习题解答7(平面应力状态 强度理论)

习题解答7(平面应力状态 强度理论)

平面应力状态 强度理论习题解答

习题解答7(平面应力状态 强度理论)

P92 49-1 危险截面(内力分布) 危险点(应力分布)AB C

F

T0

2T0

3T0

T0 T0

tF

BF

A

Mn Mn = 2T0 tmax = 3 p d WP W = P 16横截面位置 应力方向

tC

sFN s= FN = F A

F s= A Mn tmax = WP

sMn = T0 pd3 WP = 16

习题解答7(平面应力状态 强度理论)

P92 49-2(a) 应力表示(方向 数值)50 MPa 52.32

20 MPa 30 MPa

sx = 30 MPa

sy = 50 MPa

30º

18.66

tx = - 20 MPa2

a = 30ºcos2a – tx sin2a= 52.32 MPa

sa =ta =

sx + sy sx - sy2

+

sx - sy2

sin2a + tx sin2a

= - 18.66 MPa

习题解答7(平面应力状态 强度理论)

P92 49-2(b)20 MPa 27.32 40 MPa 27.32

sx = - 40 MPa tx = 20 MPa2+

sy = 0 a = 60º

30º

sa = ta =

sx + sy sx - sy2

sx - sy2

cos2a – tx sin2a = - 27.32 MPa

sin2a + tx sin2a

= - 27.32 MPa

习题解答7(平面应力状态 强度理论)

P93 49-3(a) 解析法 sx = 50 MPa20 MPa

11.65 54.82

50 MPa

sy = 0 tx = - 20 MPa a = 30º sx + sy sx - sy sa = + cos2a – tx sin2a2 2 = 54.82 MPa sx - sy ta = sin2a + tx sin2a = 11.65 MPa 2 = 0.8 a0 = 19.33º a0' = 109.33º

- 2t x tan2a0 = sx - sy 57.02 MPa sx - sy 2 sx + sy ± ( ) + tx 2 = s max = - 7.02 MPa 2 2 min s1 = 57.02 MPa s2 = 0 s3 = - 7.02 MPa 57.02 MPa 32.02 MPa sx - sy 2 ) + tx 2 = t max = ± ( 19.33º - 32.02 MPa 2 min 7.02 MPa sx - sy tan2a1 = = 1.25 a1 = 64.33º 2t x a1' = - 25.67º 64.33º sx + sy s a1 = = 25 MPa 25 MPa 32.02 MPa 2

30º

习题解答7(平面应力状态 强度理论)

P93 49-3(c) sx = - 40 MPa = tx sy = - 20 MPa a = - 60º 20 MPa sx + sy sx - sy 40 MPa s = + cos2a – tx sin2a a 59.64 2 2 28.66 40 MPa = - 59.64 MPa sx - sy ta = sin2a + tx sin2a = 28.66 MPa 30º 2 - 2t x = - 4 a0 = 52.02º a0' = - 37.98º tan2a0 = sx - sy 11.23 MPa sx - sy 2 sx + sy 2 ± ( ) + tx = s max = - 71.23 MPa 2 2 min 71.23 MPa s1 = 11.23 MPa s2 = 0 s3 = - 71.23 MPa 11.23 MPa 41.23 MPa sx - sy 2 2 ) + tx = 52.02º t max = ± ( - 41.23 MPa 2 min s - sy 30 MPa tan2a = x = 0.25 a1 = 7.02º 1 2t x a1' = 97.02º sx + sy 7.02º s = = - 30 MPa 41.23 MPa a1 2

习题解答7(平面应力状态 强度理论)

P93 49-3(e)sx = - 20 MPa sy = 30 MPa tx = 20 MPa a = - 30º 30 MPa sx + sy sx - sy + cos2a – tx sin2a 20 MPa sa = 2 2 30º 20 MPa = 9.82 MPa s s x y 9.82 ta = sin2a + tx sin2a = 31.65 MPa 31.65 2 - 2t x tan2a0 = = 0.8 a0 = 19.33º a0' = 109.33º sx - sy 37.02 MPa sx - sy 2 sx + sy 2 ± ( ) + tx = s max = - 27.02 MPa 2 2 min s1 = 37.02 MPa s2 = 0 s3 = - 27.02 MPa 27.02 MPa 32.02 MPa smax + smin )= 19.33ºt max = ±( - 32.02 MPa 2 min 37.02 MPa 5 MPa a1 = a0 + 45º= 64.33º a1' = - 25.61º sx + sy 64.33º s a1 = = 5 MPa 2 32.02 MPa

习题解答7(平面应力状态 强度理论)

2 bh P93 49-4 M = 10 kN· m Wz = = 100 cm3 6 h/ 2 a M M smax = = 100 MPa = sd = - sa = 2sc Wz FS 3 bh b h/4 FS = 120 kN Iz = = 500 cm4 c 12 * = 75 cm3 Sz*3 = 56.25 cm3 d Szmax * * F · S FS· Szmax z S = 22.5 MPa = 30 MPa = tb tc = 3 tmax = Iz· b Iz· b s = 100 MPa t = 30 MPa b a

s1 = s2 = 0 s3 = - 100 MPa s = 30 MPa = - s s = 0 1 2 3 t = 22.5 MPa s = 100 MPa s = 50 MPa cd

s1 = 58.6 MPa s2 = 0 s3 = - 8.6 MPa

s1 = 100 MPa

s2 = s3 = 0

习题解答7(平面应力状态 强度理论)

P94 50-1(b) 图解法sx = 0 = sy25 MPa 25

tx = 25 MPa sC = 0

a = 45º

tD 0 Dx 22 a a 0 B C O Da Dy 45º A

R = 25 MPa

45º25 MPa

s

sa =

- 25 MPa s1 = 25 MPas3 = - 25 MPa tmax = 25 MPa

ta = 0s2 = 0 a0 = 45º a1 = 0º几何关系

25 MPa

25 MPa

s a1 = s C = 0对应关系

25 MPa

习题解答7(平面应力状态 强度理论)

P94 50-1(d) sx = 0 sy = - 80 MPa tx = 20 MPa80 MPa 20 MPa 77.32 Da 24.64

a = 60º

tD0

sC = - 40 MPa R = 44.72 MPa1 a0 = - 13.28º tan2a0 = 2 s sa = - 77.32 MPa

30º

2a 2a 1 Dx 33.44º 2a0 A B C O 20 MPa Dy

ta = 24.64 MPa s2 = 0

84.72 MPa 13.28º 4.72 MPa

s1 = sC + R = 4.72 MPa s3 = sC - R = - 84.72 MPa

31.72º 40 MPa 44.72 MPa

tmax = R = 44.72 MPa a1 = 31.72º sa1 = sC = - 40 MPa

习题解答7(平面应力状态 强度理论)

P94 50-2

E = 200 GPa

n = 0.25 正应变 线应变Mn = - T0 = - 2.5 kN· m pD3 WP = = 42.41 cm3 16 Mn tx = = - 58.95 MPa WP sx = 0 = sy a = 30º

A

T0a

A

tx

单元体 而 sz = 0

s30º= -t sin(60º ) = 51.05 MPa s-60º= -t sin(-120º ) = - 51.05 MPa1 e30º= (s30º -ns-60º ) = 319×10–6 E无量纲

广义虎克定律:

习题解答7(平面应力状态 强度理论)

P95 50-3a

A 60º

F

3 s-30º= + cos(-60º ) = sx A 2 4 2 sx sx 1 sx s60º= + cos(120º ) = sx 2 4 2 而 sz = 0 广义虎克定律: 1 3- n e-30º= (s-30º -ns60º )= sx = 540×10–6 E 4E 4E sx = e-30º= 160 MPa 3- n 则 F = sx· A = 50 kN

F sx = A

s y = 0 = txsx

a = - 30º

sx

习题解答7(平面应力状态 强度理论)

P95 50-4 全面校核

危险截面: E A(B) C左 (D右) q 3 危险点: 1 2 F F 查表 : 导学篇 附录B-3 2 A C B D z I = 5278 cm4 W = 422.72 cm3 E z 3 * z FA =210 kN= FB * l y 1 Szmax = 246.3 cm3 Sz3 = 183.4 cm30.1l 210 2088 FS 图(kN) 8 208 M 图(kN· m) 210

I25 b

0.1l

h =250 mm b =118 mm tw =10 mm t =13 mm

内力图2

41.8 41.8 45 s r 3 s 1 s 3 =169.7MPa [s ] >

= 210 kN F A 截面 : Smax * FSmax · Szmax = 98.0 MPa < [t ] tmax = t Iz· tw m s C 截面: M = 41.8 kN· 3 左 FS = 208 kN = 72.35 MPa = 88.70 MPa st s1 =s129.2 MPa 3 = - 40.5 MPa=153.5 MPa 不满 …… 此处隐藏：2586字，全部文档内容请下载后查看。喜欢就下载吧 ……