity, Oxford University Press, London, 1950

8. H. Verguts and R. Sowerby, The Pure Plastic Bending of Laminated Sheet Metals, Int. J. Mech. Sci.,

Vol 17, 1975, p 31

Stress-Strain Behavior in Bending

P. Dadras, Wright State University (retired)

Residual Stress and Springback

When a specimen that has been bent beyond the elastic limit is unloaded, the applied moment M becomes zero,

and the radius of curvature increases from R

to R′

. For a fiber at distance y from the neutral axis, this produces

a strain difference:

(Eq 28)

The removal of the bending moment, which is an unloading event, is assumed to be elastic. Therefore:

(Eq 29)

The change in bending moment for complete unloading is ΔM = M, where M is the applied bending moment

prior to unloading. Therefore:

(Eq 30)

which reduces to

(Eq 31)

The distribution of the residual stresses can be found from either of the following equations:

(Eq 32)

(Eq 32a)

It is important that the correct signs for σ

and y be used when applying these equations.

As an example, the springback and the residual stress distribution of a strip of annealed 1095 steel was

examined (Ref 9). For this material, yield strength is σ

= 308 MPa (44.7 ksi), Poisson's ratio is ν = 0.28, and

the approximate constitutive equation for σ in metric units of measure is:

= (2 × 10

MPa)ε for ε ≤ 0.00154

(Eq 33)

= (896 MPa)ε

0.16

for ε ≥ 0.00154

(Eq 34)

In English units of measure, σ is:

= (29 × 10

psi)ε for ε ≤ 0.00154

= (126 ksi)

0.16

for ε ≥ 0.00154

The width of the strip, b, is 50 mm (2 in.), and its thickness, 2h, is 5 mm (0.2 in.). It is assumed that the strip is

bent to R

= 100 mm (4 in.). Because (b/2h = 10, plane-strain deformation prevails. Because of this, the elastic

modulus in plane-strain

(Eq 35)

is employed, and the plastic flow stresses (Eq 34) are multiplied by (2 / ). These approximate plane-strain

adjustments are considered adequate when the simplified elastic-plastic analysis, which was discussed earlier in

this article, is used. The thickness of the elastic core in metric units of measure is:

(Eq 36)

In English units of measure, the thickness of the elastic core is:

which is 6.6% of the total plate thickness. The final radius of curvature in metric units of measure after

springback is found from Eq 31:

(Eq 37)

which results in R′

= 116.9 mm (4.60 in.). A more elaborate analysis of springback (Ref 9) for this case

predicts R′

/2h = 23.41 (or R′

= 117.05 mm, or 4.608 in.). Also, for bending the same strip to R

= 40 mm (1.6

in.) and R

= 500 mm (19.7 in.), the final radii of curvature from Eq 31 are 42.8 mm (1.68 in.) and 1150 mm

(45.3 in.), respectively. The corresponding results from Ref 9 are 42.9 mm (1.69 in.) and 1075 mm (42.3 in.).

The distribution of residual stresses after bending to R

= 100 mm (3.937 in.) is obtained from Eq 32a(a).

Therefore, in this case, in metric units:

σ′

= σ

+ yE(0.0014457)

(Eq 38)

In English units:

σ′

= σ

+ yE(0.03672)

At R = R

, y = h = 2.5 mm (0.098 in.):

= -0.025

and

For this location:

σ′

= -556 + 2.5(2.17 × 10

)

× (0.0014457) = +228 MPa

σ′

= -80.63 + 0.098 (31465)(0.03672) = +32.6 ksi

Similarly, the magnitude of σ

for other values of y can be determined. Figure 11 shows the distribution of

applied and residual stresses.

Fig. 11 Distribution of applied and residual stresses

Reference cited in this section

9. O.M. Sidebottom and C.F. Gebhardt, Elastic Springback in Plates and Beams Formed by Bending, Exp.

Mech., Vol 19, 1979, p 371–377

Stress-Strain Behavior in Bending

P. Dadras, Wright State University (retired)

References

1. J.M. Gere and S.P. Timoshenko, Mechanics of Materials, 4th ed., PWS Publishing Co., 1997

2. A.C. Ugural and S.K. Fenster, Advanced Strength and Applied Elasticity, 3rd ed., Prentice Hall, 1995

3. D. Horrocks and W. Johnson, On Anticlastic Curvature with Special Reference to Plastic Bending: A

Literature Survey and Some Experimental Investigations, Int. J. Mech. Sci., Vol 9, 1967, p 835–861

4. J. Chakrabarty, Theory of Plasticity, McGraw-Hill, 1987

5. P. Dadras, Plane Strain Elastic-Plastic Bending of a Strain-Hardening Curved Beam, Int. J. Mech. Sci.,

in press Nov 2000

6. P. Dadras and S.A. Majlessi, Plastic Bending of Work Hardening Materials, ASME Trans. J. Eng. Ind.,

Vol 104, 1982, p 224–230

7. R. Hill, The Mathematical Theory of Plasticity, Oxford University Press, London, 1950

8. H. Verguts and R. Sowerby, The Pure Plastic Bending of Laminated Sheet Metals, Int. J. Mech. Sci.,

Vol 17, 1975, p 31

9. O.M. Sidebottom and C.F. Gebhardt, Elastic Springback in Plates and Beams Formed by Bending, Exp.

Mech., Vol 19, 1979, p 371–377

Fundamental Aspects of Torsional Loading

John A. Bailey, North Carolina State University;Jamal Y. Sheikh-Ahmad, Wichita State University

Introduction

TORSION TESTS can be carried out on most materials, using standards specimens, to determine mechanical

properties such as modulus of elasticity in shear, yield shear strength, ultimate shear strength, modulus of

rupture in shear, and ductility. Torsion tests can also be carried out on full-size parts (shafts, axles, and twist

drills) and structures (beams and frames) to determine their response to torsional loading. In torsion testing,

unlike tension testing and compression testing, large strains can be applied before plastic instability occurs, and

complications due to friction between the test specimen and dies do not arise.

Torsion tests are most frequently carried out on prismatic bars of circular cross section by applying a torsional

moment about the longitudinal axis. The shear stress versus shear strain curve can be determined from

simultaneous measurements of the torque and angle of twist of the test specimen over a predetermined gage

length.

Certain shear properties of materials can also be determined by single or double direct shear tests. In these

types, of tests loads are applied to bars, usually of circular section, in such a way as to produce failure (shear)

on either one (single) or two (double) transverse planes perpendicular to the axis of the bar. The shear strength

of the bar is determined by dividing the shear load by the cross-sectional area of the bar. Such tests provide

little fundamental information on the shear properties of materials and are primarily used in the design of rivets,

bolts keyway systems, and so forth, that are subjected to shearing loads in service (Ref 1) (see also the article

“Shear, Torsion, and Multiaxial Testing” in this Volume).

The following sections discuss the torsional deformation of prismatic bars of circular cross section. Discussion

of the torsional response of prismatic bars of noncircular cross section (rectangular, elliptical, triangular) in the

elastic range can be found in Ref 2.

References cited in this section

1. C.L. Harmsworth, Mechanical Testing, Vol 8, ASM Handbook, American Society for Metals, 1985, p 62

2. S.P. Timoshenko and J.N. Goodier, Theory of Elasticity, 3rd ed., McGraw-Hill, 1970, p 291

Fundamental Aspects of Torsional Loading

John A. Bailey, North Carolina State University;Jamal Y. Sheikh-Ahmad, Wichita State University

Prismatic Bars of Circular Cross Section

Elastic Deformation (Solid Bars). In torsional testing of prismatic bars of circular cross section it is assumed

that:

• Bar material is homogeneous and isotropic.

• Twist per unit length along the bar is constant.

• Sections that are originally plane to the torsional axis remain plane after deformation.

• Initially straight radii remain straight after deformation.

Figure 1 shows the torsional deformation of a long, straight, isotropic prismatic bar of circular section.

Assuming the above-mentioned constraints, the displacements are given by:

(Eq 1)

where dθ/dz is the angle of twist per unit length (θ/L) and L is the gage length of the test specimen. The strains

are given by:

(Eq 2)

For an isotropic material that obeys Hooke's law, the corresponding stress state is given by:

= 0

θθ

= 0

rθ

= 0

zθ

= Gγ

zθ

(Eq 3)

where G is the shear modulus that is related to Young's modulus (E) and Poisson's ratio (ν) by:

(Eq 4)

The stress distribution across a prismatic bar of circular cross section is given by:

(Eq 5)

Thus, the shear stress is zero at the center of the bar (r = 0) and increases linearly with radius. The maximum

value of the shear stress occurs at the surface of the bar (r = a) and is given by:

(Eq 6)

Fig. 1 Torsion of a solid circular prismatic section

The torque (T) transmitted by the elemental section shown in Fig. 1 is given by:

dT = (τ

zθ

r) dA

(Eq 7)

zθ

2πr

(Eq 8)

Combining Eq 5 and 8 gives:

(Eq 9)

Integration gives:

(Eq 10)

or on rearrangement:

(Eq 11)

where J = πa

/2 is the polar moment of inertia of a prismatic bar of circular section about its axis of symmetry.

Thus, the angle of twist can be calculated from knowledge of the applied torsional load, shear modulus, and bar

geometry. Combining Eq 6 and 10 gives:

(Eq 12)

(Eq 13)

Thus, the maximum shear stress can be calculated from knowledge of the torsional loading and bar geometry.

Plastic Deformation (Solid Bars) of Non-Work-Hardening Material. When the surface shear stress (τ

zθ

)

max

of a

solid bar during torsional loading reaches the yield shear stress (k) of the test material, plastic deformation

(flow) occurs. The deformation zone begins at the surface of the bar and advances inward as an annulus

surrounding an elastic core. The stress distributions are shown schematically in Fig. 2 for a non-work-hardening

and a work-hardening material.

Fig. 2 Section through prismatic bar of circular section

For a non-work-hardening material, the total torque transmitted by the bar, according to Ref 3, is given by:

(Eq 14)

The first term on the right side of Eq 14 is the torque transmitted by the elastic core, where the shear stress

varies linearly with r. The second term on the right side of Eq 14 is the torque transmitted by the plastic annulus

, where the shear stress is constant and independent of r. The elastic-plastic boundary occurs at r = r

Integration of Eq 14 gives:

(Eq 15)

Compatibility at the elastic-plastic boundary requires that:

(Eq 16)

Combining Eq 15 and 16 and rearranging gives:

(Eq 17)

(Eq 18)

where θ

is the angle of twist at which yielding begins. When θ is very large compared to θ

, then:

(Eq 19)

where T

is the torque required for fully plastic flow. Equation 18 can now be rewritten as:

(Eq 20)

When the bar becomes fully plastic, the torque becomes independent of the angle of twist. In the elastic regime,

the shear stress at the surface of the bar is given by Eq 6. In the elastic-plastic and fully plastic regimes, the

shear stress at the surface of the bar is k. The shear strain at the surface of the bar is:

(Eq 21)

for all regimes.

Plastic Deformation (Solid Bars) of Work-Hardening Material. In practice, most materials work harden when

tested at temperatures below 0.5 T

, where T

is the homologous temperature and is given by T

= T

(where T

is the testing temperature and T

is the melting point temperature of the material). The result is that

the torque continues to increase up to fracture.

The shear stress versus shear strain curve in the plastic range can be computed from the torque-twist curve

using the procedure given below. It is important to note that the computed values of stress and strain are those

that occur at the surface of the bar and that the material is insensitive to the rate of deformation.

The torque, according to Ref 3 and 4, is given by:

(Eq 22)

where the subscripts on the shear stress are dropped. Changing the variable from r to γ gives:

(Eq 23)

where θ

is the twist per unit length. In general, the shear stress versus shear strain curve can be written as:

τ = f(γ)

(Eq 24)

Thus, Eq 24 becomes:

(Eq 25)

Differentiating Eq 25 with respect to θ

gives:

d( ) = 2πf(γ

) dγ

(Eq 26)

At the specimen surface:

= f(γ

)

(Eq 27)

and

= aθ

(Eq 28)

Substituting Eq 27 and 28 into Eq 26 gives:

d( ) = 2πτ

dθ

(Eq 29)

Expanding Eq 29 gives: