基礎(chǔ)工程力學(xué)2作業(yè)

基礎(chǔ)力學(xué)2作業(yè)（7-11章）P153

7-1（b）試作桿的軸力圖，并指出最大拉力和最大壓力的值及其所在的橫截面（或這類(lèi)橫截面所在的區(qū)段）。10kN20kN10kN最大拉力為20kN，在CD段；最大壓力為10kN，在BC段。1m1m1m10kN20kN30kN20kNABCD解：P153

7-2試求圖示直桿橫截面1-1、2-2和3-3上的軸力，并作軸力圖。如橫截面面積A=200mm2，試求各橫截面上的應(yīng)力。=

-100MPa-

20

·103s

1

=

200·10-6-10

·103=

-50MPas

2

=

200

·10-610

·103=

50MPas

3

=

200

·10-6a20kN

10kN120kN13a2a3

210kN解：10kN20kNP154

7-5鉸接正方形桿系如圖所示，各拉桿所能安全地承受的最大軸力為[FNt]=125kN,壓桿所能安全地承受的最大軸力為

[FNc]=150kN,試求此桿系所能安全地承受的最大荷載F的值。FFABCDaaF2=

F2FAC

=

FBCFACAC、BC、AD、BD均為拉桿，故22

=

176.75kN2

F

￡

125kN

F

￡

125AB為壓桿，故

F

￡

150kN所以Fmax

=150kN解：根據(jù)對(duì)稱(chēng)性只分析A、C點(diǎn)C點(diǎn)BCFA點(diǎn)FACFADFAB由靜力平衡方程得所以AB由靜力平衡方程得

F

=

FP155

7-8橫截面面積A=200mm2的桿受軸向拉力F=10kN作用，試求斜截面m-n上的正應(yīng)力及切應(yīng)力。F=10kN300mna020=

37.5MP32

2

(

)=

50

·-300s

=

s

cos

-

30=

50MPaF

F

10

·103=A

A

200

·10-6s

0

=

N

=322sin

600=

-50

· =

-21.7MPa4sin(-

2

·

300

)=

-s

0t

=

s-300

0解：P155

7-10

等直桿如圖示，其直徑為d=30mm。已知F=20kN,l=0.9m,E=2.1×105MPa，試作軸力圖，并求桿端D的水平位移ΔD以及B、C兩橫截面的相對(duì)縱向位移ΔBC。20kN-20kN20kN2F

2F

F1l/3

1323

2EAEA

EAF

l

F

l

F

lD211=

0.04mm2.1·10

·

4

·

0.0320

·103

·

0.3=

N

1

1

+

N

2

2

+

N

3

3

=Dp112BCF

lEAp-20

·103

·

0.3D=

N

2

2

==

-0.04mm2.1·10

·

4

·

0.03l/3l/3AB

CD解：P156

7-1312BDl2B``

B`解：1、以B點(diǎn)為研究對(duì)象畫(huà)受力圖2、列靜力平衡方程3、2桿應(yīng)力與s

p

比較4、求出2桿變形Dl25、畫(huà)變形圖即是B的位移。6、BB``P156

7-14

直徑為d=0.3m,長(zhǎng)為l=6m的木樁，其下端固定。如在離樁頂面高1m處有一重量為P=5kN的重錘自由落下，試求樁內(nèi)最大壓應(yīng)力。已知木材E=10×103MPa，如果重錘驟然放在樁頂上，則樁內(nèi)最大壓應(yīng)力又為多少？參照P138例題7-10APl5

·103

5

·103

·

6=

1

+

1

+2hEA

P1

+=

1

+

p

·

0.1522

·1·10

·109

·p

·

0.152

=

Kds

st=

15.4MPa當(dāng)h=0時(shí)(A)P2=

0.14MPa5

·103=

2

·p

·

0.15sd

=

1

+

1解：

sd7-15ddpds圓桿周向線應(yīng)變e

=p

d

+Dd

)-pd

=Dd

=e(1)(2)30e

=

d

-

30

=

-en

=

-0.0003d所以d

=

29.991mmP156

7-16試判定圖示桿系是靜定的，還是超靜定的；若是超靜定的，試確定其超靜定次數(shù)，并寫(xiě)出求解桿系內(nèi)力所需的位移相容條件（不必具體求出內(nèi)力）。圖中的水平桿是剛性的，各桿的自重均不計(jì)。解：aaa12Δl11.5aaΔl2δ2δ1ααβ1

1

15Dl

=

d

sina

=

3

d2Dl2

=

d2

sin

b

=

2

d221d

=

2d52222=

6

22

d3

·

2d=

5DlDl1F

lN

1

11Dl

=2EADlF

l=

N

2

2EAl2

=

6

2 2a

=

24l1

Dl2

5

2.5a

25=

Dl1FN

2FN

1P156

7-16解：11

15Dl

=

d

sina

=

3

d2

2222

dDl

=

d

sin

b

=21d

=

2d56

22522= 1

= 2

d3

·

2dDl2DlF

lN

1

11Dl

=2EADlF

l=

N

2

2EAl2

=

6

2 2a

=

24l1

Dl2

5

2.5a

25=

Dl1FN

2FN

1aa121.5aaαaΔl1αΔl21d2d50·103

50·103t

== =

66.35MPa

<[t]=100MPapdh

p

0.02

0.0124bsA

=

p

(D2

-

d

2

)4450

·10350

·103s

bs

=

p=

p=

102.1MPa<

s

bs

=

240MPa(D2

-

d

2

)(0.0322

-

0.022)拉桿頭部滿(mǎn)足剪切強(qiáng)度條件擠壓力：Fbs=F拉桿頭部滿(mǎn)足擠壓強(qiáng)度條件。hdD50kNP1567-18

試校核圖示拉桿頭部的剪切強(qiáng)度和擠壓強(qiáng)度。已知：D=32mm，d=20mm，h=12mm，材料的許用切應(yīng)力[t]=100Mpa，許用擠壓應(yīng)力[sbs]=240Mpa。解：（1）剪切面：A=πdh；剪力：Fs=F（2）擠壓面：P1577-20

矩形截面木拉桿的接頭如圖所示，已知b

=250mm，PPahcPPFFFFballbl

0.25l=

0.2m

=

200mm50

·103

l

?0.25

·1·106F

50

·103=￡

[t]=

1MPat

=bsbsab

a

0.25=

0.02m

=

20mm50

·103

a

?0.25

·10

·106=

10MPaF

50

·103￡

[s

]s

=

=F=50KN，木材的順紋許用擠壓應(yīng)力[σbs]=10MPa,順紋許用切應(yīng)力[τ]=1MPa

。試求接頭處所需的尺寸l和a。解：P1838-1(c)

作圖示桿的扭矩圖，并指出最大扭矩的值及其所ABCD在的橫截面。1.5kN.m1kN.m1.5kN.m2kN.m3kN.mE11.50.5++-單位：KN.m3-P1838-4

實(shí)心圓軸的直徑d=100mm，長(zhǎng)l=1m，作用在兩個(gè)端面上的外力MeABABC25100解:(1)p16=

71.3MPa·

0.13T

14

·103=

ptmax

=

W0·

0.148

·1010

·14

·103

·1=

0.0178rad

=

1.02

Tl

pp32j

=

GI

=(2)tA

=

tB

=

tmax

=

71.3MPaaC2=

35.7MP=

tmaxt偶之矩均為Me=14kN.m，但轉(zhuǎn)向相反。材料的切變模量G=8×104MPa。試求：（1）橫截面上的最大切應(yīng)力，以及兩個(gè)端面的相對(duì)扭轉(zhuǎn)角；（2）圖示橫截面上A、B、C三點(diǎn)處切應(yīng)力的大小及指向。MeTτAτBO

τCP1838-5

空心鋼圓軸的外直徑D=80mm，內(nèi)直徑d=62.5mm，外力偶之矩為Me=10N.m，但轉(zhuǎn)向相反。材料的切變模量G=8×104MPa。試求：（1）橫截面上切應(yīng)力的分布圖；（2）最大切應(yīng)力和單位長(zhǎng)度扭轉(zhuǎn)角。MeMedD解:(1)31062.516pTtmax

=

W=p4

=

0.16MPa

80

·

0.08

·

1

-

maxAaDt

=

d

t=

0.125MP10TGI

pp32q

===

4.90

·10-5

rad

m

62.5

4

8

·1010

··

0.084

·

1

-

80

(2)τAτmaxAT184P

8-10

軸的許用切應(yīng)力[t

]=20MPa

，切變模量G

=8·104

MPa

，單位長(zhǎng)度桿的許用扭轉(zhuǎn)角

[q

]=

0.25

0)m。試按強(qiáng)度條件及剛度條件選擇此實(shí)心圓軸的直徑。M1M2M3M4M5nPM

=

9.55M1

=

0.86kN

m,M2

=

2.86kN

m,M3

=

0.57kN

m,M

4

=

1.05kN

m,M5

=

0.38kN

mpTmax=

7.99

·10

m

=

79.9mm-216Tmax[t]p￡

[t]

d

?

3tmax

=

Wmaxp32TmaxqGI

pG

[q

]p

2T=

max180

￡

[q

]

d

?

4180=

8.74

·10-2

m

=

87.4mm因此，d

?87.4mm解：?P2289-1

試求圖示各梁指定橫截面上的剪力和彎矩。ABF=ql/2ql/4l/2l1C

21

23

D32Fs1=

ql18M

=

1

ql

2s

2F

=

028M

=

1

ql

2s3F

=

038M

=

1

ql

2(c)FBFA2qlFA

=

FB

=解：求得支座約束力(

c

)B2aqFBa

FAaCqa2C121

A23344Fs1

=

02M1

=

qaFs

2

=

02M

2

=

qa6s3F

=

7

qa3M

=

qa2FA解：求得支座約束力6

6FB=

7qa

=

11qa(f)6s3F

=

1

qa33M

=

5

qa2(

f

)?P2289-3

試寫(xiě)出圖示各梁的剪力方程和彎矩方程，并作出剪力圖和4kN/mACB1m2m(d)彎矩圖。指出最大剪力和最大彎矩的值以及它們各自所在的橫截面。80kN解：求得支座約束力2=

80

+

4

·8

=

56kNFA

=

FBFs

(x1

)=

56

-

4x1Fs

(x2

)=

-24

-

4x221

1

1M

(x

)=

56x

-

2x(

)

(

)2

2222M

x

=

56

x-

80

x

-

4

-

2

x56kN40kN40kN56kN192kN.mtmax

=

56kNM

max

=192kN

mA支座右側(cè)截面C截面FBFA2kN/mAC1m4mB20kN.mFs

(x2

)=

9

-

2

x2(

)21

1

1M

x=

9

x

-

x(

)222

2M

x=

9

x

-

x

-

20tmax

=

9kNMmax

=

12kN

mA支座右側(cè)截面C右側(cè)截面解：求得支座約束力FA

=

9kN FB

=

1kNFs

(x1

)=

9

-

2

x11kN8kN.m12kN.m0.25kN.m(e)FBFA9kN0.5mP229:9-4,9-5,9-69-4

繪出圖示各梁的剪力圖和彎矩圖。1m5KN15KN10

KN.m15

KN.m10KN1m+10KNFs圖—5KN15

KN.m5

KN.m5

KN.m+—M圖（a）+80KN80KN剪力圖+單位：KN.m彎矩圖q=100kN/mACD0.2mE1.6m1m2m12(b)80KN80KN1m2mABC1m0.5kN1.5kN2kN/m0.75m1.5kN+-0.5kN+0.56kN.m

0.5kN.m(c)aABCDqqaaa2qaqaq+-qa+qaqaqa2/2-+qa2/29-5

(a)ABCqqa22aqa/35qa/3+-qa/35a./33a5qa/3+25qa2/184qa2/3qa2/39-5

(b)9-5

(c)qllqql2qlql2/2++ql2/2qllqql22qlqlql2—ql+——BlqFBx+8ql2l

2ql22ql++ql2/4lF=qlABCl/2FAFBql/2+-ql/2FA9-6A9-7(a)槽鋼平放Iz=

-162.8MPa=

-176·10-8M

y

5·103

·57.3·10-3s

A

=

As

D

=

s

A

=

-162.8MPaBzIsM

y

5

·103

·13.7

·10-3=

B

==

38.9MPaCzIs176

·10-8M

y

5

·103

·

20.7

·10-3=

C

==

58.8MPa176

·10-8ADBzy

CyA

=

yD

=

78

-

20.7

=

57.3mmyB

=

20.7

-

7.0

=

13.7mmyC

=

20.7mmCBDyAz9-7(b)槽鋼豎放AzM

ys5·103

·125·10-3=

A=

-=

-18.5MPaI

3370·10-8s

D

=

-s

A

=

18.5MPas

B

=

s

C

=

09-83m120kN.m5m11m2215kNzyAD18030050CB1-1截面M1

=

20kNAz

12M

yIs20·103

·150·10-3=

A=

-0.180·0.33Bz

12IsM

y

20·103

·100·10-3=

B

==

4.9MPa0.18·0.33s

C

=

0sD

=

-sA

=

7.4MPa2-2截面M2

=

-25kNAz

12I=

-7.4MPa

sM

y

25·103

·150·10-3=

A==

9.26MPa0.180·0.33Bz

12M

yI25·103

·100·10-3s

=

B

=

-=

-6.2MPa0.18·0.33s

C

=

0sD

=

-sA

=

-9.26MPaP2309-9

試求圖示組合截面對(duì)于水平形心軸z的慣性矩Iz。120×10120×10工22a212+

0.12

·0.01·(0.11+

0.005)=

3400

·10

+

2

·

=

6.6

·10-5

m4=

6.6

·107

mm4120

·103

·10-12-823I

z

=

I

z1

+

I

z

2

+

I

z

319-11矩形截面外伸梁如圖所示。①試求點(diǎn)1、2、3、4、5五個(gè)點(diǎn)處橫截面上的應(yīng)力；②以單元體分別表示各該點(diǎn)處的應(yīng)力狀態(tài)。ll/2l/2lFFⅠⅠⅡⅡⅢⅢ1245h/4

3h/4zbh2

A3Ft1

=

-2

2

4Fl

h3Fls

1

=

0

bh3

=

2bh2s

2

=

12t

=

02bh23Fl2

A3F323Fl23Fls

=bh

=

bh2

6t

=

03Flbh2243Fl2bhs

4

=

s

2

=t

=

05323Flbhs

=

s

=t5

=

02bh23Fl23FlbhFl/2FF9-12

由兩根No.36a槽鋼組成的梁，如圖所示。已知：F=44kN，q=1kN/m；鋼的許用應(yīng)力s

]=170MPa,t]=100MPa

。試校核此梁的強(qiáng)度。AB6×1m=6m113kN113kNF

F

F

F

Fqyz1M

=

113

·

3

-

44

·

2

-

44

·1

-

·1·

32

=

202.5kN

m2maxFs

max

=

113kN2S

*z

max=

385224

·10-9

m

3+

(180

-

16)·

9

·

180

-

16=

96

·16

·(180

-

8)h

=

360mm,

b

=

96mmd

=

9.0mm,

t

=

16mm[

]=Iz=

153.15MPa

<

s2

·11900

·10-8202.5

·103

·

0.18Mmax

ymaxs

max

=[]tFz=

20.3MPa

<

tI

d

11900

·10-8

·

9

·10-3

·

2113

·103

·

3.85224

·10-4S

*=

s

max

z

max

=max因此，該梁滿(mǎn)足正應(yīng)力和切應(yīng)力強(qiáng)度條件。9-13

由工字鋼做的簡(jiǎn)支梁受力如圖所示。已知材料的許用應(yīng)力s

]=

170MPa,

t]=

100MPa20kN/mACD0.5mE2.5m4m80kN

60kNB1m113.125kN76.875kN試選擇工字鋼號(hào)碼。113.125kN33.125kN16.875kN76.875kN2.156m1M

=

113.125·2.156-

80·1.656-

·20·1.6562

=

83.994kN

m2maxFs

max

=

113.125kNzM

83.994

·103W

=

max

=

494.08

·10-6m

3[s

]

170

·106選擇28a號(hào)工字鋼3zW

=

508cmmaxzFtS*113.125·103=

smax

zmax

==

54.1MPa

<[t]I

b

24.6·10-2

·8.5·10-3故選擇28a號(hào)工字鋼。P194：9-15

試求圖示等截面梁的轉(zhuǎn)角方程和撓度方程，并求外力偶AC2l/3作用著的C截面處的撓度。MeBl/3Me/lMe/lxM

e

lM

(x)=12MeEIq1

=

-

2l

x

+

C1

（1）3MeEIw1

=

-

6l

x

+

C1

x

+

D1

（2）(

)2eM

e

lM

x

=x

-

M222eeM2lEIq

=-x

+

M

x

+

C（3）32222

22M6lEIw

=-

Me

x

+x

+

C

x

+

D（4）21x

=

0,

w

=

0;

x

=

l,

w

=

0邊界條件1

2

1

23x

=

2

l,

w

=

w

,q

=

q連續(xù)條件連續(xù)條件代入(1)、(3)和(2)、(4)，得229231

e1

eD

+

M

l

2

=

DC

=

2

M

l

+

C

,邊界條件代入(2)、(4)，得21292992

ee1

eD

=

M

lD

=

0,C

=

1

M

l,

C

=

-

5

M

l21eeM

12l

9EIq

=-x

+

M

l1

e

M

x

2l

2EIw

=-

x2

6l

32259ee

eM2lEIq

=-x

+

M

x

-

M

l2225

299e

e6l

2EIw

=-

Me

x3

+

Mex

-

M

lx

+

M

lMwC

2Mel6EIl2

l222

2

3

l

-

3

l

=

81EI=

e

3

23C截面處，x

=2

l9-17圖示雙跨等截面梁。（1）試求支座約束力；（2）繪剪力圖和彎矩圖ABRC基本靜定系（3）試求支座截面轉(zhuǎn)角qA和qBq幾何方程：w

C=05q(2l

)

5ql44物理方程：

wcq

=

384EI

=

24EIccRC48EI

6EIR

(2l

)R

l

33=

c

w

=

-物理方程代入幾何方程，cwc

=

wcq

+wcR

=

0得：C4R

=

5

ql聯(lián)立靜力平衡方程得，A

B3R

=

R

=

3

qlABqql83ql385RC

=

4

qlql383

ql8ql853

l8l381289ql

21289ql

25

ql81

ql

28（2）繪剪力圖和彎矩圖c48EI16EI524EIql

32q(2l

)3=ql

?(2l

)4-qA

=

qAq

+qAR

=（3）試求支座截面轉(zhuǎn)角qA和qBql

3qB

=

-q-

A

=

-

48EIP252：10-1

求圖中所示應(yīng)力狀態(tài)下單元體斜截面ab上的應(yīng)力，并用分離體在該面上示出。30060MPaa

2300300t

=

-

40

-

60

sin600

+

(-

30)cos600

=

-25 3

-15

=

-58.3MPas

=

-

40

+

60

+

-

40

-

60

cos600

-

(-

30)sin600

=

10

-

25

+15 3

=

11.0MPa2

20s

x

=

-40MPa,s

y

=

60MPa,tx

=

-30MPa,a

=

3011.0MPa40MPa30MPab

58.3MPaP252：10-3

題10-1圖中所示應(yīng)力狀態(tài)下，分別求①主應(yīng)力的值；②用圖示出不等于零的兩個(gè)主應(yīng)力的作用面。–

-

47.7MPa

67.7MPa22502

+

302

=

+

(-

30)2

=

10

–-

40

+

60

-

40

-

60

21

=s

2

s

-15.480

74.520-

40

-

602

·(-

30)=

-0.6

a

=

s

-s2ttan

2ax

y=

-

x

=

-s

2

=

0,

s

3

=

-47.7MPas1

=

67.7MPa,（1）

若

sx

>

sy

，則

‰a1‰<

450(2) 若

sx

<

sy

，

則‰a1‰

>

450(3) 若

sx

=

sy

，則

{xt

>

0

，a1

=