Table 5.2.2 Beams of Uniform Cross Section, Loaded Transversely (Continued )

Một phần của tài liệu Strength of Material (Trang 25 - 29)

Concentrated load W

Uniformly dist . load Wwl; cc1 R1⫽Wc1

lW2

2 R2⫽Wc

lW 2 (a) W

Wc1⫺c 2c Mmax⫽R2

x1 2⫽R22l

2W,冉x1⫽RW2l

(b) WWc1⫺c

2c

Mmax⫽冉W⬘ ⫹W2冊ccl1, (x1⫽c1)

Deflection of beam under W⬘: f⫽冉W⬘ ⫹l28cccc11

Wc3EIl2c21

cc1

R1⫽W⬘(3cc1)c21 l3 ⫹W

2 R2⫽W⬘(c⫹3c1)c2

l3 ⫹W 2 Mmax⫽M1⫽Wcc21

l2 ⫹Wl 12 Deflection under W

f⫽ 1

EIWc3l3c331⫹Wc24l2c21冊

Table 5.2.3 Uniformally Distributed Loads on Simply Supported Rectangular Beams 1-in Wide*

(Laterally Supported Sufficiently to Prevent Buckling)

[Calculated for unit fiber stress at 1,000 lb/in2(70 kgf/cm2): nominal size]

Total load in pounds ( kgf )† including the weight of beam

Depth of beam, in (cm)§

Span, ft

(m)‡ 6 7 8 9 10 11 12 13 14 15 16

5 800 1,090 1,420 1,800 2,220 2,690 3,200 3,750 4,350 5,000 5,690

6 670 910 1,180 1,500 1,850 2,240 2,670 3,130 3,630 4,170 4,740

7 570 780 1,010 1,290 1,590 1,920 2,280 2,680 3,110 3,570 4,060

8 500 680 890 1,120 1,390 1,680 2,000 2,350 2,720 3,130 3,560

9 440 600 790 1,000 1,230 1,490 1,780 2,090 2,420 2,780 3,160

10 400 540 710 900 1,110 1,340 1,600 1,880 2,180 2,500 2,840

11 360 490 650 820 1,010 1,220 1,450 1,710 1,980 2,270 2,590

12 330 450 590 750 930 1,120 1,330 1,560 1,810 2,080 2,370

13 310 420 550 690 850 1,030 1,230 1,440 1,680 1,920 2,190

14 290 390 510 640 790 960 1,140 1,340 1,560 1,790 2,030

15 270 360 470 600 740 900 1,070 1,250 1,450 1,670 1,900

16 250 340 440 560 690 840 1,000 1,170 1,360 1,560 1,780

17 230 320 420 530 650 790 940 1,100 1,280 1,470 1,670

18 220 300 400 500 620 750 890 1,040 1,210 1,390 1,580

19 210 290 380 470 590 710 840 990 1,150 1,320 1,500

20 200 270 360 450 560 670 800 940 1,090 1,250 1,420

22 180 250 320 410 500 610 730 850 990 1,140 1,290

24 160 230 290 370 460 560 670 780 910 1,040 1,180

26 150 210 270 340 420 520 610 720 840 960 1,090

28 140 190 250 320 390 480 570 670 780 890 1,010

30 130 180 240 300 370 450 530 630 730 830 950

* This table is convenient for wooden beams. For any other fiber stress S⬘, multiply the values in table by S⬘/1,000. See Sec. 12.2 for properties of wooden beams of commercial sizes.

† To change to kgf, multiply by 0.454.

‡ To change to m, multiply by 0.305.

§ To change to cm, multiply by 2.54.

the section with respect to the X axis. When this principal axis has been found, the other principal axis is at right angles to it .

Calling the moments of inertia with respect to the principal axes Ix

and Iy, the unit stress existing anywhere in the section at a point whose coordinates are x and y (Fig. 5.2.33) is SMy cos␣/IxMx sin␣/Iy, in which M⫽bending moment with respect to the section in question,

␣⫽the angle which the plane of bending moment or the plane of the

loads makes with the y axis, M cos␣⫽the component of bending moment causing bending about the principal axis which has been desig- nated as the X axis, M sin␣ ⫽the component of bending moment causing bending about the principal axis which has been designated as the Y axis. The sign of the two terms for unit stress may be determined by inspection in the usual way, and the result will be tension or com- pression as determined by the algebraic sum of the two terms.

Table 5.2.4 Approximate Safe Loads in Pounds (kgf) on Steel Beams,*Simply Supported, Single Span

Allowable fiber stress for steel, 16,000 lb/in2(1,127 kgf/cm2) (basis of table) Beams simply supported at both ends.

L⫽distance between supports, ft (m) a⫽interior area, in2(cm2) A⫽sectional area of beam, in2(cm2) d⫽interior depth, in (cm) D⫽depth of beam, in (cm) w⫽total working load, net tons ( kgf )

Greatest safe load, lb Deflection, in

Shape of section Load in middle Load distributed Load in middle Load distributed Solid rectangle 890AD

L

1,780AD L

wL3 32 AD2

wL3 52 AD2 Hollow rectangle 890(ADad )

L

1,780(ADad ) L

wL3 32(AD2⫺ad2)

wL3 52(AD2⫺ad2)

Solid cylinder 667AD

L

1,333AD L

wL3 24AD2

wL3 38AD2 Hollow cylinder 667(ADad )

L

1,333(ADad ) L

wL3 24(AD2⫺ad2)

wL3 38(AD2⫺ad2)

I beam 1,795AD

L

3,390AD L

wL3 58AD2

wL3 93AD2

Fig. 5.2.32 Fig. 5.2.33

In general, it may be stated that when the plane of the bending mo- ment coincides with one of the principal axes, the other principal axis is the neutral axis. This is the ordinary case, in which the ordinary formula for unit stress may be applied. When the plane of the bending moment does not coincide with one of the principal axes, the above formula for oblique loading may be applied.

Internal Moment Beyond the Elastic Limit

Ordinarily, the expression MSI/c is used for stresses above the elastic limit , in which case S becomes an experimental coefficient SR, the modulus of rupture,and the formula is empirical. The true relation is obtained by applying to the cross section a stress-strain diagram from a tension and compression test , as in Fig. 5.2.34. Figure 5.2.34 shows the side of a beam of depth d under flexure beyond its elastic limit; line 1 – 1 shows the distorted cross section; line 3 – 3, the usual rectilinear

Table 5.2.5 Coefficients for Correcting Values in Table 5.2.4 for Various Methods of Support and of Loading, Single Span

Max relative deflection under Max relative max relative safe

Conditions of loading safe load load

Beam supported at ends:

Load uniformly distributed over span 1.0 1.0

Load concentrated at center of span 1⁄2 0.80

Two equal loads symmetrically concentrated l/4c

Load increasing uniformly to one end 0.974 0.976

Load increasing uniformly to center 3⁄4 0.96

Load decreasing uniformly to center 3⁄2 1.08

Beam fixed at one end, cantilever:

Load uniformly distributed over span 1⁄4 2.40

Load concentrated at end 1⁄8 3.20

Load increasing uniformly to fixed end 3⁄8 1.92

Beam continuous over two supports equidistant from ends:

Load uniformly distributed over span

1. If distance a⬎0.2071l l2/(4a2)

2. If distance a⬍0.2071l l

l⫺4a

3. If distance a⫽0.2071l 5.83

Two equal loads concentrated at ends l/(4a)

NOTE: l⫽length of beam; c⫽distance from support to nearest concentrated load; a⫽distance from support to end of beam.

BEAMS 5-27 Table 5.2.6 Properties of Various Cross Sections*

(I⫽moment of inertia; I/c⫽section modulus; r⫽√I/A⫽radius of gyration)

N.A.

Ibh3 12

bh3 3

b3h3 6(b2⫹h2)

bh

12(h2cos2ab2sin2a) I

cbh2 6

bh2 3

b2h2 6√b2⫹h2

bh

6 冉h2h cos acos2a⫹⫹bb sin a2sin2a

rh

√12⫽0.289h h

√3⫽0.577h bh

√6(b2⫹h2) √h2cos2a12⫹b2sin2a

Ib

12(H3⫺h3) H4⫺h4

12

H4⫺h4 12

bh3 36; c⫽2

3h I

cb 6

H3⫺h3 H

1 6

H4⫺h4 H

√2 12

H4⫺h4 H

bh2 24

r⫽√12(HH3⫺⫺hh)3 √H212⫹h2 √H212⫹h2 √18h

N.A.

Ibh3 12

5√3

16 R4 1⫹2√2

6 R4

I cbh2

12

5⁄8R3 5√3

16 R3 0.6906R3

rh

√6 √245 R 0.475R

NOTE: Square, axis same as first rectangle, side⫽h; Ih4/12; I/ch3/6; r⫽0.289h.

Square, diagonal taken as axis: Ih4/12; I/c⫽0.1179h3; r⫽0.289h.

Fig. 5.2.34

relation of stress to strain; and line 2 – 2, an actual stress-strain diagram, applied to the cross section of the beam, compression above and tension below. The neutral axis is then below the gravity axis. Theouter material may be expected to developgreater ultimate strengththan in simple stress, because of the reinforcing action of material nearer the neutral axis that is not yet overstrained. This leads to anequalization of stress over the cross section. SRexceeds the ultimate strength SMin tension as follows: for cast iron, SR⫽2 SM; for sandstone, SR⫽3 SM; for concrete, SR⫽2.2 SM; for wood (green), SR⫽2.3 SM.

In the case of steel I beams, failure begins practically when the elastic limit in the compression flange is reached.

Because of the support of adjoining material, theelastic limit in flexure Spis also greater than in tension, depending upon the relation of breadth to depth of section. For the same breadth, the difference decreases with

Table 5.2.6 Properties of Various Cross Sections* (Continued ) Equilateral Polygon

A⫽area

R⫽rad circumscribed circle

r⫽rad inscribed circle n⫽no. sides a⫽length of side Axis as in preceding

section of octagon

IA

24(6R2⫺a2)

A

48(12r2⫹a2)

AR2 4 (approx)

I cI

r

I R cos180°

n

AR 4 (approx)

AI⫽√6R224⫺a2⬇R2

⫽√12r248⫹a2

I6b2⫹6bb1⫹b21 36(2bb1) h3 c⫽1

3 3b⫹2b1

2bb1

h

I

c⫽6b2⫹6bb1⫹b12

12(3b⫹2b1) h2 rh√12b2⫹12bb1⫹2b21 6(2bb1)

IBH3⫹bh3 12 I

cBH3⫹bh3 6H

r⫽√12(BHBH3⫹⫹bhbh)3

IBH3⫺bh3 12 I

cBH3⫺bh3 6H

r⫽√12(BHBH3⫺⫺bhbh)3

I⫽1⁄3(Bc31⫺B1h3⫹bc33⫺b1h31) c1⫽1

2

aH2⫹B1d2⫹b1d1(2Hd1) aHB1db1d1

r⫽√Bdbd1⫹Ia(hh1)

I⫽1⁄3(Bc31⫺bh3⫹ac32) c1⫽1

2

aH2⫹bd2 aHbd c2⫽Hc1

r⫽√Bda(HId)

I⫽␲d4 64 ⫽␲r4

4 ⫽A 4r2

⫽0.05d4(approx)

I c⫽␲d3

32 ⫽␲r3 4 ⫽A

4r

⫽0.1d3(approx)

AIr2⫽d4

increase of height . No difference will occur in the case of an I beam, or with hard materials.

Wide plates will not expand and contract freely, and the value of E will be increased on account of side constraint . As a consequence of lateral contraction of the fibers of the tension side of a beam and lateral swelling of fibers at the compression side, the cross section becomes distorted to a trapezoidal shape, and the neutral axis is at the center of gravity of the trapezoid. Strictly, this shape is one with a curved perime- ter, the radius being rc/␮, where rcis the radius of curvature of the neutral line of the beam, and␮is Poisson’s ratio.

Deflection of Beams

When a beam is subjected to bending, the fibers on one side elongate, while the fibers on the other side shorten (Fig. 5.2.35). These changes in length cause the beam to deflect . All points on the beam except those directly over the support fall below their original position, as shown in Figs. 5.2.31 and 5.2.35.

Theelastic curveis the curve taken by the neutral axis. The radius of curvature at any point is

rcEI/M

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