Mohr's circle of stress 15

B

t

cy (-120N/mrn2) Fig. 1.10 Stress system for Example 1 .I.

The stress system at the point in the material may be represented as shown in Fig. 1.10 by considering the stresses to act uniformly over the sides of a triangular element ABC of unit thickness. Suppose that the direct stress on the principal plane AB is U. For horizontal equilibrium of the element

U~~ cos

e

^{= U x}

+

T x . , , ~ ~ ^{~ ~} which simplifies to

rXy tan 8 = u - C J ~

Considering vertical equilibrium gives

uAB sin

e

⁼ uyAC

+

^T,,BC

(ii) or

Txy cot

e

^{= -}

Hence from the product of Eqs (i) and (ii)

2

T x y = (. - S x ) ( U -

(4

Now substituting the values nX = 160N/mm2, gy = -120N/mm 2 and u = c1 = 200 N/mm2 we have

T~~ = f 113 N/mm2

Replacing cot B in Eq. (ii) by 1/ tan 8 from Eq. (i) yields a quadratic equation in u

(iii) and 74,;

2

^{- U ( U x - U y )}

+

^{UxUy - Try 2 =} 0

The numerical solutions of Eq. (iii) corresponding to the given values of ox,

are the principal stresses at the point, namely

aI = 200 N/mm2 (given), aJr = - 160 N/mm 2

16 Basic elasticity

Fig. 1.11 Solution of Example 1 . I using Mohr's circle of stress.

Having obtained the principal stresses we now use Eq. (1.15) to find the maximum shear stress, thus

The solution is rapidly verified from Mohr's circle of stress (Fig. 1.11). From the arbitrary origin 0, OP1 and 0 P 2 are drawn to represent o . ~ = 160N/mm2 and ay = -120N/mm2. The mid-point C of PIP2 is then located. OB = uI = 200N/

mm2 is marked out and the radius of the circle is then CB. OA is the required principal stress. Perpendiculars PIQl and P2Q2 to the circumference of the circle are equal to

&T~,, (to scale) and the radius of the circle is the maximum shear stress.

The external and internal forces described in the previous sections cause linear and angular displacements in a deformable body. These displacements are generally defined in terms of strain. Longitudinal or direct strains are associated with direct stresses ⁰ and relate to changes in length while shear strains define changes in angle produced by shear stresses. These strains are designated, with appropriate suffixes, by the symbols ^E and y respectively and have the same sign as the associated stresses.

Consider three mutually perpendicular line elements OA, OB and OC at a point 0 in a deformable body. Their original or unstrained lengths are Sx, Sy and Sz respec- tively. If, now, the body is subjected to forces which produce a complex system of direct and shear stresses at 0, such as that in Fig. 1.6, then the line elements will deform to the positions O'A', O'B' and O'C' shown in Fig. 1.12.

The coordinates of 0 in the unstrained body are (x, y , z ) so that those of A, B and C are (x

+

^Sx, y , z ) , (x, y

+

^{by, z ) and (x, y ,} z

+

Sz). The components of the displacement

1.9 Strain 17

Fig. 1.12 Displacement of line elements OA, OB and OC.

of 0 to 0’ parallel to the x , y and z axes are u, v and w. These symbols are used to designate these diplacements throughout the book and are defined as positive in the positive directions of the axes. We again employ the first two terms of a Taylor’s series expznsion to determine the components of the displacements of A, B and C.

Thus, the displacement of A in a direction parallel to the x axis is zi

+

(dzr/dx)Sx.

The remaining components are found in an identical manner and are shown in Fig. 1.12.

We now define direct strain in more quantitative terms. If a line element of length L at a point in a body suffers a change in length AL then the longitudinal strain at that point in the body in the direction of the line element is

. AL

E = lim -

L - 0 L

The change in length of the element OA is (O’A’ - OA) so that the direct strain at 0 in the x direction is obtained from the equation

(1.16) O‘A‘ - OA O’A’ - Sx

- -

OA SX

E, = Now

d V dw

- u>’

+

^{(V - v ) ~}

+

( W + ~ S X - w or

18 Basic elasticity

which may be written when second-order terms are neglected O’A’ = Sx( 1

+

^{2 g ) ’}

Applying the binomial expansion to this expression we have

O’A’ = Sx (1

+E)

^(1.17)

in which squares and higher powers of &/ax are ignored. Substituting for O‘A’ in Eq. (1.16) we have

It follows that

E, = -

& =-

aY

i1

^(1.18)

The shear strain at a point in a body is defined as the change in the angle between two mutually perpendicular lines at the point. Therefore, if the shear strain in the x-7 plane is T,~ then the angle between the displaced line elements O’A’ and O’C‘ in Fig. 1.12 is 7r/2 - yxz radians.

T,=. From the trigonometrical relationships for a triangle

Now cos A’O’C’ = cos(7r/2 - yxz) = sin yxz and as yxz is small then cos A 1 “ 0 C = (O’A’)2

+

^{(O‘C’)2 - (A’C’)2}

2 (0’ A’ ) (O’C’)

COS A’O’C’ = (1.19)

We have previously shown, in Eq. (1.17), that O’A‘= Sx(l+$) Similarly

o’c‘

^{= Sz( 1}

+E)

But for small displacements the derivatives of u, w and w are small compared with 1, so that, as we are concerned here with actual length rather than change in length, we may use the approximations

O’A’ M Sx, O’C’ M Sz Again to a first approximation

( A C ) “ 2 =

(

S z - - S x ) 2

+ (

S x - - S z ^{) ?} Substituting for O’A’, O’C‘ and A’C‘ in Eq. (1.19) we have

(ax2)

+ ^(q2

^{- [Sz -} (aw/ax)SxI2 - [Sx - (au/az)Sz]2

COS A‘O‘C’ =

2SxSz

1 .IO Compatibility equations 19

Expanding and neglecting fourth-order powers gives

2( aw/ax)SxSz

+

2( du/dz)SxSz

COS A'O'C' =

2sxsz or

Similarly

dw au

"ixz = - _ax

+

^- _{a z}

a v au

Y x y = - _ax

+

^- _{a y}

aw a v 7,; = -

+

^-

ay dz

J

(1.20)

It must be emphasized that Eqs (1.18) and (1.20) are derived on the assumption that the displacements involved are small. Normally these linearized equations are adequate for most types of structural problem but in cases where deflections are large, for example types of suspension cable etc., the full, non-linear, large deflection equations, given in many books on elasticity, must be employed.

...~ , ,.

.

. - . * , .,. , . . ,

p- -=J--y'*y-