Lecture 3 The Concept of , Generalized Stresses and Equilibrium

Problem 3-1: Cauchy’s Stress Theorem

Cauchy’s stress theorem states that in a stress there is a traction vector t that linearly depends on the outward unit n:

a. Express Cauchy’s stress theorem in index form.

b. Suppose the state of stress at a in x, y, z is given by the

below. Determine the normal stress n and the shear stress on the defined by .

Problem 3-1 Solution:

(a)

tnii ji

(b)

A normal vector to a specified by

fx,,yz axbyczd0 is given by

a Nf b c where f denotes the .

1 The of the surface

23xy z9

is

2 N 1 n 1 N 2122(3)2 3

2 1 n 1 14 3

The magnitude of normal stress n can be calculated by

n tn

where t is the surface tra ction vector

20 10 10 2 80 11 tn 10 30 0 1 50 14 14 10 0 50 3170

Then

80 2 11720 tn 50 1 n 14 14 14 170 3

n 51Force / area

The direction of the normal stress vector n is n

2 51 n 1 nn 14 3

2 Shear stress vector can be simply calculated by subtract the normal stress vector n from the traction vector t

80 2 160 / 7 13601 t 50 1 10 / 7 14 7 14 14 1703110 / 7

3 Problem 3-2: Three invariants of a stress tensor

Supposethestateofstressatapointinax,y,zcoordinatesystemisgivenby

100 1 180 0200 180 0 20

a. Calculate the three invariants of this stress tensor.

b. Determine the three principal stresses of this stress tensor.

Problem 3-2 Solution: a) Invariants of stress tensor

Recall: these values do not change no matter the coordinate system selected

I1 xyz100 20 20

I1 60

I 22 2 2 xy yz zx xy yz zx (10020)(10020)20220080

I2 10000

I 2 22 2 3 x yz xyyzzx xyz yzx zxy (1002020)20020(80)2 0

I3 168000 b) Principal stresses

The eigenvalue problem for a stress tensor

4 100 0 80 0200 80 0 20 is given by

100 0 80 det 0 20 0 0 80 0 20

Solve for , we have three eigenvalues

1 140

2 60

3 20

The principal stresses are the three eigenvalues of the stress tensor

xp 140

yp 60

zp 20

5 Problem 3-4:

Suppose the state of stress at a point relative to a x, y, z coordinate system is given by:

15 10 10 5

Try to find a new coordinate system (x’, y’) that corresponds to the principal directions of the stress tensor.

a. Find the principal stresses.

b. Determine the transformation matrix .

c. Verify .

Problem 3-4 Solution: a) Determine principal stresses

The eigenvalue problem for a stress tensor

15 10 10 5 is given by

15 10 det 0 10 5

Solve for , we have two eigenvalues

19.14 1 2 9.14

The principal stresses are the two eigenvalues of the stress tensor

xp 19.14

yp 9.14

6

b) Determine transformation matrix

Calculate the eigenvectors for the stress tensor

For 1 19.14

15 19.14 10 x1 0 10 5 19.14 x2

Set x1 1, use the first

x2 0.414

Normalize eigenvector

1 1 1 122( 0.414) 0.414

0.92 1 0.38

For 1 9.14

15 ( 9.14) 10 x1 0 10 5 (9.14) x2

Set x1 1, use the first equation

x2 2.414

Normalize eigenvector

1 1 2 1(22 2.414)2.414

0.38 1 0.92

Check 12 ?

7 0.920.38 0.380.92 0 12 12!

Transformation Matrix

L 12 0.92 0.38 L 0.38 0.92

b) Verify

T ' L L

If we transform the given stress tensor, will we arrive at the principal stresses?

Note: use 4 decimal places to get principal stresses

T 0.9238 0.3827 15 10 0.9238 0.3827 19.174 0 LL 0.3827 0.9238 10 5 0.3827 0.9238 0 9.14

19.174 0 ' 09 .14

8 Problem 3-5: Beam Equilibrium

Derive the equation of force and moment equilibrium of a beam using the equilibrium of an infinitesimal beam element of the length dx.

Problem 3-5 Solution:

Consider an element of a beam of length dx subjected to

1. Distributed loads q

2. Shear forces V and Vd V

3. Moments M and M dM

Equilibrium of forces in x and y direction

dN FN00dNN 0 x dx dV FV0(qdxVdV)0 q () y dx

Moment equilibrium of the element at point O

dx MM0( qdx)(VdV)dxMdM0 2

From equation (*),

dV q dx dV dV dx qdx dx

10 Substitute the above equation into the moment equilibrium equation, we have

q dM ()dx 22Vdx q()dx 0 2

Ignore the second-order terms, we have

dM Vdx dM V dx

To sum up, the of force and moment equilibrium of a beam are

dN 0 dx dV q dx dM V dx

11 Problem 3-6: Moderately large deflections in beams

Explain which of the three equilibrium equations below is affected by the finite rotations in the theory of the moderately large deflections of beam.

1. Moment equilibrium

2. Vertical force equilibrium

3. Axial force equilibrium

Problem 3-6 Solution:

Consider an element of a beam of length dx subjected to

1. Distributed loads q

2. Shear forces V and Vd V

3. Moments M and M dM and under moderately large deflection

Equilibrium of forces in t direction

FVt 0cosqdxcos(VdV)cos0 dV q dx (1)

Note: All terms involve cos

12 Moment equilibrium of the element at point O

dx MM0( qdx)(VdV)dxMdM0 2

From equation (1),

dV q dx dV dV dx qdx dx

Substitute the above equation into the moment equilibrium equation, we have

q dM ()dx 22Vdx q()dx 0 2

Ignore the second-order terms, we have

dM Vdx dM V dx

The moment equilibrium equation is not affected by moderately large deflection.

Force Equilibrium in x direction

FNx 0(dN)cosNcos0 dN 0 dx (2)

The axial force equilibrium equation is not affected by moderately large deflection.

13 Define effective shear force V * as the sum of the cross-sectional shear V and the projection of the axial force into the vertical direction

dw VV* Nsin VN dx (3)

Force equilibrium in y direction

Vd** VVq*dx0 dV * q 0 dx

Combining equation (3), we have

dV d dw dV dN dw d 2w NqNq2 0 dx dx dx dx dx dx dx use equation(2), we have the vertical force equilibrium equation

dV d 2w Nq0 dx dx2

The vertical force equilibrium equation is affected by moderately large deflection.

14 Problem 3-7:

At full draw, and archer applies a pull of 150N to the bowstring of the bow shown in the figure. Determine the bending moment at the midpoint of the bow.

Problem 3-6 Solution:

P is the pulling force at point A and T is the tension in the string. Force balance at point A

Fx 0 PT 2cos70 0

15 P 150 T219.3N 2cos702cos70

Free body diagram

where the tension T is projected in x and y direction as Tx and Ty.

Moment balance at point B

M 0 B TTxy0.7 0.35 MB0

Finally, the bending moment at the midpoint of the bow is

MTBx0.7 Ty0.35 TTcos 700.7 sin 70 0.35

219.3 cos700.7 219.3sin 70 0.35 124.6Nm

16 Problem 3-8:

A curved bar ABC is subjected to loads in the form of two equal and opposite force P, as shown in the figure. The axis of the bar forms a semicircle of r.

Determine the axial force N, shear force V, and bending moment M acting at a cross defined by the angle

Problem 3-8 Solution:

The free body diagram and force balance at angle :

We can determine axial force N, shear force V according to the force balance diagram

NP sin VP cos

Moment balance at point B

M 0 B Pr sin M 0

We have the bending moment M acting at a cross section defined by the angle

MPrsin

17 Problem 3-9:

The centrifuge shown in the figure rotates in a horizontal plane (the xy plane) on a smooth surface about the z axis (which is vertical) with an angular acceleration . Each of the two arms has weight w per unit length and supports a weight Ww 2.5 L at its end.

Derive formulas for the maximum shear force and maximum bending moment in the arms, assuming bL /9and cL /10

Problem 3-9 Solution:

Shear force density distribution and free body diagram:

Where weight at the tip of arm W is considered as lumped mass

18 Force balance at any point of radius r

bL wW Vr **drbLc r gg

WW 2 bLr2 bLc 2gg

Moment balance at any point of radius r

bL wW Mr **rrdr*bLcbLcr r gg bL WW11*3 *2 rrr bLcbLcr gg32 r

W1132W bLr32rbLrbLcbLc r g32g

By observation maximum shear force and bending moment occurs at r=b (closest point to the

center )

WW 2 VbLb2 bLc max 2gg 22 WL LWLL LL 29gg9 9 10

wL2 V 3.64 max g

W113232W Mbmax Lbrb LbbL c b L c b g32g 33 22 WL11L LLWLLL LrLLL g39929 9g9 1010

wL3a M 3.72 max g

19 Problem 3-10:

A simple beam AB supports two connective loads P and 2P that are d apart (see figure). The wheels may be placed at any distance x from the left support of the beam.

(a) Determine the distance x that will produce maximum shear force in the beam, and also

determine the maximum shear force Vmax

(b) Determine the distance that will produce the maximum bending moment, and also draw the corresponding bending moment diagram. (Assume P=10kN, d=2.4m, and L=12m)

C D

Problem 3-10 Solution: a) Free body diagram:

C D

RAB R are reaction forces.

Moment balance at point A:

M A 0

20 RB LPx2P xd 32Px Pd 32Px Pd R B LL

Force balance in y direction

FY 0

RAB 3PR

23dx RPA 2 LL

Shear force distribution

C D

32Px Pd 23dx There are two possible maximum shear forces, and P 2 , depending on LL LL the of the wheels x:

22dd If x 0, the two possible maximum shear forces are VPmax max 3 , LL

dd If xLd, the two possible maximum shear forces are Vmax Pmax 3 , LL

In the range of 0 d L , by observation, the maximum shear force occurs when x L d , at the left hand side of point B.

d 2.4 VPmax 31 03 28kN at x L d Note :0 x L d L 12

21 b) From the shear force diagram, we know the bending moment diagram could be either i orii

The maximum bending moment occurs at either point C or D. Thus we need only calculate moment at these two points to determine the maximum bending moment.

23dx 32Px Pd Knowing RPA 2 and RB , LL LL

23dx2 MRCAxP3 x LL

MRDAxdPd 23d dx 3x2 2d Px33d Pd LLLL 25dd3x2 Pd33xPd LLL

Calculate maximum M C and M D

d If xM0, CD0, M2Pd1 L

d If xLd, MCDPd1 ,M0 L

By observation, the maximum moment occurs at x=0, where

d 2.4 M D 21Pd 2102.4138.4Nm L 12 MN38.4 m max

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2.080J / 1.573J Structural Mechanics Fall 2013

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