The center of gravity is at the intersection of the line joining the centers of gravity of the tri- angles, and the middle line FG... 482 TRONG TAM CENTER OF GRAVITY Segment
Trang 1`
ge 25 _ SỐ TAY
ENGINEERING ENGLISH
(With key to pronunciation - Alustrations)
Trang 2g Ộsenda ẤEM dị ỔLOT Nor DAU: aul 4B
voub név CE vd aãi 888 tài sa8ek ,ậ đãi E
gilt ii sat ages wh ods
Mau cdu hoc fap va sử đây đền Anh ti eisai nã tng Hag tang! những năn:gì4iđây) :đễ, đáp ứng: phần nào nhủ sầu đố, skũng đât (bia soạw :' 6ềỪẶhy ¡ếng Ỉ A-Í Íậ sêẶ' Nội: dụng suấn
sách gom Phần mở dau, Phần thuật ngữ và Phan số liệu -
- Trong Phần mở dau chúng tôi giới thiậu lối phiên âm Quốo tế mới
nhất [dựa 0/1/1170) M041) 10) eda Daniel Jone 1992) duce
dùng để phiên âm các thuật ngữ kỹ thuật, giới thiệu sách doe cae thuật ngữ và câu thông thường trong khoa học ký thuật 1
- Phần sáo thuật ngữ được phân loại theo từng ohủ đẾ bao quất trong ngành eơ khắ, từ vẽ ký thuật, nguyên lý máy, Bến sáo phương pháp gia sông co khi Mỗi thuật ngữ Ếược trình bay băng tiếng Việt, tiếng
Anh có phiên âm kém theo hình minh họa Để tiện tra sứu, sắp thuật
ngữ và sác hình minh họa Ếược Sanh số thứ tự
- Phần số liệu gầm cáo hình vẽ, cáo bảng tiêu chuẩn, sáo sông thức - tắnh toán phổ biến trong các ngành kỹ thuật, phần nầy Ếược sắp xếp theo cáo mục: Nguyên lý máy, truyền động ồai, xắch, bánh răng, ổ lăn, cáo mối ghép
ứua quyển sách này, bạn đọc không những chỉ tra sứu sáo thuật ngữ tiếng Anh kỹ thuật mã oồn tra sứu các bảng tiêu chuẩn sẵn thiết về kắch thước, dung sai, lấp gháp, vật liệu v.v về mặt dữ liậu lan mat thuật ngữ gốc tiếng Ảnh
Biên soạn sách này chúng tôi dựa trên việo chuyan dich cd chon loc quyển The Concise Illustrated Ruseiẩn - English Dictionary of Mechanical Engineering cua Vladimir V Shvarts (Moscow Russian
Trang 3
IV
Language Publishers 1980); Dữ liệu Bược shọn lạo từ Machinery’s Handbook của Erik Oberg vi F D Jones tai bản lẫn thứ 17, vốn duge
xem là sách gối Öẩu eho kỹ sự eơ khí vì người lầm công táo kỹ thuật
Chúng tôi rất vai mững và sắm kích khi nhận ðược những ehỉ dẫn sửa ban doc xa gần, góp phần nang cao chat lượng oho những lần tái bản
NHÓM BIÊN SOẠN
Trang 4V
KEY TO PHONETIC SYMBOLS
KỸ HIỆU PHÁT ÂM
Vowels and diphthongs Nguyên âm rả nguyên âm đơi
1 i: asinsee /Si⁄/ 11 3: asinfur /fa:(r)/
2 1 asinsit /stt/ 12 9 asinago /2'ga0/
3 e asinten /ten/ 13 er asin page /peIidz '4 œ asinhat /het/ 14 ao asin home /hàm/
5 a: asinarm /œm/ l5 ar asinfive /farv/ -
6 p asingot /gpt/ 16 ab asinnow = /nvv/
7 2: asinsaw /so⁄/ = 17 or asin join /dz2m
8 ư asinput /pot/- 18 1a asinnear /nro(r)/
9 u:asintoo /tu:/ 19 eo asinhair /hea(r)/
10.A asincup /kAp 20 co asin pure /pjuo(r)/
l p asinpen /pen 13 s asinso /Sa0/
2 b asinbad /bzd/ 14 z asinzoo /2u⁄
3 t asintea (/tU⁄ 15 f asinshe = /fi:/
4 d asindid /did/ 16 4 asin vision /’vi3n/
5 k asincat /kœU 17 h asinhow = /hav/
6 g asingot /got/ 18 m asinman /mzn/
7 tf as.in-chin Afin/ 19 n asin no /nav/
8 d3 asin June /dju:n/ 20 n asinsing /smy
9 f asinfal /:V 21.1 asinleg /leg/
10.v asin voice /vois/ 22.r asinred /red/
11.8 asinthin /O1n/ 23 j asinyes /jes/
12.6 asinthen /den/ 24 w asinwet /wet/
wo
/ dấu trọng âm ‹vd: about /a'bao
Trang 6CƠ HỌC 474 MAT PHANG NGHIENG - CHEM
INCLINED PLANE - WEDGE
If friction is taken into account, then force P to pull body up is:
P= TW (u cosa+ sina)
Force F1 to pull body down is:
Py = W (@ cos a@— sin a) Force P: to hold body stationary:
P,= W (sin a—p cos œ)
in which z is the coefficient of friction
Neglecting friction: | With friction: Neglecting friction: } With friction:
+, Sina Coefficient of fric- A Coefficient of fric-
W=Px cos P=Wx sin (a+) | yy p xŠ- Pxcotal P= tan (œ+† ở)
h
O=Px=‡PXxcotưa 26
With friction: ¬ Coefficient of friction = w = tan ở
P=2Qtan(a+¢)
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P required for moving a body on an inclined plane The
friction on the plane is not taken into account The
- 11w column headed “ Tension P in Cable per Ton of 2000
o ¢ = Pounds” gives the pull in pounds required for moving
a # T one ton along the inclined surface The fourth column
1 ý gives the perpendicular or normal pressure If the co-
“~ efficient of friction is known, the added pul? required
~_ ‘Tbe table below makes it possible to find the force]
Q X coefficient of friction = additional puil required
đêm Rise, Ft.| Angle a |Cable per sure @ om | Rise, Ft.| Anglea |Cable per sure O or | tpn [aie [areca] | net [alr be
Trang 8—-—-——ø—-—-_—.—a† A pull of 80 pounds is exerted at the
<i — end of the lever, at W;
) im 12 inches and L = 32 inches ˆ
- balance the lever
EÐ:W =i:L Fx L=oWxil / 80% 12 960
| Fe = —— w= zo pounds
how long must Z be made to secure
La Xe WX? } Fxa FX L| equilibrium?
W+FP F’° W+F W Le 180%3 Coy 20
t - Total length FE of a lever is 25 inches
-¬->*+—-—-a——-—> A weight of go pounds is supported at
F F = 22X19 56 pounds
F= wx! W = txE and a= 5 feet, what should L cqual to Ƒ + i secure equilibrium?
Wxa Wxlil Fxa FXL 2200% §
W x2 FXL Let F= 12 tons; W = 4.5 tons;
= E W= i @=16 feet Find £ and §
5X 16
L Xa Wxil, Fxa FXL La TC S95 feats
| pounds; a= 4,6 = 7, and ¢ = 10 inches
&\ fp RA F Tf x = 6 inches, find F
When three or more forces act en 20X% 4+ 30X7+15 X10
„aWxsz†+Px?+Qxc above, how long must lever arm x be
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477
KHỚP KHUỶU TOGGLE-JOINT
Toggle-joint — If arms ED and EH are
of unequal length:
Fa P= F
The relation between P and F changes
If arms ED and EH are equal: |
sponding to the angle found The coeffi-
cient is the ratio of the resistance to the force applied, and multiplying the force applied by the coefficient gives the resist- ance, neglecting friction
Coeffi- Coeffi- Coeffi- Coeffi-
Trang 10475 l
BÁNH XE - RÒNG RỌC WHEELS AND PULLEYS
FXR=Wxr wound the lifting rope of a windlass is
R at the periphery of a gear of 24 inches FMR diameter, mounted on the same shaft
W = as the drum and transmitting power
r to it, if one ton (2000 pounds) is to be
one-half the veloc- ity of the force ap- plied at F
In the illustration is shown a com-
bination of a double and triple block
The pulleys each turn freely on a pin
as axis, and are drawn with different diameters, to show the parts of the rope more clearly There are 5 parts
of rope Therefore, if 200 pounds is
to be lifted, the force F required at the
end of the rope is:
rxXnX rs Let the pitch diameters of gears A,
B, Cand D be 30, 28, r2 and ro inches,
m=6; and r=5 Let Reet, and fo= 4 Then the force F required to lift a weight W of 2000 pounds, friction
being neglected, is:
Fu
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| 479
RONG ROC VI SAI - VIS
DIFFERENTIAL PULLEY - SCREW
Waar
‘Force Moving Body on Horizontal Plane — F
tends to move B along line CD; @Q is the component which actually moves B; P is the pressure, due to
F, of the body on CD
Q= FX cosa: Pua V Fi— Qi
Screw —- F = force at end of handle or wrench;
R = lever-arm of F; r = pitch radius of screw; p =
lead of thread; Q = load Then, neglecting friction:
M Vu g=z#x
If » is the coefficient of friction, then:
For motion in direction of load Q which assists it:
it is a€ the geometric center The center of gravity of a uniform rdund rod, for
example, is at the center of its diameter halfway along its length; the center of
gravity of a sphere is at the center of the sphere For solids, areas, and arcs that are not symmetrical, the determination of the center of gravity may be made
experimentally or may be calculated by the use of formulas
‘The tables that follow give such formulas for some of the more important shap¢s
For more complicated and unsymmetrical shapes the methods outlined on page 313 may be used
Example: A piece of wire is bent into the form of a semi-circular arc of 10-inch radius How far from the center of the arc is the center of gravity located? _ Accompanying the third diagram on page 308 is a formula for the distance from the center of gravity of an arc to the center of the arc: ¢ = 2r +, Therefore, in thiscase, ›
8=2 Xa TÔ + 3.1416 = 6,366 inches.
Trang 12ở of the center of gravity from side a is:
h (b+) , 2(z+b+c}
where & is the height perpendicular to đ
is equal to one-third the height perpendicular to
that side Hence, ¢= h+ 3
Circular Arc — The center of gravity is on the
line that bisects the arc, at a distance
2 2
a - _ Π+48 ) from the center of the circle
For an arc equal to one-half the periphery:
the line joining the middle points of parallel lines
The trapezoid can also be divided into two tri-
angles The center of gravity is at the intersection
of the line joining the centers of gravity of the tri-
angles, and the middle line FG
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Ayea of « Parallelogram — The center of gravity
is at the intersection of the diagonals
Any Four-sided Figure — Two cases are possible,
as shown in the illustration To find the center
of gravity of the four-sided figure ABCD, each of
the sides is divided into three equal parts A line is then drawn through each pair of division
points next to the points of intersection A, B, C,
and D of the sides of the figure ‘These lines form
& parallelogram EFGH; the intersection of the
diagonals EG and FA locates the required center
Part of Circle Ring — Distance 6 from center of
gravity to center of circle is:
(R* — r#) sin œ
(Rt—-r)œ Angle a is expressed in degrees
È = 38.197
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482
TRONG TAM CENTER OF GRAVITY
Segment of an Ejlipse.— The center of gravity
of an elliptic segment ABC, symmetrical about one of the axes, coincides with the center of gravity
of the segment DBF of a circle, the diameter of
which is equal to that axis of the ellipse about 3 which the elliptic segment is symmetrical
xŒ
+ C ‡ _ For the compleroent area ABC:
- ! 5 pherical Surface of Segments and Zones of
` Spheres - Distances ø and š which determine the
center of gravity, are:
cylinder (or prism) with parallel end surfaces, is
located at the middle of the line that joins the
centers of gravity of the end surfaces
The center of gravity of a cylindrical surface or
shell, with the base or end surface in one end, is found from:
ahi
o= Thad
The center of gravity of.a cylinder cut of by an
inclined plane is located by:
Trang 15If the cylinder is bollow, the center of gravity
of the solid shell is found by:
H+— pe H*— }a
tance from the base equal to one-quarter of the
| height; ora = 3&4
| he center of gravity of the triangular surfaces forming the pyramid is located on the line joining the apex with the center of gravity of the base surface, at a distance from the base equal to one-
third of the height; or a = ‡ À
Cone —- The same rules apply as for the pyramid
For the solid cone:
of the end surfaces If A1= area of base surface,
| and 42 area of top surface,
Frustum af Cone — The same rules apply as for
‘the frustum of a pyramid For a solid frustum of 2) 4 circular cone the formula below is also used:
_ h{(R1!-+L + Rr + 3r?)
_— 4(Œ?*.t+rRr+r?)
‘ Ậ cal surface of a frustum of a cone is determined by:
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484 TRONG TAM
CENTER OF GRAVITY
joining the center of gravity of the base with the middle point of the edge, and is located at:
a (25+)
Spherical Segment — The center of gravity of a
solid segment is determined by:
Spherical Sector — The center of gravity of a
solid sector is at:
a= § (1+ cosa) r =f (27—h)
Segment of Ellipsoid or Spheroid.— The center
of gravity of a solid segment ABC, symmetrical
m about the axis of rotation, coincides with the center
—p— of gravity of the segment DBF of a sphere, the
diameter of which is equal to the axis of rotation
Pareboloid — The center of gravity of a solid
paraboloid of rotation is at:
a=th
Center of Gravity of Two Bodies — If the weights
of the bodies are P and Q, and the distance between
their centers of gravity is 4, then:
Trang 17485
MECHANICS
Center of Gravity of Figures of any Outline — If the figure is symmetrica!
about a center line, as in Fig 1, the center of gravity will be located on that line
To find the exact location on that Jine, the simplest method is by taking moments with reference to any convenient axis at right angles to this center line Divide the area into geometrical figures, the centers of gravity of which can be easily found
In this case, divide the figure into three rectangles KLMN, EFGH and OPRS
Call the areas of these rectangles 4, B and C, respectively, and find the center of gravity of each Then select any convenient axis, as XX, at right angles to the center line YF, and determine distances 2, 5 and c The distance y of the center
of gravity of the complete figure from the axis XX is then found from the equation:
gravity is determined by the equations:
x Aa + Bh +Ca Aa+t Bb+Ce
A+B+CO 7”"A+B+C
As an example, let A = 14 square inches, B = 18 square inches, and C = 20 square inches Let a= 3 inches, b= 7 inches, and ¢= 11.5 inches Let a: = 6.5 inches,
& = 8.5 inches, and ¢; = 7 inches Then:
pe EEK OS+IBXBS+20X7 _ a = 7.38 inches
14+ 18+ 20 4X 3+18X% 7+ 20X 11.5 398
¥ 144 184-20 sa 7.65 inches
In other words, the center of gravity is located at a distance of 7.65 inches from
the axis XX and 7.38 inches from the axis YY.
Trang 18(Af = mass of body = weight 4.32.16)
† TT Prisms — With reference to axis A — A:
j L _ With refererce to axis 8 — :
an, 3 I=M (? + my
Trang 19-MOMENT QUAN TINH
Trang 20
BAN KINH HO! CHUYEN
RADIUS OF GYRATION
Bar of Small Diameter
- ” Axis through center Cylinder
Axis through center
bent to Circular Shape
Axis, a diameter of the
Trang 21Axis, diameter at mid- Cylinder Parallelepiped
length Axis at a distance Axis at distance from
Longitudinal Axis Axis through center Aik a A
Trang 22B(R+sR+6A) 3 (Ea k~vettậr
| Sphere
Axis its diameter Ellipsoid
Axis its diameter Paraboloid
Axis through center
L
k= 0.6325 Bur Ry k=o 5773 r
Trang 23
_ MATERIALS —
SỨC BÊN VAT LIEU
Trang 24
Aweiion Moment of Section Modulus | Radius of Gyration
Trang 25
** 5 Btsinta) ieee) =o289 x 8 ff
tbsine | aaa Ranta
Trang 2712a 4costa29° 6L 400s? 224° 48 cost 32§°
Trang 29Ls t— 19 |
ai” == 19
|e a@~ 3 ale (@~h) 4
Trang 31499
_ MOMENT QUAN TINH : MODUL MAT CAT
_ MOMENTS OF INERTIA, SECTION MODULUS etc., OF SECTION —
Trang 32
500
MOMENT QUAN TINH : MODUL MAT CAT
MOMENTS OF INERTIA, SECTION MODULUS etc., OF SECTION
Distance from Neutral
Section Area of Sect ion, Axis to Extreme Fiber,
Trang 33Am"
MOMENT QUÁN TÍNH : MODUL MAT CAT
MOMENTS OF INERTIA, SECTION MODULUS etc OF SECTION
Trang 34
502
TINH CHAT CAC TIET DIEN KHUNG CAT
PROPERTIES OF SECTIONS FOR PUNCH AND SHEAR FRAMES
Trang 35+1 + Z, = Section Modulus for Compression;
fog 7 2, = Section Moduius for Tension;
Trang 36
504
MODUL TIET DIEN CHU NHAT
SECTION MODULUS FOR RECTANGLES
Length | Section || Length | Section |{ Length | Section [| Length | Section
of Side | Modulus |] of Side [ Modulus |] of Side | Modulus || of Side | Modulus
MODUL TIET DIEN VA MOMENT QUAN TINH
CUA TRUC TRON
SECTION MODULUS AND MOMENTS OF INERTIA
Diam.) Modulus | of Inertia |] Di#™-|Modutue] of Inertia |] >18™-/Modulus| of Inertiz |
543 | 0.00037 ] 0.00003 32964 {| 0.0091 | 0.00207 S4 0.0413 | O.OT55O
He | 0.00065 | 0.00006 8iệa | o.oIIT Ì o.ooz7o 2363 | 0.0467 0.08825
In this and succeeding tables, the Polar Section Modulus fora shaft of given diameter
Trang 37
SECTION MODULUS AND MOMENTS OF INERTIA
FOR ROUND SHAFTS
Diam, aces of Inertia || 9#-| sodutus fof Inertia {| P#™-| Modulus lof Inertia
3.00 | 0.0981 | 0.0490 |] 1.50 | 0.3323 | 0.2485 |] 2.00 | 0.7854 | 0.7854
I1.OI | O.1OII | ©.OSIG || r.51 | 0.3380 | 0.2552 || 2.02 | 0.7972 | 0.8012
1.02 | 0.1041 | 0.0538 || 1.52 | 0.3447 | 0.2620 |] 2.02 | 0.8092 | 0.8172 4.03 | 0.1072 | 0.0552 |] r.53 | 0.3516 | 0 2689 || 2.03 | 0.8212 | 0.8335 1.04 | 0.1104 | 0.0574 I] x.54 | 0.3585 | 0.276r |] 2.04 | 0.8334 | o.8sor
r.0§ | 0.1136 | c.0596 || t.55 | 0.3655 |] 0 2833 || 2.05 | 0.8457 | 0.8669
3.06 | 0.1169 | 0.0619 || 1.56 | 0.3727 | 0.2907 || 2.06 | 0.8582 | 0 8839 1.07 | 0.1202 | 0.0643 || 1.57 | 0.3799 | 0.2982 || 2.07 | 0.8707 | 0.gor2 1.08 | 09.1236 | 0.0667 [| 1.58 [| 0.3872 | 0.3059 |] 2.08 | 0.8834 | 0.9788 1.09 | 0.1272 | 0.0692 || 1.59 | 0.3946 } 0.3137 || 2.09 | 0.8962 | 0.9366 1.10 | 60,1307 | 0.0718 |} 1.60 | 0.4021 | 0.3217 || 2.10 | o.go92 | 0.9547 1.21 | 0.1347 | 0.0745 | 1.65 | 0.40907 | 0.3298 }) 2.11 | 0.9222 | 0.9729 1.12 | 0.1379 | 0.0772 || 1.62 | 0.4173 | 0.3380 || 2.12 | 0.9354 | 0.9915 1.13 | 0.1416 | 0.0800 }} 1.63 | 0.4251 | 0.3465 |] 2.13 | 0.9487 | r.0103 z.¥4 | 0.1454 | 0.0829 ¡| 1.64 | 0.4330 | 0.3550 || 2.14 | 0.9621 | 1.02795 I.25 | 0.1493 | 0.0859 |] 1.65 | 0.4410 | 0.3638 || 2.15 | 0.9757 | 1.0488
1.36 | 0.1532 | 0.0888 fl 1.66 | 0.4490 | 0.3727 || 2.16 | 0.9894 {| 1.0685
1.T17 | O.IS72 | o.ogr9 || 1.67 | 0.4572 | 0.3828 I] 2.17 | 1.0032 | 1.0884 2.318 | o 1613 | 0.0951 |} 1.68 | 0.4655 | 0.3910 || 2.18 | r.orzx | 1.1086 1.19 | 0.1654 | 0.0984 || 1.69 | 0.4738 | 0.4004 |} 2.19 | r.0gzr | 1.1291 1.20 | 0, 1696 | o 1018 || 1.70 | 0.4823 | 0.4100 |f 2.20 | r.04gq | 1.1499 1.2E | 0.1739 | o 1052 || r.7r | 0.4908 [| 0.4797 || 2.21 | 1.os06 | 1.1709 1.22 | 0.1782 | 0.1087 |! 1.72 | 0.4995 | 0.4296 || 2.22 | 1.Oy41 | 1.1923
1.23 | 0, 1826 | 0.1123 |/ 1.73 | 0.5083 | 0.4397 || 2.23 |] 1.0887 | 1.2739
1.24 | 0.1871 | o 1160 |/ 1.74 | o.517r | 0.4499 |} 2.24 | 1.1034 | 1.2358 |
1.25 | o, 1917 | 0.1198 || 1.7S | o.s26t | o.4694 || 2.25 | x.1783 | 1.2580
1.26 | 0.1963 | 0.1237 || 1.76 | 0.5352 | 0.4710 || 2.26 | 1.1442 | 1.2806 1.27 | O.2OII | O.1277 || 1.77 | 0.5444 | o.4ð18 || 2.2y | 1.1483 | 1.3034 1.28 | o.2os8 | o.1317 | 1.78 | o.5536 | 0.4927 || 2.28 | 1.1636 | 2.3265 '1,.29 | 0.2107 | 0.1359 jl 1.79 | © 5630 | 0.5039 || 2.29 | r.3790 | 1.3499 | 1,30 | 0.2157 | 0.1402 |! 1,80 | 0.5726 | 0.5153 || 2.30 | 3.1945 | 1.3737
-4.31 | 0.2207 | 0.1445 |] 1.82 | 0.5821 | 0.5268 || 2.37 | x 2202 | 1.3977
1.32 | 0.2258 | 0, 1490 || 1.82 | 0.5918 | 0.5385 || 2.32 | 1.2250 | 1.4234 1.33 | 0.2309 ] 0.1535 |} 1.83 | 0.6016 | 0.5505 |} 2.33 | 1.2418 | 1.4468 1.34 | 0.2362 | o 1582 -|| 1.84 | o.6rrs | o.s626 || 2.34 | 2.2579 | 1.4728 1.35 | 0.2475 | 0 2630 || 1.85 | 0.6216 | 0.5749 || 2.35 | 3.2745 | 1.497%
1.36 | 0.2469 | 0.1679 || 1.86 [ 0.6317 | 0 5875 || 2.36 | 1.2904 | 1.5227 1.37 | 0.2524 } 0.1729 |} 1.87 | 0.6419 | 0.6002 || 2.37 | 1.3O6g | 1.5487 -2.38 | 0.2580 | 0.1780 || 4.88 | 0.6524 | 0.6132 || 2.38 | 1.3235 | r.S7SO 1,39 | 0.2636 | 0.1832 || 1.89 | 0.6628 | 0.6263 || 2.39 | 1.3403 | x.6016 1.40 | 0.2694 | 0.1886 |/ 1.90 | 0.6734 | 0.6397 || 2.4o | 1.4572 | 1.6286 1.41 | 0.2752 | 0.1940 || x.91 | 0.6840 | 0.6532 || 2.47 | x.3742 | 1.6559
1.42 | 6.2811 | 0.1995 || 1.92 | 0.6948 | 0.6670 || 2.42 | x 3924 | 1.6836 1.43 | 0.2870 | o.2os2 || 1.93 | 0.7057 | 0.6810 || 2.43 | 1.4087 | 1.7176
Trang 38
506
MODUI, TIẾT ĐIỆN VÀ MOMENT QUÁN TÍNH
CUA TRỤC TRON
SECTION MODULUS AND MOMENTS OF INERTIA
FOR ROUND SHAFTS
| Section | Moment Section | Moment Section | Moment
Diam | stodulus lot Inertia || O!8™-] Modutus lof Inertia || O*™-| modulus fot Inertia
2.51 | 1.5525 | 1.9483 || 3.o1 | 2.6773 | 4.9293 || 3.51 | 4.2455 | 7.4507
2.52 | I.57II | 1.9706 || 3.02 | 2.7042 [ 4.0831 || 3.52 | 4.2818 | 7.5360 2.53 | 1.5899 | 2.0z12 || 3.03 | 2.7310 | 4.1375 || 3-53 | 4.3784 [ 7.6220 2.54 | 1.6083 | 2.0432 || 3.04 | 2.758r | 4.1924 |] 3.54 4.3558 7.7087
2.55 | 1.6279 | 2.0755 |} 3-05 | 2.7855 | 4.2478 || 3.55 | 4.3922 | 7.7962
“2.56 | 1.647 | 2.1083 || 3.06 | 2.8130 | 4.3038 |] 3-56 | 4.4294 | 7.8845
2.57 | t.666§ | 2.1414 || 3.07 | 2.8406 | 4.3604 || 3.57 | 4.4669 | 7.9734
1 2.58 | 2.6860 | 2.1749 || 3.08 | 2.8685 | 4.4175 || 3.58 | 4.5045 | 8.0632 2.59 | r.7057 } 2.2088 Il 3.09 } 2.8965 | 4.4751 || 3-59 | 4-5434 | 8.1536 Ï
2.74 | 1.9975 | 2.7266 || 3.23 | 3.3083 } 5.3430 || 3.73 | 5.0948 | 9.5018 2.74 | 2.0195 | 2.7668 || 3.24 | 3.3392 | 5.4©94 |[ 4.74 | 5-1359 | 9.6047 © 2.75 | 2.o417 | 2.8o;4 || 3.35 | 3.37ot | 5.4765 || 3.75 | 5.1771 | 9.7972
2.76 | 2.0641 | 2.8484 || 3.26 | 3.4014 | 5.5442 |] 3.76 | 5.2187 | g.8112 2.77 | 2.0866 { 2.8899 || 3.27 | 3.4328 | 5.6126 H 3.77 | 5.2605 | 9.9160 2.78 | 2.1093 | 2.9319 || 3.28 | 3.4644 °{ 5.6815 | 3.78 | 5.3024 |10.0276 2.79 | 4.1321 | 2.9743 || 3.29 | 43.4061 | 5.7511 || 3.79 | 5-3444 |to.1286 2.80 | 2.1550 | 2.0172 |] 3.30 | 3.5280 | 5.8214 |! 3.80 | 5.3870 [10.2350
2.81 | 2.1783 | 3.0605 {| 3.3x | 3.5603 | 5.8923 }] 3.8x | 5.4297 [10.3436
2,82 | 2.2076 | 3.1043 || 3.32 | 3.5926 | 5.9638 || 3.82 | 5.4726 |1O.4526 2.83 | 2.2251 | 3.1486 || 3.33 | 3.6252 | 6.0363 [| 3.83 | °§ 5156 |z0 5624 1
| 2.84 | 2.2488 | 3.17933 || 3.34 | 3.6580 | 6.1088 || 3.84 | 5.5590 |r0 6732 2.85 | 2.2727 | 3.2385 |} 3.35 | 3.6909 | 6.1823 |] 3.85 | 5.6025 |to.7B48 2,86 | 2.2966 | 3.2842 || 3.36 [ 3.724: | 6.2564 |) 3.86 | 5.6462 }20.8970
2.87 | 2.3208 | 3.3304 || 3.37 | 3.7575 | 6.3322 || 3.87 | 5.6903 |r1.o1Iro
2.88 | 2.3452 | 3.3771 3-38 | 3.7909 | 6.4067 4.88 | 5.7345 |11.1280
2.89 | 2.3697 | 3.4242 || 3.39 | 3.8246 | 6.4829 i 3.89 | 5.7789 |xz 2400 2.90 | 2.3940 | 3.4779 || 3.40 | 3.8590 | 6.5597 |] 3.90 | 5.8240 |z1.356c
2.91 | 2.4192 | 3.5200 |} 3.41 | 3.8928 | 6.6372 || 3.92 | 5.8685 |z1.4730 2.92 | 2.4442 | 3.5686 || 3.42 | 3.9272 | 6.7154 || 3.92 | 5.0137 |r1.s9ro |
2.03 | 2.4695 | 3.6178 || 3.43 | 3.9617 | 6.7943 |] 3.93 | 5.9590 |x1.7100 | 2.94 | 2.4949 | 3.6674 || 3.44 | 3-9965 | 6:8739 || 3.94 | 6.oo46 |rt.820o 2.95 | 2.5204 | 3.7175 || 3.45 | 4.0314 | 6.9542 lÌ 4.95 | 6.osos |r1.g5oo
2.96 | 2.5461 | 3.7682 |/ 3.46 | 4.0666 | 7.0332 | 3-96 | 6.0966 |rz.o6ạẠo - 2.97 | 2.5720 | 3.8196 |! 3.47 | 4.1019 | 7.1168 |] 3.97 | 6.1429 [12.1930 2.98 | 2.5981 | s.8yrr |} 348 | 4.1375 | 7.1976 |} 3.98 | 6.1894 [12.3270 2.99 | 2.6243 | 3.9233 || 3.49 | 4.1732 7.2824 11 3.09 | 6.2361 12.4416 |
Trang 39
807 MODUL TIET DIEN VA MOMENT QUAN TINH
CUA TRUC TRON
SECTION MODULUS AND MOMENTS OF INERTIA
FOR ROUND SHAFTS
Section | Moment Section | Moment 3 Section | Moment
*) Modulus fof Inertia |] [#"- | Modulus lof Inertia | Diam | Modulus jof Inertia
6.2830 | 12.566 8.946 | 20.129 |] s.oo | 12.272 | 30.680 6.3304 | 12.692 9.006 | zo.3o8 || s.or | 12.345 | 30.926 6.3779 | 12.820 9.066 | 20.489 |] 5.02 [| 12.420 | 3x 173 6.4256 | 12.948 g.126 | 20.671 || 5.03 | 12.493 ] 31.423 6.4736 | 13.077 9.186 | 20.854 |] 5.04 | 12.568 | 31.673 6.5217 | 13.227 9.247 | 21.039 || 5.05 | 12.644 | 31.925 6.5701 | 13.337 g.308 | 21.224 |] 5.06 | 12.718 } 32.179
6.6188 | 13.469 9.370 | 21.4rr || 5.07 | 12.794 | 32.434
6.6677 | 13.692 || 4.58 | 9.431 | 22.599 |] 5.08 | 12.870 | 32.691
6.7169 | 13.736 || 4.59 | 9.493 | 24.788 [| 5.09 | 12.946 | 32.949
6.7660 | 13.871 1] 4.60 | 9.556 | 21.979 || 5.10 | 13.023 | 33.209 6.8159 | 14.007 [| 4.62 | 9.618 | 22.170 {| 5.12 | 13:099 | 33.470 6.8657 | 14.143 || 4.62 | 9.682 | 22.363 |} 5.12 | 13.177 | 33.733
6.9164 | 14.28x [| 4.63 | 9.744 | 22.557 i] 5.23 | 13.254 | 33.997 6.0663 | 14.420 || 4.64 | 9.807 | 22.753 || 5.14 | 13.332 | 34.263
7.0169 | 14.560 Í| 4.6s | 9.870 | 22.950 || S.15 | 13.410 | 34.530
7.o677 | 14.701 || 4.66 | 9.934 | 23.148 || 5.16 | 13.488 | 34.799 7.1188 | 14.843 || 4.67 | 9.998 | 23.347 i| 5.17 | 13.567 | 35.070
7.1702 | 14.985 jf 4.68 | 10.063 | 23.548 || 5.78 | 13.645 | 35.342 7.2217 | 15.129 {| 4.69 | 10.127 | 23.750 |} §.29 | 13.725 | 35.615 7.2740 | 15.274 |f 4.70 | 10.193 | 23.953 [| 5.20 | 13.804 | 35.892 7.3256 |] 15.420 If 4.72 | 10,258 | 24.157 || 5.22 | 13.884 | 36 x68 7.3779 | 15.568 |] 4.72 | 10.323 | 24.363 [| 5.22 | 13.964 | 36: 446
7-4305 | 15.715 || 4.73 | 10.389 | 24.570 || 5.23 | 14.045 | 36.726 7-4833 | 15.865 [| 4.74 | 10.455 | 24.779 || §.24 | 14.125 | 37.008
7-5364 | 16.075 |} 4.75 | 10.522 | 24.989 || 5.25 | 14.206 | 37.291 7.5898 | 16.366 || 4.76 ] 10.588 | 25.200 || 5.26 | 14.287 | 37.576 7:6433 | 16.319 |) 4.77 | fo.65§ | 25.412 || 5.27 | 14.369 | 37.863 7-6972 | 16.472 || 4.78 | 10.722 | 25.626 || 5.28 | r4.45x | 38.151 7.7813 | 16.626 j} 4.79 | 10.790 | 25.841 || 5.29 | 14.534 | 38.440
7.8060 | 16.782 || 4.80 | 10.857 | 26.058 1] 5.30 | 14.616 | 38.732
7.8602 | 16.998 || 4.81 | 10.925 | 26.275 [| 5.32 | 14.609 | 39.025 7.9149 | 17.096 I] 4.82 | 10.094 | 26.495 |] 5.32 | 14.782 | 39.320 7.9701 | 17.255 || 4.83 | 12.062 | 26.778 f] 5.33 | 14.866 | 39.617
8.0254 | 17.415 || 4.84 | jt.131 | 26.937 || 5.34 | 14.949 | 39.915
8.oBio | 17.576 |] 4.85 | Ir.200 | 27.160 || sý.35 | 1s.O34 | 40.215 8.1369 | 17.738 || 4.86 | rr.zôg | 27.385 || 5.36 | rs.rr8 | 40.516 8.1930 | 17.902 || 4.87 | 12.339 | 27.613 || 5.37 | rg.202 | 40.819 8.2494 | 18.066 || 4.88 | 11.409 | 27.839 [| 5.38 | 15.288 | 4x 124 8.3060 | 18.231 |] 4.89 | 11.479 | 28.067 | 5.39 | t5.373 | 42.432 8.3630 | 18.398 3) 4.90 | rz 550 | 28.298 || s.4o | 15.459 | 41.739 8.4200 | 18.566 || 4.91 | rr.621 | 28.530 |] 5.4z | 15.545 | 42.049 8.4775 | 18.735 {| 4.92 | rr.692 | 28.763 || 5.42 | r5.632 | 42.362
8.535r | 18.905 || 4.03 | 11.763 | 28.997 || 5.43 | 15.728 | 42.674 8.5930 | 19.077 I 4.94 | 12.835 | 29.233 || 5.44 | 15.805 | 42.990 8.6513 } 19.249 || 4.95 | Ir.907 | 209.471! |} 5.45 | 15.893 | 43.307
8.7097 | 19.423 || 4.96 ] rz.979 | 29.720 || 5.46 | 15.980 | 43.626 8.7685 | 19.598 |] 4.97 | 12.052 | 29.950 |] 5.47 | 16.068 | 43.946 8.8274 | 19.773 || 4.98 | 12.124 | 30.192 || 5:48 | 16.157 | 44.268 8.8867 | 19.950 || 4.99 | 12.198 | 30.435 |{-5.49 | 16.245 | 44.592
Trang 40
TIEU CHUAN M¥ CHO THEP XAY DUNG TIET DIEN U
AMERICAN STANDARD STRUCTURAL CHANNELS
These channels are commonly designated
by nominal depth, group symbol, and weight | 4 x
: tạp 9LI13.4 wa: min T | ‡
t Data from American Institute of Steel Construction Manual, Fifth Edition, 19so
_ *Car and Shipbuilding Channel; not an American Standard
Meaning of symbols: {= moment of inertia; Z = section modulus; r = radius of