SECTION-6

SECTION 6 COMBINED STRESSES

ECCENTRIC LOADING (NORMAL STRESSES)

DESIGN PROBLEM

361.It is necessary to shape a certain link as shown in order to prevent interference with another part of the machine. It is to support a steady tensile load of 2500 lb. with a design factor of 2 based on the yield strength. The bottom edge of the midsection is displaced upward a distance a = 2 in. above the line of action of the load. For AISI C1022, as rolled, and h 3b, what should be h and b?

Solution:

For AISI C1022, as rolled, (Table AT7).

By trial and error method:

362. A tensile load on a link as described in 361 varies from 0 to 3000 lb.; it is machined from AISI 1045, as rolled, and the lower edge of the link is a = 0.5 in. above the center line of the pins; h 3b. Determine the dimensions of the link for N = 2 based on the Soderberg line.

Solution:

Soderberg Line:

For AISI 1045, as rolled (Table AT 7).

Size factor = 0.85

Load factor (axial) = 0.80


Use b x h = 9/16 in x 1 5/8 in363.The same as 362, except that the load continuously reverses, 3 kips to -3 kips.

Solution:


Use b x h = 5/8 in x 1 13/16 in364.A circular column (See Fig. 8.3, Text), the material of which is SAE 1020, as rolled, is to have a length of 9 ft. and support an eccentric load of 16 kips at a distance of 3 in. from the center line. Let N = 3. (a) What should be the outside diameter Do if the column is hollow and Di = 0.75Do? (b) What should be the diameter if the column is solid?

Solution:

a. Try J.B. Johnson

For SAE 1020, as rolled,

Transition point

By trial and error method

Therefore use Eulers equation

To check:

Use Do = 3 13/16 in, Di = 2 13/16 inb. For solid, also using Eulers equation.

By trial and error method.

Use D = 3 in.365. The same as 364, except that the length is 15 ft.Solution:

Eulers Equation:

a.

To check:

Use Do = 3 5/8 in, Di = 2 5/8 inb.


use 3 3/16 iin

Use D = 3 3/16 in.

366. A link similar to one shown is to be designed for: steady load F = 8 kips, L = 20 in. = 30o; aluminum alloy 2024-T4; N = 2.6 on the yield strength. It seems desirable for the dimension b not to exceed 1 3/8 in. Determine b and h and check their proportions for reasonableness. The support is made so that the pin at B carries the entire horizontal component of F.

Solution:

Aluminum alloy (2024-T4),

Therefore, use ,

367.A column 15 ft. long is to support a load F2 = 50,000 lb. Acting at a distance of e = 8 in. from the axis of the column as shown (with F1 = 0). Select a suitable I-beam for a design factor of 3 based on yield strength. The upper end of the column is free. See handbook for the properties of rolled sections.

Solution:

Use C1020, structural steel, sy = 48 ksi

Secant Formula

From Strength of Materials, 3rd Edition by F.S. Singer and A. Pytel, Table B-2, pg. 640, select Wide-Flange Sections by trial and error.

Then selecting W360 x 51, properties are

A = 6450 mm2 = 10 in2Depth = 355 mm = 14 in

k = 148 mm = 5.83 in

I = 1.41 x 108 mm4 = 338.8 in4Substitute,

Therefore suitable wide flange I-beam is W14 x 34 lb. (English units)A = 10 in2Depth = 14 in

k = 5.83 in

I = 338.8 in4368. The same as 367, except that F1 = 50,000 lb.

Solution:

Use C1020, structural steel, sy = 48 ksi

Transition Point

Check J.B. Johnson Formula

From Strength of Materials, 3rd Edition by F.S. Singer and A. Pytel, Table B-2, pg. 640, select Wide-Flange Sections by trial and error.

Then selecting W310 x 21, properties are

A = 2690 mm2 = 4.17 in2k = 117 mm = 4.61 in

Substitute,

Check for validity of JB Johnson Formula

Therefore, JB Johnson formula is valid and suitable wide flange I-beam is W12 x 14 lb. (English units)

A = 4.17 in2k = 4.61 in

CHECK PROBLEMS

369.A cam press, similar to that of Fig. 19-1, Text, exerts a force of 10 kips at a distance of 7 in. from the inside edge of the plates that make up the frame. If these plates are 1 in. thick and the horizontal section has a depth of 6 in., what will be the maximum stress in this section?Solution:

370. A manufacturer decides to market a line of aluminum alloy (6061-T6) C-clamps, (see Fig. 8.4, Text). One frame has a T-section with the following dimensions (letters as in Table AT 1): H = 1 1/16 , B = 17/32, a = 1/8, and t = 1/8. The center line of the screw is 2 3/8 in. from the inside face of the frame. (a) For N = 3 on the yield strength, what is the capacity of the clamp (gripping force)? (b) Above what approximate load will a permanent deformation of the clamp occur?

Solution:

AA 6061-T6, (Table AT 3)

See Fig. 8.4, (C-clamp) T-section

From Table AT 1 (T-section)

For :

a.

b.

371. A C-frame (Fig. 8.5 Text) of a hand-screw press is made of annealed cast steel, ASTM A27-58 and has a section similar to that shown. The force F acts normal to the plane of the section at a distance of 12 in. from the inside face. The various dimensions of the sections are: a = 3 in., b = 6 in., h = 5 in., d = e= f = 1 in. Determine the force F for N = 6 based on the ultimate strength.

Solution:

For :

For ASTM A27-58 Annealed Cast Steel

372. In the link shown (366), let b = in., h = 2 in., d = 2 in., L = 18 in., and = 60o. The clearance at the pins A and B are such that B resists the entire horizontal component of F; material is AISI C1020, as rolled. What may be the value of F for N = 3 based on the yield strength?

Solution:

Refer to Prob. 366.

For AISI C1020, as rolled, .

373. The link shown is subjected to a steady load F1 = 2.1 kips; b = 0.5 in., h = a = d = 2 in., L = 18 in.; material AISI 1040, cold drawn (10% work). The dimensions are such that all of the horizontal reaction from F2 occurs at A; and F2 varies from 0 to a maximum, acting towards the right. For N = 1.5 based on the Soderberg line, what is the maximum value of F2? Assume that the stress concentration at the holes can be neglected.

Solution:

For :

Let

For SAE AISI 1040 Cold Drawn (10% Work)

376. A free-end column as shown, L = 12 ft. long, is made of 10-in. pipe, schedule 40, (Do = 10.75 in., Di = 10.02 in., k = 3.67 in., Am = 11.908 in2., I = 160.7 in4., Z = 29.9 in3.). The load completely reverses and e = 15 in.; N = 3; material is similar to AISI C1015, as rolled. (a) Using the equivalent-stress approach, compute the safe (static) load as a column only. (b) Judging the varying loading by the Soderberg criterion, compute the safe maximum load. (c) Determine the safe load from the secant formula. (d) Specify what you consider to be a reasonable safe loading.

Solution:

For AISI C1015, as rolled.

a. As a column only (static)

Use J.B. Johnson Formula:

b. Varying load: ,

Axial load factor = 0.80

c. Secant Formula

d. 6740 lbs.

377. A bracket is attached as shown (367) onto a 14-in. x 193-lb., wide flange I-beam (A = 56.73 sq. in., depth = 15.5 in., flange width = 15.710 in., Imax = 2402.4 in4., Imin = 930.1 in4., kmin = 4.05 in.). The member is an eccentrically loaded column, 40 ft. long, with no central load (F1 = 0) and no restraint at the top. For e = 12 in. and N = 4, what may be the value of F2?

Solution:

Using secant formula:

For C1020, as rolled, structural steel,

378. A 14-in. x 193-lb., wide flange I-beam is used as a column with one end free (A = 56.73 sq. in., depth = 15.5 in., Imax = 2402.4 in.4, Imin = 930.1 in.4, kmin = 4.05 in., length L = 40 ft.). If a load F2 is supported as shown on a bracket at an eccentricity e = 4 in. (with F1 = 0), what may be its value for a design factor of 4? Flange width = 15.71 in.

Solution:

Using secant formula:

379. The same as 378, except that F1 = 0.5 F2.

Solution:

By Trial and error:

380. The cast-steel link (SAE 080) shown (solid lines) is subjected to a steady axial tensile load and was originally made with a rectangular cross section, h = 2 in., b = in., but was found to be too weak. Someone decided to strengthen it by using a T-section (dotted addition), with h and b as given above. (a) Will this change increase the strength? Explain. (b) What tensile load could each link carry with N = 3 based on yield?

Solution:

For SAE 080,

(a) This change will not increase the strength because of increased bending action that tends to add additional stress.

(b)

Rectangular cross section:

T-section:

,

COPLANAR SHEAR STRESSES

381. The figure shows a plate riveted to a vertical surface by 5 rivets. The material of the plate and rivets is SAE 1020, as rolled. The load F = 5000 lb., b = 3 in., = 0, and c = 5 in.; let a = 3D. Determine the diameter D of the rivets and the thickness of plate for a design factor of 3 based in yield strengths.

Solution:

For SAE 1020, as rolled.


For thickness of plate, .

382. The same as 383, except that = 30o.

Solution:


Say (same as 381).

For .

383. Design a riveted connection, similar to that shown, to support a steady vertical load of F = 1500 lb. when L = 18 in. and = 0o. Let the maximum spacing of the rivets, horizontally and vertically, be 6D, where D is the diameter of the rivet; SAE 1020, as rolled, is used for all parts; N = 2.5 based on yield. The assembly will be such that there is virtually no twisting of the channel. The dimensions to determine at this time are: rivet diameter and minimum thickness of the plate.

Solution:

,

For SAE 1020, as rolled, .


Solution:

,

say

say .

385. The plate shown (381) is made of SAE 1020 steel, as rolled, and held in place by five in. rivets that are made of SAE 1022 steel, as rolled. The thickness of the plate is in., a = 2 in., c = 6 in., b = 4 in., and = 0. Find the value of F for a design factor of 5 based on the ultimate strength.

Solution:

Plate, SAE 1020, as rolled (Table AT 7)

Rivets, SAE 1022, as rolled (Table AT 7)

use

From 381.


Solution:

387. The plate shown is made of AISI 1020 steel, as rolled, and is fastened to an I-beam (AISI 1020, as rolled) by three rivets that are made of a steel equivalent to AISI C1015, cold drawn. The thickness of the plate and of the flanges of the I-beam is in., the diameter of the rivets is in., a = 8.5 in., b = 11.5 in. and c = 4.5 in., d = 4 in. For F2 = 0, calculate the value of F1 for N = 2.5 based on yield strength.

Solution:

Plate, AISI 1020 Steel, as rolled,

Rivet, AISI C1015, cols drawn,

Use

,

.

388. The same as 387, except that F1 = 0, and the value of F2 is calculated.

Solution:

NORMAL STRESSES WITH SHEAR

DESIGN PROBLEMS

389. The bracket shown is held in place by three bolts as shown. Let a = 5 in., = 30o, F = 1500 lb.; bolt material is equivalent to C1022, as rolled. (a) Compute the size of the bolts by equation (5.1), Text. (b) Assuming that the connecting parts are virtually rigid and that the initial stress in the bolts is about 0.7sy, compute the factor of safety by (i) the maximum shear stress theory, (ii) the octahedral shear theory. (c) Compute the maximum normal stress.

Solution:

(a) Eq. 5-1, ,

For C1022, as rolled,

Select

Say , UNC,

(b)

(i) Maximum shear theory

(ii) Octahedral shear theory

(c) Maximum normal stress = 39,764 psi.

390. For the mounted bracket shown, determine the rivet diameter (all same size) for N = 3, the design being for the external loading (initial stress ignored); F = 2.3 kips, = 0, c = 17 in., a = 1 in., b = 14 in.; rivet material is AISI 1015, as rolled. Compute for (a) the maximum shear theory, (b) the maximum normal stress theory, (c) the octahedral shear theory.

Solution:

For AISI 4015, as rolled.

(a) Maximum shear theory

say

(b)

say

(c)

say

392. The same as 390, except that the two top rivets are 2 in. long and the bottom rivet is 1 in. long.

Solution:

For AISI 4015, as rolled.

(b) Maximum shear theory

say

(b)

say

(c)

say

393. The same as 390, except that the load is applied vertically at B instead of at A; let AB = 8 in. The two top rivets are 12 in. apart.

Solution:

(a)

From Problem 390.

say

(b)

say

(c)

say

394. The bracket shown is made of SAE 1020, as rolled, and the rivets are SAE 1015, cold drawn. The force F = 20 kips, L = 7 in., and = 60o. Let the design factor (on yield) be 2. (a) Determine the thickness t of the arm. (b) Compute the rivet diameter by both maximum shear and octahedral shear theories and specify a standard size. (c) Decide upon a proper spacing of rivets and sketch the bracket approximately to scale. Is some adjustment of dimensions desirable? Give suggestions, if any. (No additional calculations unless your instructor asks for a complete design.)

Solution:

Bracket: SAE 1020, as rolled,

Rivets: SAE 1015, cold drawn,

(a) Bracket.

say

(b)

Max. shear:

say

Octahedral shear,

say

(c) Spacing

,

use adjust to 2 in

Adjust spacing to 2 in from 7/8 in as shown.CHECK PROBLEMS

396. (a) If the rivets supporting the brackets of 390 are 5/8 in. in diameter, = 0, c = 14 in. a = 2 in., and b = 18 in., what are the maximum tensile and shear stresses in the rivets induced by a load of F = 10 kips. (b) For rivets of naval brass, hard, compute the factor of safety by maximum shear and octahedral shear theories (initial tension ignored).

Solution:

(a)

(b) Naval Brass, hard, .

Max. shear theory;

Octahedral shear theory;

397. The same as 396, except that the two top rivets are in. in diameter and the bottom one is in. in diameter.

Solution:

(a)

(b)

Max. shear theory;


398. What static load F may be supported by the -in. rivets shown, made of cold-finished C1015, with N = 3; = 0, a = 1 , b = 9, c = 14, f = 9, g = 12 in.? Count on no help from friction and ignore the initial tension. Check by both maximum shear and octahedral shear theories.

Solution:

For cold-finished, C1015, .

Max. shear theory;


399.The 2-in., UNC cap screw shown has been subjected to a tightening torque of 20 in-kips. The force F = 12 kips, = 60o, and Q = 0; L = 24 in., a = 20 in., b = 15 in.; screw material is AISI C1137 as rolled. (a) What is the approximate initial tightening load? (b) What is the increase in this load caused by the external force F if the bar is 8 in. wide and 2 in. thick and the unthreaded shank of the screw is 2 in. long? (See 5.9, Text.) (c) What are the maximum tensile and shear stresses in the bolt? (d) Compute the factor of safety from maximum normal stress, maximum shear, and octahedral shear theories.

Solution:

(a)

(b)

For 2-UNC

Width across flat = 3 in.

(c)

(d) For C1137, as rolled,

Maximum Shear:

Octahedral shear,

400. The plate shown is attached by three -in., UNC cap screws that are made of ASTM A325, heat-treated bolt material; L = 26 in., a = 6 in., b = 4 in., = 0. The shear on the screws is across the threads and they have been tightened to an initial tension of 0.6sp (sp = proof stress, 5.8, Text). Which screw is subjected to (a) the largest force, (b) the largest stress? What safe static load can be supported by the screws for N = 1.5 based on the Hencky-Mises criterion?

Solution:

For ASTM A325, Heat-Treated, -in. UNC

(a) Largest force, at A

(b) Largest stress, at A

For in UNC,

By Hencky-Mises Criterion.

401. The same as 400, except that the cap screw A is in. in diameter.

Solution:

(3/4 in UNC)

(1/2 in UNC)

For in UNC.

For in. UNC,

(a) Max. force, at in.

(b) Max. stress, at in.

NORMAL STRESSES WITH TORSION

DESIGN PROBLEMS

402. A section of a machined shaft is subjected to a maximum bending moment of 70,000 in-lb., a torque of 50,000 in-lb., and an end thrust of 25,000 lb. The unsupported length is 3 ft. and the material is AISI C1030, normalized. Since the computations are to be as though the stresses were steady, use N = 3.3. Compute the diameter from both the maximum-shear and the octahedral-shear theories and specify a standard size.Solution:

For AISI C1030, normalized,

Maximum shear

By trial and error,

Octahedral Shear

By trial and error,

use standard

403. The same as 402, except that the unsupported length is 15 ft. Do not overlook the moment due to the weight of the shaft, which acts in the same sense as the given bending moment.Solution:

Maximum shear

By trial and error,

Octahedral Shear

By trial and error,

use standard

404. A shaft is to be made in two sections, I and II, of diameters D1 and D2, somewhat as shown, machined from AISI 1045, annealed. It is expected that a = 8 in., b = 24 in., L = 20 in., and the load Q = 2 kips, so seldom repeated that the design is for steady load. The factor of safety is to be 2.2 on the basis of the octahedral-shear theory and closely the same in each section. The ends A and B are restrained from twisting, but they are designed to support the balancing reactions from Q without other moments. Decide upon standard size for D1 and D2.

Solution:

For AISI 1045, annealed, ,

Octahedral Shear

By trial and error,

Trial

Actual

12.2681.8201.25

1.252.3721.7301.37

1.372.3981.7141.40

1.402.4031.7121.40

Use standard ,

But , use

say

405. The shaft shown overhangs a bearing on the right and has the following dimensions: a = 5 in., b = in., and e = 10 in. The material is AISI C1040, annealed. This shaft is subjected to a torque T = 10,000 in-lb., forces F1 = 10,000 lb., and F2 = 20,000 lb. Using a static-design approach, determine the diameter D for N = 2.5, with computations from the maximum-shear and octahedral-shear theories.

Solution:

Bending due to and load

Bending due to

Tension,

since

For AISI C1040, annealed (Fig. AF 1)

Maximum shear,

Octahedral Shear

use

406. The same as 405, except that F2 = 0.

Solution

,

Maximum shear,

Octahedral Shear

use

CHECK PROBLEMS

407. The shaft shown overhangs a bearing at the right and has the following dimensions: D = 2 in., a = 4 in., b = in., c = 2 in., d = 6 in., e = 8 in., r = in. This shaft is subjected to a torque T = 8000 in-lb. and forces F1 = 8000 lb., and F2 = 16,000 lb. Determine the maximum-shear and normal stresses, and the octahedral-shear stress: (a) at points A and B ( = 45o), (b) at points M and N, (c) at point G.

Solution:

Bending due to :

Tension

Bending due to and weight of beam

from the free end.

(A)

at A

Max. Shear

Max. Normal

Octahedral Shear Stress

At B:

Max. Shear

Max. Normal


(B)

at M

Max. Shear

Max. Normal


At N:

Max. Shear

Max. Normal


(C) At G.

Max. Shear

Max. Normal


411. A 4-in. shaft carries an axial thrust of 20 kips. The maximum bending moment is 2/3 of the twisting moment; material is AISI 8630, WQT 1100 F, and N =3. Use the steady stress approach and compute the horsepower that may be transmitted at 2000 rpm?

Solution:

For AISI 8630, WQT 1100oF,

By maximum shear.

412. The same as 411, except that the shaft is hollow with an inside diameter of 2 in.

Solution:

413. A hollow, alloyed-steel shaft, AISI 4130, OQT 1100 F, has an OD of 3 in. and an ID of 2 in. It is transmitting 1500 hp at 1200 rpm, and at the same time is withstanding a maximum bending moment of 40,000 in-lb. and an axial compressive force F = 10 kips. The length of the shaft between bearings is 10 ft. Using a steady stress approach, determine (a) the maximum shearing stress in the shaft, (b) the maximum normal stress, (c) the factor of safety in each case. (d) Also compute N from the octahedral shear theory.

Solution:

For alloy-steel shaft, AISI 4130, WQT 1100oF, (Table AT 7)

(Table AT 8)

(A)

(B)

(C)

(D)

VARYING STRESSES COMBINED

DESIGN PROBLEMS

414. The force F on the lever in the illustration (in the plane of the lever) varies from a maximum of 424.2 lb. to a minimum of -141.4 lb.; L = 20 in., a = 15 in., D2 = 1.2D1, r = 0.125D1, = 45o; the material is cold-drawn SAE 1040, 10% worked, the design factor N = 1.5. Compute the diameter D1 using the Soderberg-line approach with both the maximum-shear and octahedral-shear theories; indefinite life.

Solution:

For SAE 040, 10% Worked

Strength Reduction Factors

With

Fig. AF 12

Assume

Forces:

Maximum shear,

Octahedral Shear.

say

say

417. A hollow steel shaft, SAE 1045, as rolled, has an inside diameter of one half of the outside diameter and is transmitting 1600 hp at 600 rpm. The maximum bending moment is 40,000in-lb. Determine the diameter for N = 3 by both the maximum-shear and octahedral shear theories. Specify a standard size. Use the Soderberg line for obtaining the equivalent stresses.Solution:

For SAE 1045, as rolled, ,

Assume

For bending:

For torsion:

Maximum shear,

say ,

Octahedral Shear.

say ,

Standard Size ,

418. A section of a shaft without a keyway is subjected to a bending moment that varies sinusoidally from 30 to 15 then to 30 in-kips during two revolutions, and to a torque that varies similarly and in phase from 25 to 15 to 25 in-kips; there is also a constant axial force of 40 kips; the material is AISI 2340, OQT 1000 F; N = 1.5. Determine the diameter by (a) the maximum-shear-stress theory; (b) the octahedral-shear-stress theory.

Solution:

For AISI 2340, OQT 1000oF. ,

Assume

(a) Maximum shear,

say ,

Octahedral Shear.

say

419. The same as 418, except that the shaft has a profile keyway at the point of maximum moment.Solution:

(a) Maximum shear,

say

(b) Octahedral Shear.

say

CHECK PROBLEMS

420. A 2-in. shaft made from AISI 1144, elevated temperature drawn, transmits 200 hp at 600 rpm. In addition to the data on the figure, the reactions are B = 4.62 kips and E = 1.68 kips. Compute the factor of safety by the maximum-shear and octahedral-shear theories.

Solution:

For AISI 1144, Elevated Temperature, drawn, ,

,

Table AT 13

,

,

,

Maximum shear,

Octahedral Shear.

421. In the figure (399), the bar supports a static load Q = 3000 lb. acting down; L = 16 in., a = 12 in., b = 7 in. The force F = 2500 lb. is produced by a rotating unbalanced weight and is therefore repeated and reversed in both the horizontal and the vertical directions. The 1-in. cap screw, with cut UNC threads, is made of AISI C1137, annealed, and it has been subjected to a tightening torque of 4600 in-lb. The thickness of the bar is 2 in. (a) Compute the factor of safety for the load reversing in the vertical direction, and (b) in the horizontal direction (maximum-shear theory), with the conservative assumption that friction offers no resistance.

Solution:

For AISI C1137, annealed, ,

(Table AT 12)

For 1-in cap screws, UNC

Nut:

,

(a) Vertical, Moment at Edge = 0,

say , , Factor for tension = 0.80

(b) Horizontal:

Shear:

, maximum shear

422. The load Q, as seen (404), acts on the arm C and varies from 0 to 3 kips. The ends A and B of the shaft are restrained from turning through an angle but are supported to take the reactions A and B without other moments. The shaft is machined from AISI 1045, as rolled; D1 = 2, D2 = 2.5, L = 15, a = 10, b = 20 in. For calculation purposes, assume that the shaft size changes at the section of application of Q. Determine the factor of safety in accordance with the maximum-shear and octahedral-shear theories. Investigate both sections I and II. Would you judge the design to be 100% reliable?

Solution:

Use (1)

,

,

,

, (Figure AF 12)

Profile Keyway

,

Net

For AISI 1045, as rolled, ,

,

Maximum shear,

Octahedral Shear.

Not 100% reliable, N < 100.423. A rotating shaft overhangs a bearing, as seen in the illustration. A -in. hole is drilled at AB. The horizontal force F2 varies in phase with the shaft rotation from 0 to 5 kips, but its line of action does not move. A steady torque T = 8 in-kips is applied at the end of the shaft; D = 2, D2 = 2.5, a = 2, b = 5, e = 0.5, r = in. The material is AISI C1040, annealed. What steady vertical load F1 can be added as shown if the design factor is to be 2.5 from the octahedral-shear theory? Assume that the cycling of F2 may be such that the worst stress condition occurs at the hole.

Solution:

AISI C1040, annealed, (Fig. AF 1),

For hole:

,

(annealed)

-

At hole

Bending :

Octahedral Shear Theory

POWER SCREWS

424. Design a square-thread screw for a screw jack, similar to that shown, which is to raise and support a load of 5 tons. The maximum lift is to be 18 in. The material is AISI C1035, as rolled, and N 3.3 based on the yield strength.

Solution:

AISI C1035, as rolled,

say 1 in,

Transition:

Use column formula, Eulers

use 1 in,

425. (a) For the screw of 424, what length of threads h will be needed for a bearing pressure of 1800 psi? (b) Complete the design of the jack. Let the base be cast iron and the threads integral with the base. Devise a method of turning the screw with a round steel rod as a lever and fix the details of a nonrotating cap on which the load rests. (c) What should be the diameter of the rod used to turn the screw? If a man exerts a pull of 150 lb. at the end, how long must the rod be?

Solution:

(a)

Th/in = 2.5

say

(b) Assume ASTM 20. , ,

Use proportions from figure based on diameter.

Method: Manual, normal pull.

(c) (Based on proportion)

426.A screw jack, with a 1 -in. square thread, supports a load of 6000 lb. The material of the screw is AISI C1022, as rolled, and the coefficient of friction for the threads is about 0.15. The maximum extension of the screw from the base is 15 in. (a) Considering the ends of the screw restrained so that Le = L, find the equivalent stress and the design factor. (b) If the load on the jack is such that it may sway, the screw probably acts as a column with one end free and the other fixe. What is the equivalent stress and the factor of safety in this instance? (c) What force must be exerted at the end of a 20-in. lever to raise the load? (d) Find the number of threads and the length h of the threaded portion in the cast-iron base for a pressure of 500 psi on the threads. (e) What torque is necessary to lower the load?

Solution:

From Table AT 7,AISI C1022, as rolled, sy = 52 ksi

F = 6000 lb = 6 kips

For 1 in square thread, Dr = 1.0 in, Th/in. = 3.5 f = 0.15

(a) With Le = L = 15 in

Transition for AISI C1020;


Equivalent stress

Design factor

(b) With Le = 2L = 30 in

Transition for AISI C1020;


Equivalent stress

Design factor

not safe(c) For force exerted at the end of 20-in. lever to raise the load = Fa

(d) Let p = pressure = 500 psi, W = 6000 lb, Do = 1.25 in, Di = 1.00 in.Nt = number of threads, h = length of threaded portion.

Then

(e) Torque necessary to lower the load.

427.A square-thread screw, 2 in. in diameter, is used to exert a force of 24,000 lb. in a shaft-straightening press. The maximum unsupported length of the screw is 16 in. and the material is AISI C1040, annealed. (a) What is the equivalent compressive stress in the screw? Is this a satisfactory value? (b) What torque is necessary to turn the screw against the load for f = 0.15? (c) What is the efficiency of the screw? (d) What torque is necessary to lower the load?Solution:

For 2 in. square thread screw, Do = 2 in, Dr = 1.612 in, Th/in. = 2.25 from Table 8.1

W = 24,000 lb = 24 kips, L = 16 in

(a) For unsupported length, Le = L = 16 in.

For AISI C1040, annealed, Figure AF-1, sy = 47.5 ksi

Transition,

Then

Use column formula, JB Johnson Formula,

satisfactory

(b) Torque to turn the screw against the load

(c) Torque necessary to lower the load.

428.(a) A jack with a 2-in., square-thread screw is supporting a load of 20 kips. A single thread is used and the coefficient of friction may be as low as 0.10 or as high as 0.15. Will this screw always be self-locking? What torque is necessary to raise the load? What torque is necessary to lower the load? (b) The same as (a) except that a double thread is used. (c) The same as (a) except that a triple thread is used.

Solution:

Table 8.1, 2 in. square thread, Do = 2 in, Dr = 1.612 in, Th/in = 2.25

(a) Self-locking? And Torque necessary to raise the load.

If f = 0.10

If f = 0.15

Since is always greater than , the screw is always self-locking.

Torque necessary to lower the load.

(b) Self-locking? And Torque necessary to raise the load.

If f = 0.10

If f = 0.15

Since is always less than , the screw is always not self-locking.

Torque necessary to lower the load = 0(c) Self-locking? And Torque necessary to raise the load.

If f = 0.10

If f = 0.15

Since is always less than , the screw is always not self-locking.

Torque necessary to lower the load = 0

429.The conditions for a self-locking screw are given in 8.23, Text. Assume that the coefficient of friction is equal to the tangent of the lead angle and show that the efficiency of a self-locking screw is always less than 50%.Solution:

For self-locking, > , then + > 2Then,

CURVED BEAMS

430.It is necessary to bend a certain link somewhat as shown in order to prevent interference with another part of the machine. It is estimated that sufficient clearance will be provided if the center line of the link is displaced e = 3 in. from the line of action of F, with a radius of curvature of R 5.5 in., L = 10 in., material is wrought aluminum alloy 2014 T6; N = 2 on the basis of the maximum shear stress; F = 2500 lb. with the number of repetitions not exceeding 106. (a) If the section is round, what should be its diameter D? (b) If the link is bend to form cold, will the residual stresses be helpful or damaging? Discuss.

Solution:

(a) Table AT3. Wrought aluminum alloy 2014 T6

At 106 cycles

With size factor.

Equation:

Using Trial and error and Table AT 18:

By trial and error D = 1.92 in

Table AT 18: Kc = 1.152

Use D = 2 in.

(b) Residual stress is helpful due to a decrease in total stress on tension side.431.The same as 430, except that the section is rectangular with h 3b; see figure.

Solution:

(a) Table AT3. Wrought aluminum alloy 2014 T6

At 106 cycles

With size factor.

Equation:

Using Trial and error and Table AT 18:

By trial and error b = 0.787 in

Table AT 18: Kc = 1.1736

Use b = 7/8 in. h = 3b = 2 5/8 in(b) Residual stress is helpful due to a decrease in total stress on tension side.

432.A hook is to be designed similar to that shown to support a maximum load F = 2500 lb. that will be repeated an indefinite number of times; the horizontal section is to be circular of radius c and the inside radius a is 1 in. (a) Determine the diameter of the horizontal section for N = 2 based on the Soderberg line, if the material is AISI 4130, WQT 1100 F. (b) Calculate the value of the static load that produces incipient yielding.

Solution:

(a) For AISI 4130, WQT 1100 F, Table AT 7

sy = 114 ksi, su = 127 ksi, sn = su/2 for reversed bending

Soderberg line:

repeated load

For curved beam

Table AT 18,

Substitute:

By trial and error: c = 0.633

,

Use c = 11/16 = 0.6875 inDiameter = 2c = 1.375 in = 1 3/8 in(b) Static load that produces incipient yielding.

sd = sy = 114 ksi

,

433. The same as 432, except that the hook is expected to be subjected to 100,000 repetitions of the maximum load.

Solution:

(a) For AISI 4130, WQT 1100 F, Table AT 7

sy = 114 ksi, su = 127 ksi, sn = su/2 for reversed bending

At 100,000 repetitions

Soderberg line:

repeated load

For curved beam

Table AT 18,

Substitute:


,

Use c =5/8Diameter = 2c = 1.25 in = 1 1/4 in

(b) Static load that produces incipient yielding.

sd = sy = 114 ksi

,

434.A hook, similar to that shown with a horizontal circular section of diameter 2c, is to be designed for a capacity of 2000 lb. maximum, a load that may be applied an indefinite number of times. A value of a = 2 in. should be satisfactory for the radius of curvature of the inside of the hook. Let N = 1.8 based on the modified Goodman line. At the outset of design, the engineer decided to try AISI C1040, OQT 1100 F. (a) Compute the diameter of the horizontal section, (b) If the 45o circular section is made the same diameter, what is its design factor (modified Goodman)? Could this section be made smaller or should it be larger?

Solution:(a) For AISI C1040, OQT 1100 F, Figure AF 1su = 100 ksi, sn = su/2 for reversed bending

sn = SF x sn = 0.85(0.5)(100) = 42.5 ksi

Kf = 1.0

Modified Goodman line:

repeated load

For curved beam

Table AT 18,

Substitute:


,

Use c = 11/16 inDiameter = 2c = 1.375 in = 1 3/8 in(b) sus = 0.6su = 0.6 x 100 ksi = 60 ksisns = 0.6sn = 0.6 x 42.5 ksi = 25.5 ksi

Equivalent stress (Modified Goodman Line)

(assuming constant diameter)

Table AT 18,

Then

Since N > 1.8, this section could be made smaller.435.A C-frame hand press is made of annealed cast steel (A27-58) and has a modified I-section, as shown. The dimensions of a 45o section CD are: a = 3, b = 6, h = 4, t = 1 in., radius r = 1 in.; also g = 12 in.; and the maximum force is F = 17 kips, repeated a relatively few times in the life of the press. (a) Applying the straight-beam formula to the 45o section, compute the maximum and minimum normal stresses. (b) Do the same, applying the curved-beam formula. (c) By what theory would you judge this section to have been designed? If the radius r were increased several times over, as it could have been done, would the stress have been materially reduced? Give reasons for your conclusions.

Solution:

(a) Straight-beam formula

Consider only normal stresses, relatively static.

Then

in tension

in compression

(b) Curved-beam formula

Using Table AT18

in tension

in compression

(c) This section must be designed based on straight beam formula. Maximum stress is higher.

Increasing the radius r.

Table A-18.

in tension

in compression

The stress is reduced using by increasing the radius r in Curved Beam Formula.

Reason: As the radius r increased the stress factor for curved beam decreases thence the maximum stress is reduced.436.A heavy C-clamp, similar to the figure, is made of normalized cast steel (A27-58) and has a T-section where t= 7/16 in.; q= 2 , a =1 in. What is the safe capacity if N = 2 based on yield?

Solution:

Table AT 1

Table AT 18

For Normalized cast steel, A27-58,

Moment of Inertia

437.The same as 436, except that the section is trapezoidal with b = in. (see figure). Ignore the effect of resounding off the corners.

Solution:

From other sources.

Table AT 18

For Normalized cast steel, A27-58,

THICK-SHELL CYLINDERS; INTERFERENCE FITS

438.Special welded steel pipe, equivalent in strength to SAE 1022, as rolled, is subjected to an internal pressure of 8000 psi. The internal diameter is to be 4 in. and the factor of safety is to be 3, including an allowance for the weld. (a) Find the thickness of the pipe according to the distortion-energy theory. (b) Using this thickness find the maximum normal and shear stresses and the corresponding safety factors. (c) Compute the thickness from the thin-shell formula and from the Barlow formula.Solution:

, ,

SAE 1022, as rolled,

(a) Distortion-Energy Theory

(b) Maximum normal stress

Maximum shear stress

(c) From thin-shell formula

From Barlow formula

439.The internal diameter of the cast-steel cylinder, SAE 0030, of a hydraulic press is 12 in. The internal working pressure is 6000 psi, N = 2.5. Find the thickness of the cylinder walls (a) from the maximum-shear-stress theory, (b) from the octahedral-shear theory. (c) Compute the thickness from the thin-shell and Barlow formulas. What do you recommend?Solution:

Table AT 6. SAE 0030 = A27-58, sy = 35 ksi


(b) Octahedral Sheat Theory

(c) Thin shell formula

Barlow formula

Recommended: Maximum shear theory , t = 9.8745 in thick.440.The same as 439, except a higher-strength material is selected. Try cast-steel SAE 0105.Solution:

Table AT 6. SAE 0105 = A148-58, sy = 85 ksi


(b) Octahedral Sheat Theory

(c) Thin shell formula

Barlow formula

Recommended: Maximum shear theory , t = 1.459 in thick.

441. A 2 in. heavy-wall pipe has the following dimensions: OD = 2.875, ID = 1.771, t = 0.552 in.; inside surface area per foot of length = 66.82 in.2, outside surface area per foot of length = 108.43 in.2. The material is chromium-molybdenum alloy, for which the permissible tangential tensile stress is 15 ksi at temperatures between 700 800 F. (a) Compute the maximum internal working pressure for this pipe from Lames formula, by the maximum-shear and octahedral-shear theories. (b) What is the stress at an external fiber? (c) A higher design stress would be permitted for an external pressure alone. Nevertheless, compute the external pressure corresponding to a maximum tangential stress of 15 ksi.Solution:

(a) Lames Equation

Maximum shear theory

Octahedral shear theory

(b) Stress at external fiber,

(c) External pressure alone.

442.A cast-steel hub is to be shrunk on a 1.5-in., SAE 1035, as-rolled, steel shaft. The equivalent diameter of the hub is 2.5 in., its length is 4 in. (a) What must be the interference of metal if the holding power of this fit is equal to the torsional yield strength of the shaft? Use Baughers recommendations. (b) What are the corresponding tangential and radial stresses in the hub?Solution:

Table AT 7, SAE 1035, as rolled, sy = 55 ksi.

sys = 0.6 sy = 33 ksi

Es = 30,000 ksi

s = 0.3

For hub, Cast steel, Eh = 30,000 ksi, h ~ 0.3

(a) Interference of metalFor solid shaft, same E and .

For pi:

But

Then

as per Baughers recommendation

Then

- answer.

(b) Tangential and radial stresses in the hub

Tangential stress

Radial stress

443.The same as 442, except that the hub is ASTM 20, cast iron. Will the resulting tensile stresses be safe for cast iron?

Solution:

Table AT 6, ASTM 20, cast iron, suc = 83 ksi, su = 20 ksi (hub)

Table AT 7, SAE 1035, as rolled, sy = 55 ksi.

sys = 0.6 sy = 33 ksi

(a) Interference of metal

For hub of cast iron and the shaft is steel.

For pi:

But

Then

as per Baughers recommendation

Then

- answer.

(b) Tangential and radial stresses in the hub

Tangential stress

> 20 ksi. Not safe for cast iron ASTM 20.Radial stress

444.A cast-steel gear is pressed onto a 2-in. shaft made of AISI 3140, OQT 1000 F. The equivalent hub diameter is 4 in., and the hub length is 4 in. (a) What are the maximum tangential and radial stresses in the hub caused by a class FN 2 interference fit? Compute for the apparent maximum value of i (but recall the probability of this event). (b) What axial force F in tons will be required to press the gear on the shaft if f1 is assumed to be 0.2? (c) What torque may the force fit safely transmit? (d) Is the holding capacity of this fit large enough to transmit a torque that produces a simple torsional stress of 0.6sys in the shaft?Solution:Cast steel, E = 30 x 106 psi, = 0.27 or approximately 0.3

AISI 3140, OQT 1000 F, E = 30 x 106 psi, = 0.3, sy = 133 ksi (Fig. AF 2).

Di = 2 in, Do = 4 in, L = 4 in.For Class FN 2 interference fit.

Table 3.2, page 85, 2 in diameter.

(a) Maximum value of i = 0.0027 0.0000 = 0.0027 in

(b) For same material and same Poissons ratio

Tangential stress

Radial stress

(c) Axial force F in tons.

(d) Torque safely transmit.

f = 0.1 as recommended by Baugher.

(e) With simple torsional stress of 0.6sys.

No. The holding capacity of this fit is not large enough to transmit a torque that produces a simple torsional stress of 0.6sys in the shaft.

445.The same as 444, except that a class FN 4 fit is investigated and the computation is made for the average i.

Solution:

Cast steel, E = 30 x 106 psi, = 0.27 or approximately 0.3

AISI 3140, OQT 1000 F, E = 30 x 106 psi, = 0.3, sy = 133 ksi (Fig. AF 2).

Di = 2 in, Do = 4 in, L = 4 in.

For Class FN 4 interference fit.

Table 3.2, page 85, 2 in diameter.

Maximum value of i = 0.0042 0.0000 = 0.0042 inMinimum value of i = 0.0035 0.0012 = 0.0023 in

Average value of i = 0.5 (0.0042 + 0.0023) = 0.00325 in(a) For same material and same Poissons ratio

Tangential stress

Radial stress

(b) Axial force F in tons.

(c) Torque safely transmit.

f = 0.1 as recommended by Baugher.

(d) With simple torsional stress of 0.6sys.

No. The holding capacity of this fit is not large enough to transmit a torque that produces a simple torsional stress of 0.6sys in the shaft.

446.A No. 217 ball bearing has a bore of 3.3465 in., a width of 1.1024 in., and the inner race is approximately 3/8 in. thick. This bearing is to be mounted on a solid shaft with i = 0.0014. (a) Calculate the maximum radial and tangential stresses in the race. (b) Estimate the force required to press the bearing onto the shaft.Solution:

Di = 3.3465 in, Do = 3.3465 + 2(3/8) = 4.0965 in, i = 0.0014 in.

(a) Maximum radial stress in the race

Tangential stress

(b) Force required to press the bearing onto the shaft

, use f1 = 0.175 on the average

447.A steel disk of diameter Do and thickness L = 4 in. is to be pressed onto a 2-in. steel shaft. The parts are manufactured with class FN 5 fit, but assembled parts are selected so as to give approximately the average interference. What will be the maximum radial and tangential stresses in the disk if (a) Do is infinitely large; (b) Do = 10 in.; (c) Do = 4 in.; (d) Do = 2.5 in.?

Solution:(a) Maximum radial stress if .

Maximum tangential stress if .

(b) Maximum radial stress if .


(c) Maximum radial stress if .


448.A steel cylinder is to have an inside diameter of 3 in. and pi = 30,000 psi. (a) Calculate the tangential stresses at the inner and outer surfaces if the outside diameter is 6 in. (b) It was decided to make the cylinder in two parts, the inner cylinder with D1 = 3 in. and Di = 4.5 in., the outer cylinder with Di = 4.5 in. and Do = 6 in. (see figure). The two cylinders were shrunk together with i = 0.003 in. Calculate the pressure at the interface and the tangential stresses at the inner and outer surfaces of each cylinder. (Suggestion: first derive an equation for the interface pressure).

Solution:

(a) Tangential stresses at the inner and outer surface.

Di = 3 in, ri = 1.5 in, pi = 30,000 psi

Do = 6 in, ro = 3 in, po = 0

(b) Pressure at the interface, tangential stresses at the inner and outer surface of each cylinder.

,

p1 = 30,000 psi, ro = 3 in, ri = 2.25 in, r1 = 1.5 in

Pressure at the interface, pi.

Tangential stresses:

Inner cylinder:

Inner surface:

Outer surface:

Outer cylinder:

Inner surface:

Outer surface:

449.A phosphor-bronze (B139C) bushing has an ID = in., an OD = 1 in., and a length of 2 in. It is to be pressed into a cast-steel cylinder that has an outside diameter of 2 in. An ASA class FN 2 fit is to be used with selective assembly to give approximately the interference i = 0.0016 in. Calculate (a) pi, (b) the maximum tangential stress in the steel cylinder, (c) the force required to press bushing into the cylinder, (d) the decrease of the inside diameter of the bushing.

Solution:Phosphor Bronze B139C, Es = 16 x 106 psi (Table AT3), s = 0.36 (other reference).

Cast steel, Eh = 30 x 106 psi , h = 0.27 (Table AT 6)

(a) pi

(b) Maximum tangential stress in the steel cylinder.

(c) F

, use f1 = 0.175 on the average

(d) Decrease of the inside diameter of the bushing. The bushing is phosphor bronze. Subscript is s as in shaft.

DESIGN PROJECTS

DESIGN PROJECTS

450. A jib crane similar to the one shown is to be designed for a capacity of F = ___ (say, 1 to 3 tons). The load F can be swung through 360o; L 10 ft., b 8.5 ft., c 2 ft. The moment on the jib is balanced by a couple QQ on the post, the forces Q acting at supporting bearings. The crane will be fastened to the floor by 6 equally spaced bolts on a D1 = 30-in. bolt circle; outside diameter of base D2 = 36 in. (a) Choose a pipe size (handbooks) for the column such that the maximum equivalent stress does not exceed 12 ksi. (b) Choose an I-beam for the jib such that the maximum stress does not exceed 12 ksi. (c) Compute the maximum external load on a base bolt and decide upon the size. (d) Complete other details as required by the instructor, such as: computing Q and choosing bearings (ball or roller?), the internal construction and assembly in this vicinity, detail sketches giving full information.

451.Design an air-operated punch press similar to the one shown. Let the force at the punch be 12 tons, (or other capacity as specified by the instructor), the depth of throat to the inside edge of the frame be 25 in., the diameter and stroke of the piston about 8 in. by 8 in., the mechanical advantage of the lever about 7, and the diameter of the punch 3/16 in. Determine first the horizontal section of the frame, and locate and design the cylinder. Then determine the relative arrangement of the various links and make a force analysis, from which the design of certain parts follows. Determine the actual distance of movement of the punch (not less than about 1 in.). The illustration will assist the student in settling upon the proportions of parts for which strength calculations cannot be made.

452.Design a screw press similar to that shown for a load of ___ (say, 3) tons on the screw. The depth of the throat g is to be ___ (10) in. and the height of the throat h is to be ___ (15) in. (The instructor will assign the data.) The order of procedure may be as follows: (a) Find the diameter of the screw. If Le/k > 40, check as a column. If the top of the screw is squared off for a handwheel or handle, check this section for twisting. The equation for pivot friction, if desired, is in 18.10, Text. (b) Decide upon the diameter of the handwheel or the length of handle (if one is needed to obtain the maximum pressure), assuming that the maximum force to be exerted by a man is about 150 lb. Dimensions of handwheels may be found in handbooks. The handle may be designed by the flexure formula. (c) Design the frame. The shape of the section of the frame will depend on the material used. A T-section is suitable for cast iron (say N = 6 on the ultimate strength), a hollow box or modified I-section is suitable for cast steel. The 45o section CD of the frame should be safe as a curved beam. See Table AT 18. In this connection, it will be well to make the radius r as large as practicable, since the larger r the less the stresses from a given load. Compute the dimensions of the vertical section. It is a good plan to keep t and a the same in all sections. (d) Design the bushing if one is used. The height b depends upon the number of threads in contact, which in turn depends upon the bearing pressure used in design. (Say half-hard yellow brass?) Compute the outside bushing diameter, the diameter and thickness of the collar, and decide upon dimensions to be used. (e) Fix the location and number of bolts to be used to fasten the frame to the base plate, and determine their size. Use a common bolt material. (f) Decide upon all other details of design. Make a separate sketch of each part of the machine showing thereon all dimensions necessary for manufacture. It is suggested that, first, all materials be tentatively decided upon, after which design stresses may be chosen. See that design stresses for the various parts bear a logical relation to one another. It is not necessary to follow this procedure in detail. It is likely that one will have to leave certain details unfinished from time to time, because these details depend on parts of the design not yet completed. Make sure that all parts can be assembled after they are made. Notice that the plate on the lower end of the screw must be connected in such a manner that the screw may turn while the plate does not.

453.Design a jib crane, as suggested by the illustration, to lift a load of W of ___ tons. The maximum radius of swing is to be about ___ ft. (The instructor will assign data). Suggested procedure: (a) From catalogues, select a hoist to suit the purpose, giving reasons for your choice, and noting significant dimensions. Of course, in the end, the hoist trolley has to match the size of I-beam used. (b) Let the angle that the diagonal tension rod makes with the horizontal be about 20o to 25o, and decide upon the dimensions H and L. Note that the point G does not necessarily have to be at the extreme position of the load. As a matter of fact, some advantage may result from having G inside the outermost position of the load. Make the force analyses (including weight of hoist as part of load) for (1) the condition of maximum column action, (2) the condition of maximum bending moment on the beam, and (3) the condition for maximum force on the hinge B (to be used for the design of this hinge). (c) Find the size of I-beam such that the maximum stress for any position of the load falls within the limits of 12 and 15 ksi, usually by assuming a standard beam and checking the stress. According to the arrangement of parts, it may be necessary to design the connection at G between the rod and the beam first. With the details of this connection known and with the details of beam assumed, the location of point G, the point of application of the force T, can be determined. The bending moment of a section a minute distance to the right of G is W(x dx). A minute distance to the left of G, the bending moment is W(x + dx) Txe Tydx; that is, the moment changes suddenly at G by the amount Txe. (d) Determine the size of diagonal support, including details of connections. (e) Design the connections at each end of the diagonal and the hinge at C. Settle upon the details including the method of attaching the hinge to the vertical surface, which may be wide-flange beam. (f) Design the hinge at B and the connection to the I-beam; also the details of the method of attaching the hinge to the vertical surface. Where material is not specified, make your choice clear. There should be no doubt as to your design stresses or design factor. Show a neat large sketch, fully dimensioned, of each part separately. It is unlikely that too much detail will be shown.

End -

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

Page 1 of 2

_1397409937.unknown

_1397414171.unknown

_1397514512.unknown

_1397515991.unknown

_1403880845.unknown

_1417856448.unknown

_1418577374.unknown

_1418586078.unknown

_1418588136.unknown

_1418837331.unknown

_1418840548.unknown

_1419275143.unknown

_1419278905.unknown

_1419280354.unknown

_1419315259.unknown

_1419317021.unknown

_1419317390.unknown

_1419318266.unknown

_1419322342.unknown

_1419345229.unknown

_1419345247.unknown

_1419318326.unknown

_1419318221.unknown

_1419317339.unknown

_1419315315.unknown

_1419315663.unknown

_1419315675.unknown

_1419315404.unknown

_1419315287.unknown

_1419314889.unknown

_1419315186.unknown

_1419315231.unknown

_1419314965.unknown

_1419280560.unknown

_1419314875.unknown

_1419280450.unknown

_1419280559.unknown

_1419279659.unknown

_1419280071.unknown

_1419280216.unknown

_1419280326.unknown

_1419280132.unknown

_1419279949.unknown

_1419279964.unknown

_1419279821.unknown

_1419279258.unknown

_1419279516.unknown

_1419279365.unknown

_1419279413.unknown

_1419279280.unknown

_1419279054.unknown

_1419279158.unknown

_1419278976.unknown

_1419275487.unknown

_1419275983.unknown

_1419276022.unknown

_1419278723.unknown

_1419276005.unknown

_1419275577.unknown

_1419275935.unknown

_1419275351.unknown

_1419275392.unknown

_1419275406.unknown

_1419275472.unknown

_1419275386.unknown

_1419275282.unknown

_1419275334.unknown

_1419275152.unknown

_1418844413.unknown

_1419273659.unknown

_1419274366.unknown

_1419274975.unknown

_1419273996.unknown

_1419274064.unknown

_1419273679.unknown

_1419273686.unknown

_1419273696.unknown

_1419273667.unknown

_1419272201.unknown

_1419272629.unknown

_1419272936.unknown

_1419273021.unknown

_1419273640.unknown

_1419272960.unknown

_1419272908.unknown

_1419272553.unknown

_1419271795.unknown

_1419271837.unknown

_1419271710.unknown

_1418844426.unknown

_1419271670.unknown

_1418842915.unknown

_1418844087.unknown

_1418844228.unknown

_1418844323.unknown

_1418844391.unknown

_1418844245.unknown

_1418844178.unknown

_1418844194.unknown

_1418843454.unknown

_1418843631.unknown

_1418843661.unknown

_1418843720.unknown

_1418843509.unknown

_1418842930.unknown

_1418842962.unknown

_1418842767.unknown

_1418842841.unknown

_1418842881.unknown

_1418842791.unknown

_1418840948.unknown

_1418842681.unknown

_1418842760.unknown

_1418841000.unknown

_1418840636.unknown

_1418838265.unknown

_1418840069.unknown

_1418840284.unknown

_1418840467.unknown

_1418840504.unknown

_1418840346.unknown

_1418840180.unknown

_1418840230.unknown

_1418838411.unknown

_1418839915.unknown

_1418839943.unknown

_1418839975.unknown

_1418839873.unknown

_1418838318.unknown

_1418838329.unknown

_1418838304.unknown

_1418838020.unknown

_1418838244.unknown

_1418838251.unknown

_1418838213.unknown

_1418837592.unknown

_1418837612.unknown

_1418837535.unknown

_1418588421.unknown

_1418836566.unknown

_1418836860.unknown

_1418837075.unknown

_1418837259.unknown

_1418836915.unknown

_1418836666.unknown

_1418836675.unknown

_1418836856.unknown

_1418588612.unknown

_1418835862.unknown

_1418836464.unknown

_1418836541.unknown

_1418835619.unknown

_1418588520.unknown

_1418588578.unknown

_1418588477.unknown

_1418588280.unknown

_1418588377.unknown

_1418588408.unknown

_1418588314.unknown

_1418588238.unknown

_1418588257.unknown

_1418588202.unknown

_1418586591.unknown

_1418587951.unknown

_1418588053.unknown

_1418588090.unknown

_1418587993.unknown

_1418587912.unknown

_1418587923.unknown

_1418587882.unknown

_1418586230.unknown

_1418586490.unknown

_1418586521.unknown

_1418586412.unknown

_1418586125.unknown

_1418586147.unknown

_1418586101.unknown

_1418582994.unknown

_1418585493.unknown

_1418585854.unknown

_1418585964.unknown

_1418586051.unknown

_1418585959.unknown

_1418585692.unknown

_1418585798.unknown

_1418585575.unknown

_1418583477.unknown

_1418583576.unknown

_1418585447.unknown

_1418583122.unknown

_1418583320.unknown

_1418583374.unknown

_1418583044.unknown

_1418578280.unknown

_1418582605.unknown

_1418582918.unknown

_1418582952.unknown

_1418582852.unknown

_1418582130.unknown

_1418578289.unknown

_1418582014.unknown

_1418577916.unknown

_1418578018.unknown

_1418578068.unknown

_1418577962.unknown

_1418577756.unknown

_1418577900.unknown

_1418577720.unknown

_1418568439.unknown

_1418572072.unknown

_1418575245.unknown

_1418577170.unknown

_1418577328.unknown

_1418577204.unknown

_1418577304.unknown

_1418576960.unknown

_1418577148.unknown

_1418575308.unknown

_1418572378.unknown

_1418572524.unknown

_1418572554.unknown

_1418572866.unknown

_1418574010.unknown

_1418572641.unknown

_1418572546.unknown

_1418572424.unknown

_1418572446.unknown

_1418572401.unknown

_1418572130.unknown

_1418572233.unknown

_1418571369.unknown

_1418571409.unknown

_1418571427.unknown

_1418571434.unknown

_1418572053.unknown

_1418571416.unknown

_1418571390.unknown

_1418571401.unknown

_1418571382.unknown

_1418568889.unknown

_1418569189.unknown

_1418571352.unknown

_1418571360.unknown

_1418569370.unknown

_1418569467.unknown

_1418571343.unknown

_1418569414.unknown

_1418569302.unknown

_1418569100.unknown

_1418569160.unknown

_1418569052.unknown

_1418568825.unknown

_1418568865.unknown

_1418568807.unknown

_1418405087.unknown

_1418405775.unknown

_1418406077.unknown

_1418406128.unknown

_1418568367.unknown

_1418406087.unknown

_1418405968.unknown

_1418406020.unknown

_1418405838.unknown

_1418405471.unknown

_1418405680.unknown

_1418405715.unknown

_1418405616.unknown

_1418405199.unknown

_1418405254.unknown

_1418405109.unknown

_1418402257.unknown

_1418402396.unknown

_1418404716.unknown

_1418405057.unknown

_1418404897.unknown

_1418404643.unknown

_1418402319.unknown

_1418402342.unknown

_1418402279.unknown

_1418402186.unknown

_1418402214.unknown

_1418402230.unknown

_1418402199.unknown

_1418402097.unknown

_1418402173.unknown

_1418402074.unknown

_1403891399.unknown

_1417854220.unknown

_1417855129.unknown

_1417856150.unknown

_1417856227.unknown

_1417856309.unknown

_1417856325.unknown

_1417856289.unknown

_1417856177.unknown

_1417856191.unknown

_1417855566.unknown

_1417855731.unknown

_1417855757.unknown

_1417855995.unknown

_1417856058.unknown

_1417855808.unknown

_1417855754.unknown

_1417855720.unknown

_1417855207.unknown

_1417855218.unknown

_1417855157.unknown

_1417854542.unknown

_1417854765.unknown

_1417854817.unknown

_1417854968.unknown

_1417854779.unknown

_1417854727.unknown

_1417854601.unknown

_1417854686.unknown

_1417854301.unknown

_1417854391.unknown

_1417854426.unknown

_1417854491.unknown

_1417854408.unknown

_1417854339.unknown

_1417854270.unknown

_1414736137.unknown

_1417852349.unknown

_1417853439.unknown

_1417854176.unknown

_1417854185.unknown

_1417852923.unknown

_1417853437.unknown

_1417853438.unknown

_1417853125.unknown

_1417852382.unknown

_1417850042.unknown

_1417851437.unknown

_1417851980.unknown

_1417851421.unknown

_1414764406.unknown

_1417849486.unknown

_1417849540.unknown

_1414765042.unknown

_1414764650.unknown

_1414764225.unknown

_1414764337.unknown

_1414764190.unknown

_1414736223.unknown

_1414735887.unknown

_1414735968.unknown

_1414736057.unknown

_1414735916.unknown

_1414735944.unknown

_1414706774.unknown

_1414735848.unknown

_1414706466.unknown

_1414706597.unknown

_1414706773.unknown

_1414706526.unknown

_1412138127.unknown

_1403891515.unknown

_1403888421.unknown

_1403889699.unknown

_1403890137.unknown

_1403890760.unknown

_1403891251.unknown

_1403891357.unknown

_1403890784.unknown

_1403890709.unknown

_1403890727.unknown

_1403890652.unknown

_1403889867.unknown

_1403889986.unknown

_1403890003.unknown

_1403889884.unknown

_1403889722.unknown

_1403889409.unknown

_1403889423.unknown

_1403889617.unknown

_1403888588.unknown

_1403888579.unknown

_1403884602.unknown

_1403888229.unknown

_1403888364.unknown

_1403888384.unknown

_1403888374.unknown

_1403888284.unknown

_1403888317.unknown

_1403888255.unknown

_1403888243.unknown

_1403887786.unknown

_1403887903.unknown

_1403888039.unknown

_1403888055.unknown

_1403887928.unknown

_1403887858.unknown

_1403886749.unknown

_1403887741.unknown

_1403886533.unknown

_1403883727.unknown

_1403884239.unknown

_1403884539.unknown

_1403884263.unknown

_1403884524.unknown

_1403884147.unknown

_1403884193.unknown

_1403883776.unknown

_1403883412.unknown

_1403883702.unknown

_1403882969.unknown

_1403883329.unknown

_1403883356.unknown

_1403883248.unknown

_1403882910.unknown

_1401878222.unknown

_1403847888.unknown

_1403871154.unknown

_1403879919.unknown

_1403880488.unknown

_1403880756.unknown

_1403880817.unknown

_1403880537.unknown

_1403880143.unknown

_1403880264.unknown

_1403880375.unknown

_1403880473.unknown

_1403880327.unknown

_1403880169.unknown

_1403880118.unknown

_1403871668.unknown

_1403879107.unknown

_1403879851.unknown

_1403879013.unknown

_1403871595.unknown

_1403871617.unknown

_1403871565.unknown

_1403848117.unknown

_1403870943.unknown

_1403871130.unknown

_1403871138.unknown

_1403871076.unknown

_1403848543.unknown

_1403870847.unknown

_1403848421.unknown

_1403847932.unknown

_1403847951.unknown

_1403847907.unknown

_1401878469.unknown

_1401878686.unknown

_1403683535.unknown

_1403683602.unknown

_1403683682.unknown

_1403684421.unknown

_1403847864.unknown

_1403847802.unknown

_1403684473.unknown

_1403684393.unknown

_1403684419.unknown

_1403684236.unknown

_1403683636.unknown

_1403683670.unknown

_1403683633.unknown

_1403683572.unknown

_1403683586.unknown

_1403683546.unknown

_1401892978.unknown

_1403683413.unknown

_1403683517.unknown

_1403683238.unknown

_1401880033.unknown

_1401880159.unknown

_1401887760.unknown

_1401880066.unknown

_1401879998.unknown

_1401878646.unknown

_1401878662.unknown

_1401878581.unknown

_1401878392.unknown

_1401878431.unknown

_1401878459.unknown

_1401878403.unknown

_1401878244.unknown

_1401878334.unknown

_1401878234.unknown

_1397638283.unknown

_1397638310.unknown

_1397638322.unknown

_1397638331.unknown

_1397638338.unknown

_1397638343.unknown

_1397638346.unknown

_1397638340.unknown

_1397638334.unknown

_1397638326.unknown

_1397638316.unknown

_1397638319.unknown

_1397638313.unknown

_1397638297.unknown

_1397638304.unknown

_1397638307.unknown

_1397638300.unknown

_1397638290.unknown

_1397638293.unknown

_1397638286.unknown

_1397638250.unknown

_1397638270.unknown

_1397638277.unknown

_1397638279.unknown

_1397638274.unknown

_1397638263.unknown

_1397638266.unknown

_1397638260.unknown

_1397515995.unknown

_1397515997.unknown

_1397515998.unknown

_1397515996.unknown

_1397515993.unknown

_1397515994.unknown

_1397515992.unknown

_1397515073.unknown

_1397515568.unknown

_1397515615.unknown

_1397515642.unknown

_1397515986.unknown

_1397515989.unknown

_1397515990.unknown

_1397515988.unknown

_1397515648.unknown

_1397515984.unknown

_1397515985.unknown

_1397515982.unknown

_1397515983.unknown

_1397515651.unknown

_1397515646.unknown

_1397515631.unknown

_1397515637.unknown

_1397515640.unknown

_1397515634.unknown

_1397515624.unknown

_1397515626.unknown

_1397515617.unknown

_1397515591.unknown

_1397515603.unknown

_1397515609.unknown

_1397515613.unknown

_1397515606.unknown

_1397515597.unknown

_1397515601.unknown

_1397515594.unknown

_1397515578.unknown

_1397515585.unknown

_1397515589.unknown

_1397515582.unknown

_1397515574.unknown

_1397515576.unknown

_1397515570.unknown

_1397515136.unknown

_1397515170.unknown

_1397515546.unknown

_1397515561.unknown

_1397515565.unknown

_1397515557.unknown

_1397515177.unknown

_1397515543.unknown

_1397515172.unknown

_1397515155.unknown

_1397515163.unknown

_1397515167.unknown

_1397515156.unknown

_1397515145.unknown

_1397515151.unknown

_1397515138.unknown

_1397515098.unknown

_1397515112.unknown

_1397515131.unknown

_1397515134.unknown

_1397515117.unknown

_1397515121.unknown

_1397515106.unknown

_1397515109.unknown

_1397515103.unknown

_1397515084.unknown

_1397515093.unknown

_1397515094.unknown

_1397515089.unknown

_1397515078.unknown

_1397515081.unknown

_1397515076.unknown

_1397514613.unknown

_1397515026.unknown

_1397515049.unknown

_1397515060.unknown

_1397515067.unknown

_1397515069.unknown

_1397515064.unknown

_1397515055.unknown

_1397515058.unknown

_1397515052.unknown

_1397515038.unknown

_1397515044.unknown

_1397515046.unknown

_1397515041.unknown

_1397515033.unknown

_1397515035.unknown

_1397515030.unknown

_1397515002.unknown

_1397515014.unknown

_1397515019.unknown

_1397515023.unknown

_1397515016.unknown

_1397515008.unknown

_1397515011.unknown

_1397515005.unknown

_1397514991.unknown

_1397514996.unknown

_1397514999.unknown

_1397514994.unknown

_1397514985.unknown

_1397514988.unknown

_1397514980.unknown

_1397514562.unknown

_1397514584.unknown

_1397514597.unknown

_1397514608.unknown

_1397514610.unknown

_1397514604.unknown

_1397514592.unknown

_1397514595.unknown

_1397514589.unknown

_1397514573.unknown

_1397514579.unknown

_1397514582.unknown

_1397514576.unknown

_1397514567.unknown

_1397514569.unknown

_1397514565.unknown

_1397514536.unknown

_1397514550.unknown

_1397514557.unknown

_1397514559.unknown

_1397514553.unknown

_1397514541.unknown

_1397514544.unknown

_1397514539.unknown

_1397514524.unknown

_1397514529.unknown

_1397514533.unknown

_1397514527.unknown

_1397514519.unknown

_1397514522.unknown

_1397514516.unknown

_1397499053.unknown

_1397499843.unknown

_1397500253.unknown

_1397514465.unknown

_1397514487.unknown

_1397514502.unknown

_1397514507.unknown

_1397514510.unknown

_1397514504.unknown

_1397514495.unknown

_1397514498.unknown

_1397514492.unknown

_1397514478.unknown

_1397514483.unknown

_1397514485.unknown

_1397514480.unknown

_1397514471.unknown

_1397514475.unknown

_1397514468.unknown

_1397500276.unknown

_1397514452.unknown

_1397514460.unknown

_1397514463.unknown

_1397514456.unknown

_1397500282.unknown

_1397514449.unknown

_1397500279.unknown

_1397500265.unknown

_1397500269.unknown

_1397500273.unknown

_1397500267.unknown

_1397500259.unknown

_1397500262.unknown

_1397500256.unknown

_1397500024.unknown

_1397500221.unknown

_1397500243.unknown

_1397500248.unknown

_1397500251.unknown

_1397500246.unknown

_1397500226.unknown

_1397500234.unknown

_1397500223.unknown

_1397500035.unknown

_1397500042.unknown

_1397500218.unknown

_1397500039.unknown

_1397500029.unknown

_1397500031.unknown

_1397500027.unknown

_1397499873.unknown

_1397500012.unknown

_1397500019.unknown

_1397500022.unknown

_1397500016.unknown

_1397499879.unknown

_1397500009.unknown

_1397499875.unknown

_1397499855.unknown

_1397499867.unknown

_1397499870.unknown

_1397499864.unknown

_1397499849.unknown

_1397499853.unknown

_1397499847.unknown

_1397499341.unknown

_1397499390.unknown

_1397499416.unknown

_1397499816.unknown

_1397499824.unknown

_1397499826.unknown

_1397499819.unknown

_1397499787.unknown

_1397499791.unknown

_1397499420.unknown

_1397499771.unknown

_1397499402.unknown

_1397499408.unknown

_1397499413.unknown

_1397499410.unknown

_1397499404.unknown

_1397499395.unknown

_1397499399.unknown

_1397499393.unknown

_1397499363.unknown

_1397499376.unknown

_1397499381.unknown

_1397499386.unknown

_1397499379.unknown

_1397499370.unknown

_1397499372.unknown

_1397499366.unknown

_1397499352.unknown

_1397499358.unknown

_1397499361.unknown

_1397499355.unknown

_1397499347.unknown

_1397499349.unknown

_1397499344.unknown

_1397499102.unknown

_1397499128.unknown

_1397499140.unknown

_1397499145.unknown

_1397499147.unknown

_1397499142.unknown

_1397499133.unknown

_1397499136.unknown

_1397499130.unknown

_1397499113.unknown

_1397499119.unknown

_1397499121.unknown

_1397499115.unknown

_1397499108.unknown

_1397499110.unknown

_1397499105.unknown

_1397499077.unknown

_1397499090.unknown

_1397499096.unknown

_1397499099.unknown

_1397499092.unknown

_1397499083.unknown

_1397499087.unknown

_1397499079.unknown

_1397499065.unknown

_1397499071.unknown

_1397499074.unknown

_1397499067.unknown

_1397499060.unknown

_1397499063.unknown

_1397499057.unknown

_1397415465.unknown

_1397415815.unknown

_1397498669.unknown

_1397499030.unknown

_1397499041.unknown

_1397499048.unknown

_1397499051.unknown

_1397499045.unknown

_1397499036.unknown

_1397499039.unknown

_1397499033.unknown

_1397498686.unknown

_1397499024.unknown

_1397499027.unknown

_1397498692.unknown

_1397498676.unknown

_1397498682.unknown

_1397498673.unknown

_1397416172.unknown

_1397416255.unknown

_1397416257.unknown

_1397416258.unknown

_1397416256.unknown

_1397416253.unknown

_1397416254.unknown

_1397416251.unknown

_1397416252.unknown

_1397416249.unknown

_1397416250.unknown

_1397416248.unknown

_1397416157.unknown

_1397416166.unknown

_1397416170.unknown

_1397416161.unknown

_1397416150.unknown

_1397416153.unknown

_1397416147.unknown

_1397415516.unknown

_1397415782.unknown

_1397415796.unknown

_1397415801.unknown

_1397415804.unknown

_1397415798.unknown

_1397415789.unknown

_1397415792.unknown

_1397415785.unknown

_1397415685.unknown

_1397415693.unknown

_1397415776.unknown

_1397415779.unknown

_1397415688.unknown

_1397415522.unknown

_1397415525.unknown

_1397415520.unknown

_1397415490.unknown

_1397415502.unknown

_1397415509.unknown

_1397415511.unknown

_1397415506.unknown

_1397415497.unknown

_1397415499.unknown

_1397415493.unknown

_1397415477.unknown

_1397415483.unknown

_1397415487.unknown

_1397415480.unknown

_1397415471.unknown

_1397415474.unknown

_1397415468.unknown

_1397414985.unknown

_1397415095.unknown

_1397415436.unknown

_1397415451.unknown

_1397415458.unknown

_1397415462.unknown

_1397415453.unknown

_1397415443.unknown

_1397415445.unknown

_1397415442.unknown

_1397415419.unknown

_1397415425.unknown

_1397415432.unknown

_1397415422.unknown

_1397415411.unknown

_1397415414.unknown

_1397415099.unknown

_1397415068.unknown

_1397415083.unknown

_1397415089.unknown

_1397415091.unknown

_1397415086.unknown

_1397415077.unknown

_1397415081.unknown

_1397415071.unknown

_1397414998.unknown

_1397415063.unknown

_1397415066.unknown

_1397415060.unknown

_1397414990.unknown

_1397414994.unknown

_1397414987.unknown

_1397414230.unknown

_1397414959.unknown

_1397414970.unknown

_1397414978.unknown

_1397414980.unknown

_1397414975.unknown

_1397414965.unknown

_1397414968.unknown

_1397414963.unknown

_1397414928.unknown

_1397414948.unknown

_1397414951.unknown

_1397414942.unknown

_1397414945.unknown

_1397414936.unknown

_1397414915.unknown

_1397414925.unknown

_1397414913.unknown

_1397414200.unknown

_1397414214.unknown

_1397414222.unknown

_1397414224.unknown

_1397414218.unknown

_1397414206.unknown

_1397414210.unknown

_1397414203.unknown

_1397414184.unknown

_1397414190.unknown

_1397414196.unknown

_1397414186.unknown

_1397414175.unknown

_1397414180.unknown

_1397414173.unknown

_1397412404.unknown

_1397413356.unknown

_1397413624.unknown

_1397414117.unknown

_1397414145.unknown

_1397414158.unknown

_1397414165.unknown

_1397414167.unknown

_1397414161.unknown

_1397414151.unknown

_1397414156.unknown

_1397414148.unknown

_1397414135.unknown

_1397414140.unknown

_1397414143.unknown

_1397414138.unknown

_1397414126.unknown

_1397414132.unknown

_1397414120.unknown

_1397414088.unknown

_1397414104.unknown

_1397414111.unknown

_1397414113.unknown

_1397414107.unknown

_1397414094.unknown

_1397414101.unknown

_1397414092.unknown

_1397413635.unknown

_1397413644.unknown

_1397414084.unknown

_1397413641.unknown

_1397413629.unknown

_1397413633.unknown

_1397413627.unknown

_1397413446.unknown

_1397413475.unknown

_1397413612.unknown

_1397413618.unknown

_1397413621.unknown

_1397413616.unknown

_1397413604.unknown

_1397413608.unknown

_1397413481.unknown

_1397413461.unknown

_1397413467.unknown

_1397413470.unknown

_1397413464.unknown

_1397413455.unknown

_1397413458.unknown

_1397413451.unknown

_1397413415.unknown

_1397413430.unknown

_1397413440.unknown

_1397413443.unknown

_1397413438.unknown

_1397413425.unknown

_1397413427.unknown

_1397413422.unknown

_1397413366.unknown

_1397413410.unknown

_1397413413.unknown

_1397413369.unknown

_1397413361.unknown

_1397413363.unknown

_1397413358.unknown

_1397412829.unknown

_1397413034.unknown

_1397413065.unknown

_1397413079.unknown

_1397413085.unknown

_1397413353.unknown

_1397413081.unknown

_1397413070.unknown

_1397413076.unknown

_1397413067.unknown

_1397413048.unknown

_1397413058.unknown

_1397413062.unknown

_1397413053.unknown

_1397413042.unknown

_1397413044.unknown

_1397413037.unknown

_1397412859.unknown

_1397412871.unknown

_1397412969.unknown

_1397412972.unknown

_1397413027.unknown

_1397412966.unknown

_1397412863.unknown

_1397412868.unknown

_1397412861.unknown

_1397412844.unknown

_1397412853.unknown

_1397412856.unknown

_1397412848.unknown

_1397412838.unknown

_1397412841.unknown

_1397412834.unknown

_1397412460.unknown

_1397412789.unknown

_1397412806.unknown

_1397412813.unknown

_1397412825.unknown

_1397412810.unknown

_1397412800.unknown

_1397412803.unknown

_1397412793.unknown

_1397412471.unknown

_1397412476.unknown

_1397412785.unknown

_1397412473.unknown

_1397412465.unknown

_1397412468.unknown

_1397412462.unknown

_1397412432.unknown

_1397412447.unknown

_1397412453.unknown

_1397412457.unknown

_1397412450.unknown

_1397412438.unknown

_1397412442.unknown

_1397412435.unknown

_1397412416.unknown

_1397412427.unknown

_1397412430.unknown

_1397412420.unknown

_1397412409.unknown

_1397412413.unknown

_1397412407.unknown

_1397410692.unknown

_1397411605.unknown

_1397411893.unknown

_1397412375.unknown

_1397412387.unknown

_1397412393.unknown

_1397412396.unknown

_1397412389.unknown

_1397412381.unknown

_1397412384.unknown

_1397412378.unknown

_1397411906.unknown

_1397411912.unknown

_1397411915.unknown

_1397411909.unknown

_1397411899.unknown

_1397411902.unknown

_1397411896.unknown

_1397411850.unknown

_1397411865.unknown

_1397411881.unknown

_1397411889.unknown

_1397411875.unknown

_1397411857.unknown

_1397411860.unknown

_1397411854.unknown

_1397411617.unknown

_1397411824.unknown

_1397411835.unknown

_1397411620.unknown

_1397411610.unknown

_1397411613.unknown

_1397411608.unknown

_1397411298.unknown

_1397411576.unknown

_1397411591.unknown

_1397411597.unknown

_1397411601.unknown

_1397411595.unknown

_1397411585.unknown

_1397411589.unknown

_1397411581.unknown

_1397411556.unknown

_1397411570.unknown

_1397411573.unknown

_1397411567.unknown

_1397411304.unknown

_1397411308.unknown

_1397411300.unknown

_1397411271.unknown

_1397411284.unknown

_1397411290.unknown

_1397411294.unknown

_1397411287.unknown

_1397411276.unknown

_1397411279.unknown

_1397411273.unknown

_1397411256.unknown

_1397411261.unknown

_1397411265.unknown

_1397411259.unknown

_1397411245.unknown

_1397411252.unknown

_1397410698.unknown

_1397410123.unknown

_1397410599.unknown

_1397410624.unknown

_1397410644.unknown

_1397410668.unknown

_1397410690.unknown

_1397410647.unknown

_1397410631.unknown

_1397410641.unknown

_1397410627.unknown

_1397410610.unknown

_1397410616.unknown

_1397410620.unknown

_1397410614.unknown

_1397410604.unknown

_1397410608.unknown

_1397410601.unknown

_1397410153.unknown

_1397410585.unknown

_1397410591.unknown

_1397410595.unknown

_1397410588.unknown

_1397410164.unknown

_1397410170.unknown

_1397410162.unknown

_1397410135.unknown

_1397410142.unknown

_1397410150.unknown

_1397410139.unknown

_1397410130.unknown

_1397410133.unknown

_1397410126.unknown

_1397410077.unknown

_1397410099.unknown

_1397410111.unknown

_1397410116.unknown

_1397410118.unknown

_1397410113.unknown

_1397410104.unknown

_1397410108.unknown

_1397410102.unknown

_1397410087.unknown

_1397410093.unknown

_1397410096.unknown

_1397410090.unknown

_1397410083.unknown

_1397410085.unknown

_1397410080.unknown

_1397410053.unknown

_1397410066.unknown

_1397410071.unknown

_1397410074.unknown

_1397410069.unknown

_1397410061.unknown

_1397410063.unknown

_1397410058.unknown

_1397409948.unknown

_1397409953.unknown

_1397409956.unknown

_1397409950.unknown

_1397409941.unknown

_1397409945.unknown

_1397409939.unknown

_1396967115.unknown

_1396971364.unknown

_1396972760.unknown

_1396973158.unknown

_1397409886.unknown

_1397409914.unknown

_1397409927.unknown

_1397409932.unknown

_1397409934.unknown

_1397409929.unknown

_1397409919.unknown

_1397409924.unknown

_1397409917.unknown

_1397409901.unknown

_1397409907.unknown

_1397409911.unknown

_1397409903.unknown

_1397409894.unknown

_1397409896.unknown

_1397409889.unknown

_1397409856.unknown

_1397409870.unknown

_1397409877.unknown

_1397409881.unknown

_1397409874.unknown

_1397409863.unknown

_1397409865.unknown

_1397409860.unknown

_1396973162.unknown

_1396973164.unknown

_1396973165.unknown

_1396973163.unknown

_1396973160.unknown

_1396973161.unknown

_1396973159.unknown

_1396972824.unknown

_1396973150.unknown

_1396973154.unknown

_1396973156.unknown

_1396973157.unknown

_1396973155.unknown

_1396973152.unknown

_1396973153.unknown

_1396973151.unknown

_1396973146.unknown

_1396973148.unknown

_1396973149.unknown

_1396973147.unknown

_1396973142.unknown

_1396973144.unknown

_1396973145.unknown

_1396973143.unknown

_1396973140.unknown

_1396973141.unknown

_1396973138.unknown

_1396973139.unknown

_1396973137.unknown

_1396973136.unknown

_1396972797.unknown

_1396972810.unknown

_1396972818.unknown

_1396972820.unknown

_1396972811.unknown

_1396972805.unknown

_1396972808.unknown

_1396972800.unknown

_1396972782.unknown

_1396972790.unknown

_1396972793.unknown

_1396972786.unknown

_1396972767.unknown

_1396972770.unknown

_1396972781.unknown

_1396972764.unknown

_1396971894.unknown

_1396971955.unknown

_1396972473.unknown

_1396972738.unknown

_1396972754.unknown

_1396972757.unknown

_1396972750.unknown

_1396972731.unknown

_1396972735.unknown

_1396972478.unknown

_1396972453.unknown

_1396972464.unknown

_1396972466.unknown

_1396972455.unknown

_1396972447.unknown

_1396972451.unknown

_1396972449.unknown

_1396971958.unknown

_1396971926.unknown

_1396971942.unknown

_1396971949.unknown

_1396971952.unknown

_1396971946.unknown

_1396971935.unknown

_1396971938.unknown

_1396971932.unknown

_1396971913.unknown

_1396971922.unknown

_1396971924.unknown

_1396971919.unknown

_1396971906.unknown

_1396971909.unknown

_1396971899.unknown

_1396971503.unknown

_1396971855.unknown

_1396971871.unknown

_1396971882.unknown

_1396971885.unknown

_1396971874.unknown

_1396971879.unknown

_1396971864.unknown

_1396971868.unknown

_1396971866.unknown

_1396971858.unknown

_1396971520.unknown

_1396971850.unknown

_1396971853.unknown

_1396971523.unknown

_1396971514.unknown

_1396971517.unknown

_1396971509.unknown

_1396971397.unknown

_1396971410.unknown

_1396971497.unknown

_1396971500.unknown

_1396971492.unknown

_1396971494.unknown

_1396971404.unknown

_1396971407.unknown

_1396971400.unknown

_1396971376.unknown

_1396971385.unknown

_1396971391.unknown

_1396971382.unknown

_1396971371.unknown

_1396971373.unknown

_1396971368.unknown

_1396969663.unknown

_1396970169.unknown

_1396970391.unknown

_1396970419.unknown

_1396970431.unknown

_1396971356.unknown

_1396971360.unknown

_1396970437.unknown

_1396970427.unknown

_1396970429.unknown

_1396970424.unknown

_1396970405.unknown

_1396970412.unknown

_1396970416.unknown

_1396970409.unknown

_1396970397.unknown

_1396970400.unknown

_1396970394.unknown

_1396970207.unknown

_1396970223.unknown

_1396970384.unknown

_1396970387.unknown

_1396970295.unknown

_1396970216.unknown

_1396970218.unknown

_1396970212.unknown

_1396970182.unknown

_1396970189.unknown

_1396970193.unknown

_1396970187.unknown

_1396970176.unknown

_1396970179.unknown

_1396970173.unknown

_1396969798.unknown

_1396970132.unknown

_1396970156.unknown

_1396970163.unknown

_1396970166.unknown

_1396970160.unknown

_1396970143.unknown

_1396970149.unknown

_1396970136.unknown

_1396969818.unknown

_1396969834.unknown

_1396969838.unknown

_1396969821.unknown

_1396969823.unknown

_1396969811.unknown

_1396969815.unknown

_1396969805.unknown

_1396969774.unknown

_1396969787.unknown

_1396969793.unknown

_1396969795.unknown

_1396969790.unknown

_1396969780.unknown

_1396969784.unknown

_1396969777.unknown

_1396969680.unknown

_1396969765.unknown

_1396969769.unknown

_1396969683.unknown

_1396969667.unknown

_1396969676.unknown

_1396969665.unknown

_1396969049.unknown

_1396969597.unknown

_1396969633.unknown

_1396969649.unknown

_1396969655.unknown

_1396969659.unknown

_1396969652.unknown

_1396969640.unknown

_1396969644.unknown

_1396969636.unknown

_1396969613.unknown

_1396969623.unknown

_1396969628.unknown

_139

Documents

SECTION-6