g 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 sơng thức - tắnh tốn 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
Trang 3IV
Language Publishers 1980); Dữ liệu Bược shọn lạo từ MachineryỖs Handbook của Erik Oberg vi F D Jones tai bản lẫn thứ 17, vốn duge
xem là sách gối ỷẩu eho kỹ sự eơ khắ vì người lầm công táo kỹ thuật Chúng tôi rất vai mững và sắm kắch khi nhận được những ehỉ dẫn sửa ban doc xa gần, góp phần nang cao chat lượng oho những lần tái bản
Trang 4V
KEY TO PHONETIC SYMBOLS
KỸ HIỆU PHÁT ÂM
Vowels and diphthongs Nguyên âm rả nguyên âm đôi
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 Ủ asinhat /het/ 14 ao asin home /haụm/
5 a: asinarm /Ủm/ l5 ar asinfive /farv/ -
6 p asingot /gpt/ 16 ab asinnow = /nvv/
7 2: asinsaw /so⁄/ = 17 or asin join /dz2m
8 ỏ 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)/
Consonants Phụ âm
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 /Ỗvi3n/
5 k asincat /kỦU 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
Trang 6CƠ HỌC 474 MAT PHANG NGHIENG - CHEM W = weight weight of body a of body Neglecting friction: P=Wx1~Wx sing W=Px 4 kA sine =PX coseca Eo O=Wx- =H cosa h ' W = weight of body
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@Ở sin a) Force P: to hold body stationary:
P,= W (sin aỞp cos Ủ)
in which z is the coefficient of friction iW = weight of body P Re a , oF y Neglecting friction: P= 205 = 20Xsing Z Q=Pxe =} PX coseca With friction: Coefficient of friction = x P=20 (ề4 cosa + sing)
Neglecting friction: | With friction: Neglecting friction: } With friction:
+, Sina Coefficient of fric- A Coefficient of fric-
_P=ầx cos 8 tion = yx = tang #=Wx;=H Xtand! tịon~ Ấ= tan ở
Trang 7
475 ~
BANG TINH LUC TREN MAT PHANG NGHIENG TABLE OF FORCES ON INCLINED PLANES
P required for moving a body on an inclined plane The friction on the plane is not taken into account The - 11w column headed Ộ Tension P in Cable per Ton of 2000
o ằ = PoundsỢ 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-
Ộ~ efficient of friction is known, the added pul? required
~_ ỔTbe table below makes it possible to find the force]
|<-ỞỞỞ-Ởt00ỞỞỞỞ Ở\| Tensions and Pressures in Pounds te overcome friction is thus easily determined:
Q X coefficient of friction = additional puil required
: Perpendic- Perpendic-
đêm Rise, Ft.| Angle a |Cable per sure @ om | Rise, Ft.| Anglea |Cable per sure O or | tpn [aie [areca] | net [alr be
Trang 8476 DON BAY LEVERS
Types of Levers Examples
Ở-Ở-ỞỞụỞ-Ở-_Ở.ỞaẨ A pull of 80 pounds is exerted at the
<i Ở end of the lever, at W;
) im 12 inches and L = 32 inches Ẽ
F Find the value of force F required to
- balance the lever
EĐ:W =i:L Fx L=oWxil / 80% 12 960
| Fe = ỞỞ w= zo pounds
F Wxi W FXL 32 32
=r =t IÍ F=Ư2o; W180; and f= 3;
how long must Z be made to secure
La Xe WX? } Fxa FX L| equilibrium?
W+FP FỖồ W+F W Le 180%3 Coy 20
t - Total length FE of a lever is 25 inches -ể->*+Ở-Ở-aỞỞ-Ở> A weight of go pounds is supported at
AA ỞỞẬ W7; tisroinches Find the value of F
F F = 22X19 56 pounds
f:W=i:L FXL=Wxt If F = 100 pounds, W = 2200 pounds,
F= wx! W = txE and a= 5 feet, what should L cqual to Ƒ + i secure equilibrium?
Wxa Wxlil Fxa FXL 2200% ậ
+ ý Ở F = F 3 Ậ f= W-F = W + 220oỞ roo ee 5.24 feet F = 28 pounds; L= 10 inches; ằ= e coe 3 24 inches What weight W can be 7 - supported? L l=a~+T= 24-+ 1O = 44 Ìnches : (28X10 F:Wel:L FxL=Wx!l W = = 8.23 pounds
W x2 FXL Let F= 12 tons; W = 4.5 tons; = E W= i @=16 feet Find ặ and ậ
5X 16
L Xa Wxil, Fxa FXL La TC S95 feats
F-W OF F-W OW | b= 16+9.6 = 25.6 feet
Bary Let W=Ưzo, P=3o, an Q=rs
| pounds; a= 4,6 = 7, and ằ = 10 inches
&\ fp RA F Tf x = 6 inches, find F
When three or more forces act en 20X% 4+ 30X7+15 X10
a lever: Fe 6 = 73% Ibs
Pxz=W*a+tPxi+Qxc Assuming F=20 in the example
ẤaWxszẨ+Px?+Qxc above, how long must lever arm x be
F made?
F WxXet+Pxb+QxXe =ỞỞỞỞỞỞ = 20X 44+30X7-+15 X 10 | - = 22 ins :
Trang 9477 KHỚP KHUỶU TOGGLE-JOINT Toggle-joint Ở If arms ED and EH are of unequal length: Fa P= F The relation between P and F changes _ÝỞ=Ở=Ở=Ở=Ở=Ở
constantly as F moves downwards h
If arms ED and EH are equal: |
Fe P
P.Ở 2h 6
A double toggle-joint does not increase the pressure exerted so long as the rela- tive distances moved by F and P remain the same
Toggle-joints with Equal Arms F = force applied; P = resistance! @ = given angle 3P sina = F cosa} re Se = coefficient; F 2sine or, P=F & coefficient tì Equivalent expressions: FS Fs P ae ; ah Ấ P T: 88 Per diagram =ỞỞ,
To use the table, measure angle Ủ, and find the coefficient in the table corre-
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 F:W=r:R The radius of a drum on which is
FXR=Wxr wound the lifting rope of a windlass is
F= Wxr 2 inches What force will be exerted
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 R= Wxr lifted? Here W = 2000; R= 12; r =2 F 2000 X 2 FXR Fo 5 = 333 pounds r= , F=}W
The velocity with F:W = seca :2 which weight W WX seca
w Ổwill be raised equals |
one-half the veloc- ity of the force ap- plied at F 1 a=number of strands or parts of rope (m, M8, etc.) F=u-xW n The velocity with which W will be raiser] equals - of the velocity of the force applied 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: F =} x 200 = 40 pounds of gears W = A, B, Cand D are the pitch circles Wxrxnxn #xRh:x&- PxRExRiXx ni
rxXnX rs Let the pitch diameters of gears A, B, Cand D be 30, 28, r2 and ro inches,
respectively Then Re= 1s: Rp = 14:
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:
2000 X 5 X 6X 4: 12 X14 X1 ~~ 95 pounds ]
Fu
Trang 11| 479
RONG ROC VI SAI - VIS
DIFFERENTIAL PULLEY - SCREW
Differential Pulley.-ỞỞIn the differential pul- ley a chain must be used, engaging sprockets, so as to prevent the chain from slipping over the pulley faces PxR=ẬVW(RỞr ,.V@-rn) r) Pe "oR Ẽ 2PR Waar
ỔForce Moving Body on Horizontal Plane Ở 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 FiỞ Qi
Screw Ở- 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:
? 6.2832 R
M Vu g=z#x
If Ừ is the coefficient of friction, then:
For motion in direction of load Q which assists it: 6.2832 wr Ở p j F C Xa +up Ẩ ẹ For motion apposite load Q which resists it: ` p + 6.2832 ur or 6.2832r Ởpp R -TRỞ-~> FeQ
Center of Gravity Ở The center of gravity of a body, volume, area, or line is that point at which if the body, volume, area, or line were suspended it would be perfectly balanced in all positions For symmetrical bodies of uniform material
it is aẠ 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
ỔThe tables that follow give such formulas for some of the more important shapằs 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: ằ = 2r +, Therefore, in thiscase, Ỉ
Trang 12480 TRONG TAM CENTER OF GRAVITY
Perimeter of a Triangle Ở If A, B and C are the middle points of the sides of the triangle, then the center of gravity is at the center of the circle that can be inscribed in triangle ABC The distance ở of the center of gravity from side a is: h (b+) , 2(z+b+c} where & is the height perpendicular to đ
Area of Triangle.Ở The center of gravity is at the intersection of lines AD and BE, which bisect
the sides BC and AC ỔThe perpendicular distance from the center of gravity to any one of the sides
is equal to one-third the height perpendicular to
that side Hence, ằ= h+ 3
Circular Arc Ở The center of gravity is on the
line that bisects the arc, at a distance
2 2
a - _ Ể +48 ) from the center of the circle
For an arc equal to one-half the periphery: đe 2P-++ = O,6366y For an arc equal to one-quarter of the periphery: a= arV 248 = 0.90037 For an arc equal to one-sixth of the periphery: = 37-+-e = 0.95497 Circular Arc (approximate) Ở a=mtk
This formula is very nearly exact for all arcs less than one-quarter of the periphery The error is only about one per cent for a quarter circle, and decreases for smaller arcs
Area of Trapezoid Ở ~Ở The center of gravity is on
the line joining the middle points of parallel lines AB aad DE c=*(a+2Đ) ị ỷ_*È(a+È) 3 (a+) 3 (o+5) a* + ab + 5* : 3 (6+)
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
Trang 134$1
TRONG TAM
CENTER OF GRAVITY
Perimeter of a Perallelogram Ở The center of gravity is at the intersection of the diagonals
Ayea of ề Parallelogram Ở The center of gravity is at the intersection of the diagonals
Any Four-sided Figure Ở 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 ỔThese lines form
& parallelogram EFGH; the intersection of the
diagonals EG and FA locates the required center
of gravity Ổ
Circle Segment Ở The distance of the center of gravity from the center of the circle is: be rd _2 rẬ sin? Ủ 114 3 A in which A = area of segment
Circle Sector Ở Distance 6 from center of grav- ity to center of circle is: b are re 8 y sin ce "Fi Ộ3473 -107 a in which A = area of sector, and @ is expressed degrees For the area of a half-circle: be art 37 = 0.42447 For the area of.a quarter circle: beg V 2X74 39 = 0.6002" i For the area of a sixth of # circle: : b=2r+TO.6366#
Trang 14482 TRONG TAM CENTER OF GRAVITY
Segment of an Ejlipse.Ở 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 xỂ
Area of a Porabola.Ở For the complete para- bolic area, the center of gravity is on the center line or axis, and Of 3h q om s àỞ đỞ Ở 5 For one-half of the parabola: = ~ 3h ề3% Ace +t đ s and 6 8
+ C Ậ _ For the compleroent area ABC:
Ậ c=o43b and đ=o.?ste
^ \
- ! 5 pherical Surface of Segments and Zones of ` Spheres - Distances ụ and Ọ which determine the
center of gravity, are: a: tẾ gu h bồ - a 2 2 tT Cylinder The center of gravity of a solid
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:
h #? tan? Ủ be Ấ tana
2 Sk Ở 4Ế
where a is the angle between the obliquely cut off surface and the base surface
Trang 15
-483 TRONG TAM CENTER GF GRAVITY
on Ởể ao Portion of Cylinder Ở For a solid portion of a poe ặ cylinder, as shown, the center of gravity is deter-
te we ể mined by: a
i g=ugyX3.t4r6r b= gy X 3.1416h
For the cylindrical surface only:
a=}*X 3.1416? b=Ậ}>3.t4r6 Ả
If the cylinder is bollow, the center of gravity of the solid shell is found by: H+Ở pe H*Ở }a eat -r! 3
Pyramid.ỞIn a solid pyramid the center of gravity is located on the line joining the apex with the center of gravity of the base surface, at a dis-
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 = Ậ À
Cone Ở- The same rules apply as for the pyramid
For the solid cone: a=Ậ}h For the conical surface: a=
Frustum of Pyramid Ở The center ef gravity is located on the line that joins the centers of gravity of the end surfaces If A1= area of base surface, | and 42 area of top surface, h(Ai+2VAixAr+3 4D 4(414 WAL xX Ar + Ad) ads
Frustum af Cone Ở The same rules apply as for Ổthe frustum of a pyramid For a solid frustum of 2) 4 circular cone the formula below is also used:
_ h{(R1!-+L + Rr + 3r?)
_Ở 4(Ể?*.t+rRr+r?)
ỞỞg- ' - | ỔThe location of the center of gtavity of the coni-
Ổ Ậ cal surface of a frustum of a cone is determined by:
[ a gw BR")
| fo _ g(R+n)
Trang 16484 TRONG TAM
CENTER OF GRAVITY
Wedge Ở The center of gravity is on the line
joining the center of gravity of the base with the middle point of the edge, and is located at:
a (25+)
Spherical Segment Ở The center of gravity of a
solid segment is determined by: | zẤ36rỞ#} 4(rỞ*) ễỷ2(ar-Đ 4(37Ở) For a balf-sphere, a = 8 = }r Half of a Hollow Sphere Ở The center of gravity is located at: b au 3 (RtỞ #4) ) 8 (R? Ở r)
Spherical Sector Ở The center of gravity of a
solid sector is at:
a= ậ (1+ cosa) r =f (27Ởh)
Segment of Ellipsoid or Spheroid.Ở The center
of gravity of a solid segment ABC, symmetrical
m about the axis of rotation, coincides with the center
ỞpỞ of gravity of the segment DBF of a sphere, the
diameter of which is equal to the axis of rotation of the spheroid
Pareboloid Ở The center of gravity of a solid
paraboloid of rotation is at:
a=th
Center of Gravity of Two Bodies Ở If the weights of the bodies are P and Q, and the distance between
Trang 17485
MECHANICS
Center of Gravity of Figures of any Outline Ở 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: Aa+ Bb+Ce A+B+4C As an example, assume that the area A is 24 square inches, B, 14 square inches, re J Y Y | er 0 P gl_O 4C _m~ ÔÔ F _ i c + H ỞmỞ- ra K H Qa N 1 coat) lth x { Ẩ tx xt ỞỞỞ x Y ` Fig 1 , Fig 2 and C, 16 square inches, and that a= 3 inches, 6 = 7.5 inches, and ằ = 12 inches Then: mm y= 24.% 37 3+ 14X75 xổ X12 = 369 = 6,83 inches, 24 + 14 -+ 16 4
If the figure, the center of gravity of which is to be found, is not symmetrical about any axis, then moments must be taken with relation to two axes XX and FY, as shown in Fig 2 The figure is divided into convenient geometrical figures, the centers of gravity of which can be easily found, the same as before, The center of
gravity is determined by the equations:
x Aa + Bh +Ca Aa+t Bb+Ce
A+B+CO 7Ợ"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 ằ= 11.5 inches Let a: = 6.5 inches, & = 8.5 inches, and ằ; = 7 inches Then:
pe EEK OS+IBXBS+20X7 _ a = 7.38 inches
14+ 18+ 20
4X 3+18X% 7+ 20X 11.5 398 ầ 144 184-20 sa 7.65 inches
In other words, the center of gravity is located at a distance of 7.65 inches from
Trang 18486 MOMENT QUAN TINH MOMENTS OF INERTIA
(Af = mass of body = weight 4.32.16)
Trang 20488 BAN KINH HO! CHUYEN RADIUS OF GYRATION
Bar of Small Diameter
_ Axis at end Thin Circular Disk
- Ợ Axis through center Cylinder
Jot Cylinder Axis, diameter at mid-
Axis through center tể k Ở| gn center length A k=0.57731 ott, + : A Ba dp Ì Ở>lhr ỞỞri > (fF) \ bel Axis at center A A FỞ-ỞtỞỞỞ a k = 0.289 V+ 3 r? , hm *È=ẹ.7or1f ỘTG k= 0.28861 , B= dy? È? = xử; P
Bar of Small Diameter, | thin Circular Disk Cylinder
bent to Circular Shape
Axis, a diameter of the ring A ` k=.7071F }?=Ậ r
Trang 21d&0 BÁN KÍNH HỘI CHUYÊN RADIUS OF GYRATION
Thin Hollow Cylinder ` ca
Axis, diameter at mid- Cylinder Parallelepiped
length Axis at a distance Axis at distance from end A "Ok ỘA ke Ỏ é A [er iw Ia -h Ởxe : : | ` = % wet = 0.289 Vi+6r? : ; : po, kaVa +$r ke 4P+ aa ee oe ee Pa at+ hr! 12 \ 12 2 Cone
Hollow Cylinder Rectangular Prism Axis at base
Longitudinal Axis Axis through center Aik a A 1 H_ệ#_ | A b&b ho A T 4 ' *#_ đo) | -2a- 3 kee = J2Ả*+ 3 r2 20 k= 0.7071 VW K2+r1 keois77 Vie +e #*= 3 (R? +?!) k*= (0+ c!) Âxis at apex wees k= V 125 +3 12h? + 34% 20 Hollow Cylinder
Axis Ộdiameter at mid- Paratlelepiped
, leneth Axis at one erd, central Cone
Trang 22490 BAN KINH HOI CHUYEN RADIUS OF GYRATION Frustum of Cone Axis at large end _ Sphere A f Axis at 2 distance ak, tps") pet To Ở sf ava? 4
B(R+sR+6A) 3 (Ea k~vettậr
to( tr Rr+pP 2o\K:Ở.4 b= a?+ Fr?
| Sphere
Axis its diameter Ellipsoid
A Axis through center a Fy ỘỞd 0Ộ k=o.632sr; kim grt ỉ k= 0.447 V 62408 Thin Spherical Shell Bad (te) k= 0.81657; btm ầrt Hollow Sphere
Axis its diameter Paraboloid
Trang 23
_ MATERIALS Ở
Trang 24492
MOMENT QUAN TINH ; MODUL MAT CAT
MOMENTS OF INERTIA, SECTION MODULUS, etc., OF SECTION
Aweiion Moment of Section Modulus | Radius of Gyration
y = distance from Inertia I i
Trang 25493
ỘMOMENT QUAN TINH: MODUL MAT CAT
MOMENTS OF INERTIA, SECTION MODULUS, etc., OF SECTION A Ỏ area | Section Modulus Radius of Gyration : HeỞ be-yl y = distance from _ Moment of Inertia I Ũ axis to ex- | I Z == r=\V/1 treme fiber + A Ẩ Z2 T WY ồ bo Ổ | Y | > ` a7 0.5774 eo Ẩ | Ở LAY aa Ì Yj ` _ -đj$ Ở hệt bd* Ở hk* 12 (bd Ở hk) l aren + 12 6d = 0.289 bd? Ở hk? A = bd ~ hk bd Ở hk _y=id " Ps s - : x bd? vậQỌ+đ) te - 6 (02+ 4) ậv?+ = 0.4) ẤTC Ở Ở vb:+ bd (at costa Ta r2 |
** 5 Btsinta) ieee) =o289 x 8 ff
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MOMENT QUAN TINH : MODUL MAT CAT | MOMENTS OF INERTIA, SECTION MODULUS, etc., OF SECTION
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495
_ MOMENT QUAN TINH : MODUL MAT CAT
MOMENTS OF INERT IA, SECTION MODULUS, etc., OF SECTION | Moen tof tia, "Seton Moda Radius of Gyration, : , Z=u- - _ : | I ; r 4 é (a? + 4.064 5) đề (a1 -Ẩ- 4 ab + B3) VW#t@+sa+s -36 (ụỦ+-đ) 12 (a+ 25) 18 (a+ 6) - T + 2 cost ay A a ạ + 2 cos? 30ồ) d* (1 + 2 cost? 30ồ) - 4.0088 30ồ 4 cos? 30ồ 48 cos? 30ồ si ồ = 0.06 dé = Q.12 đồ = 0.264 ằ 4 nợ] 4 đ (1-2 cost 3oồ) đề (r -}- 2 cost 30ồ) s 13 4.c0s* 30ồ 6.9 4 cos? 30ồ 48 cost 30ồ " = oo6 đt = 0.104 4? =o264d Ẩ _ Ợ = " `
_ ¡4+ Ở- 4 [d(t+ 2 cos 229ồ) q/ Zt 2 cost a2")
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MOMENT QUAN TINH : MODUL MAT CAT
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MOMENT QUAN TINH ; MODUL MAT CAT
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_MOMENT QUẦN TÍNH : MODUL MAT CẮT - on MOMENTS OF INERTIA SECTION MODULUS etc OF SECTION -
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_ MOMENT QUAN TINH : MODUL MAT CAT
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MOMENT QUAN TINH : MODUL MAT CAT
MOMENTS OF INERTIA, SECTION MODULUS etc., OF SECTION
Distance from Neutral
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MOMENT QUÁN TÍNH : MODUL MAT CAT
MOMENTS OF INERTIA, SECTION MODULUS etc OF SECTION
7 alle
| ion } Radius of Gyration,
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TINH CHAT CAC TIET DIEN KHUNG CAT
PROPERTIES OF SECTIONS FOR PUNCH AND SHEAR FRAMES
3 ẼPmớa ng -_ Z2 = Section Modulus for Compression;
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TÍNH CHAT CAC TIET DIEN KHUNG CAT
PROPERTIES OF SECTIONS FOR PUNCH AND SHEAR FRAMES KEỞỞ-ỞỞ0=10Ởy~Ở> si
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MODUL TIET DIEN CHU NHAT
SECTION MODULUS FOR RECTANGLES
Section modulus values shown afe for rectangles tinch wide To obtain section mod- ulus for rectangle of given length of side, multiply value in table by given width
Length | Section || Length | Section |{ Length | Section [| Length | Section
of Side | Modulus |] of Side [ Modulus |] of Side | Modulus || of Side | Modulus 14 0.0026 234 1.26 12 24.00 25 104.2 ta 0.0059 3 1.50 12H 26.04 26 112.7 1 0.0104 314 1.76 13 28.17 27 141.5 Ho 0.0163 3}4 2.04 1344 30.38 28 130.7 1% 0.0234 344 2.34 14 32.67 20 140.2 He 0.032 4 2.67 14}8 35.04 3o 189 44 0.042 44 3.38 15 37.5 _ 32 TT 56 0,065 5 4.17 154 42.0 34 193 4 0.094 544 5.04 16 42.7 36 216 L1 0.128 6 6.00 1644 45.4 38 241 I 0.167 614 7.04 17 48.2 4o 267 144 O.211 7 8.17 1734 51.0 42 204 1)Á 0.260 74 9.38 18 54.0 44 323 134 0.315 8 10.67 18% 57.0 46 353 14 0.373 844 12.04 19 60.2 48 384 1% 0.440 9 13.50 19}ằ 63.4 so All 1% 0.510 gts 15.04 20 66.7 52 451 1% 0.586 10 16.67 21 73.5 54 486 2 0.67 1oẬ4 18.38 22 Bo.7 số 523 214 0.84 1I 20.17 23 88.2 sẽ Sốt 24% 1,04 1154 22.04 24 g6.0 6a 6oo
MODUL TIET DIEN VA MOMENT QUAN TINH
CUA TRUC TRON
SECTION MODULUS AND MOMENTS OF INERTIA
FOR ROUND SHAFTS |
Section | Moment Section | Moment ` Section | Morfient
Diam.) Modulus | of Inertia |] Di#Ỏ-|Modutue] of Inertia |] >18Ỏ-/Modulus| of Inertiz |
16 O0.00010 | O.oOoOT ?Ở7ậa | 0.0074 | 0.00755 2349 | 0.02364 0.01308
344 | 0.00027 | 0.00002 Ha | 0.0082 [| o.ooIifo 4764 | 0.0388 0.01425
543 | 0.00037 ] 0.00003 32964 {| 0.0091 | 0.00207 S4 0.0413 | O.OT55O
IMGs | 0.00050 | 0.00004 4542 | o.ofor } 0.00237 Ộ+ | 0.0440 0.07r684
He | 0.00065 | 0.00006 8iệa | o.oIIT Ì o.ooz7o 2363 | 0.0467 0.08825
1344 | 0.0008 | o.oooo8 1á O.0I23 | 0.00306 Sigs | 0.0496 0.01976
43 1 0.00102 | 0.OOOIT tiệt | 0.0134 | 0.00346 1%{ea | 0.0526 9.02135 ihe | 0.00126 | 0.00015 14a { 0.0147 | 0.00390 5344 | 0.0557 0.62905 Mệ 0.00153 | O0.ooot9 8544 | 0.0160 | 0.00438 742 | 0.0588 0.02483 Wha | 0.00183 | 0.00024 Mie | 0.0174 | 0.00498 5564 | 0.0622 0.02673 Ms | 0.00218 | o.cc03r || 2764 | ẹo.o1Bg | 0.00547 1á 0.0656 0.02872 186a | 0.00256 | 0.00038 1944 | O.0205 | 9.00609 5744 | 0.0602 0.03083 Ma | 0.00299 | 0.00047 3964 | 0.0222 ] 0.00676 #842 | 0.0728 | 0.03305 Ộtiệt | 0.00344 | 0.00057 Bề 0.0239 | 0.00748 5Qằ4 [| 0.0767 0.03539 Thậ2 | 0.00398 | 0.00068 4lfa | 0.0258 | 0.00825 1344 | 0.0807 6.03785 2344 | 0.0045 0.00082 2343 | 0.0277 | 0.00909 Slé_ | 0.0849 | 0.04044 % 0.0052 0.00097 tia | 0.0207 | 0.00999 || 3449 | 0.0892 | 0.04316
3544 | 0.0058 | o.ooIt4 IMG | 0.0318 | o.oTIoQs 684 | O.0934 | 0.04601
lãẩệy | o.oo66 0,00133 4564 | 0.0341 | 0.01198 vas kas
In this and succeeding tables, the Polar Section Modulus fora shaft of given diameter
can be obtained by multiplying its Section Modulus by 2 Similarly its Polar Moment
of Inertia can be obtained by multiplying its Moment of Inertia by a
Trang 37505 MODUL TIET DIEN VA MOMENT QUAN TINH CUA TRUC'TRON
SECTION MODULUS AND MOMENTS OF INERTIA
FOR ROUND SHAFTS
_ | Section | Moment : | Section | Moment : Section | Moment
Trang 38506 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
Trang 39807 MODUL TIET DIEN VA MOMENT QUAN TINH 4.40 pb @ NH bah hasaee SABRE SHS BS
CUA TRUC TRON
SECTION MODULUS AND MOMENTS OF INERTIA FOR ROUND SHAFTS
Section | Moment Section | Moment 3 Section | Moment
Trang 40TIEU CHUAN Mầ 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
per foot, thus: Ị
: tạp 9LI13.4 wa: min T | Ậ
Indicates a 9x 214-inch channel weighing 13.4 x $
pounds per foot x
Flange % Axis X-X Axis ầ-Y ae | a8 3 4 3 Bà an ễ ỔBros | Đ)8| Ế oo? TER & gs g5 s1| 7 |Zz|zr |r |zl|r |z = af = In | Lb f In | In | In | In | Ie | In [In] In ] In | Int] In | In *rt8x4| s8io | 16.98 | 18.00 | 4.200 | 625 | 700 | 670.7 | 74.5 | 6.29) 18.5 | 5.6 | 1-04] 88 51.9 {15.18 | ầ8.co | 4.100 | 625 | 600 | 622.1 | 69.2 | 6.40] 17.1 | 5.3 | ặ06] 87 45.8 | 13.38 | 18.00 | 4.000 | 62ậ | S00 | 573.5 | 63.7 | 6.55 | 15.8 | 5.5 Ậ E.oo | 8g 42.7 | 14, 18.00 | 3.950 | 625 | 450 | 549.2 | 61.0} 6.64) 15.0 | 4.9 | 1.10] 90 15 x354] 50.0 | 14.64 | 15.00 | 3.716 | 650 | 716 | 401.4] 53.6 | 5.24] u.2 | 3.8 87 | 80 40.0 | 13.70 | 25.00 | 3.520 | 650 | S20 | 346.3 | 46.2] 5:44] 9.43 |3.4 | 8ò | 78 33.9 9.90 } 1ậ.00 ] 3.400 } 650 } 400 | 3x2.6 | 41.7 | 5.62} 8.2 | 3.2 ;0I | 79 1213 lãảo.o 8.9 | 14.00 | 3-17O | SOI | 519 | 161.2 | 20.0 |4.2B[ s.2 | 2.1 Ộ77 | 68 25.0 7.32 | ặ2.00 | 3.047 | Sor | 387 | 143.5 | 23-91 4.43] 4.5 | 1.9 -79 | 68 20.7 6.09 ] E2.00 | 2.940 | Sor } 280 | 128.1 | 21.4 | 4.65 | 3.9 [1.7 Br) 70 tox 256] 30.0 | 8.80] 10.00 | 3.033 | 436 | -673 | 103.0 | 20.613.42] 4.0 | 1.7 -67 | 65 25.0 | 7.33 | 10.00 | 2.886 | 436 | 526 | 90.7 | 18.1 13.52 | 3.4 [1.5 68 | 62 20.0 5.86 | 10.00 | 2.730 | 436 |.379 | 78.5 | 15.7 | 3.66 | 2.8 |] 71.3 70 | ỐT 15.4 | 4.47 | 10.00; 2.600 | 436 | 240.) 66.9 | 13.4/3.87!{ 2.4 |1.2 | 724 64 Ổgx 234 | 20.0 5.86} 9.00 | 2.648 | 413 | 448 | 60.65 13.5] 3.22) 24 | ặ2 -65ồ} 59 15.0 4.39 | 9.oo | 2.485 -285 | 3o.7 Í 11.3 | 4.40 Ì T1.0 {1.0 67 Ì 59 l 13.4 3.89 | 9.00 | 2.430 2301 47.3] f0.5 | 3.497] 1.8 07 | 67] 61 8x244 | 18.75} 5.49} 8.00] 2.527 487 | 43.7 | 10.9 12.82] 2.0 i 45.0 -60 | 57 13.78 | 4.02 | 8.00 | 2.343] -303|ồ35.8} 9.0]2.99! 1.5 | 86| 62], 56 15,5 3.36 | 8.00 | 2.260 | -220] 32.3] 82)3.10] 1.3 70| 03]-.s8 413 413 390 390 390 1123 | 14.75 | 4.32} 7.00 | 2.209] 366| 4rg | 27.1 | 7.7|2.5L| 1-4 | -79] 57] -53 12.25} 3.58| 7.00 | 2,194} 366} 314| 24.1| 6.9 |2.50 | 1.2 7t1| -58] 53 Ấ0-5 2.85 | 7.00 | 2.090 | 366 | 21O0 | 21.1 | 6.0 | 2.72 98) 63) $91 55 343 343 343 320 320 296 296 6x2 | 13.0 | 3.81 6.00] 2.157 437? | 17.3] ậ.8)2.13] 2.1 45] 53] 52 10.5 ậ 3.07] 6.oo | 2.034 3J14 | 15.1] 5.O | 2.22 87] 57| 53] 50 8.2 2.39 4 6.00 | I.920] -206 | 13.0] 4.3 | 2.34 7o| 50] 54 | 52 ậsxij(| 0o | 2/63 | 5.00 | 1.885 325| 88] 3511.83] 64| 43| 40 | 4B 6.7 1.95 | 5.00 | 1.750 Igo 7.44 3.017.905 431 38 | so| 49 4xIỷÊ| 7.25] 2.12 | 4.00] 1.720] -320 4.5] 2.3) 147 44] 3s | -46 | 46 5.4 1.56 | 4.00 1.580 1Ro 3.8 | 12.6) 3.56 -Ề32| 29| 45 | 46 4114| 6.0 1.75] 3.00 | 1.596 | 273 | 356 2.1} 1.4] 1.08 31] 27 1 42] 46 5Đ | 1.46| 3oo | 1.498 |.273|.298| I1.8| 1.2|1.12| 25] 34 | 4xX |-.44 41 | t.10| 3/00 |I.410|.273|.172| 1.6| 1I.Ẩ|I.17| 20| 2E| 4r | 44
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