1、试画出杆的轴力图。若横截面面积A=400mm2,试求各横截面上的应力。孙题孙题2-2 20kN20kN10kNFN图+102010要求:要求:上下对齐,标出大小,标出正负上下对齐,标出大小,标出正负单位:kN)MPa(5011AN112233)MPa(2522AN)MPa(2533AN试画出杆的轴力图。试画出杆的轴力图。刘题刘题2.1a 40kN20kN30kN要求:要求:上下对齐,标出大小,标出正负上下对齐,标出大小,标出正负332211FN图+502010单位:kN试作轴力图,试作轴力图,单题单题2-1(c)2kN3kN2kN3kNFN+3kN2kN1kN刘题刘题2.12 图示结构,BC为
2、钢杆,AB为木杆,木杆的横截面面积 A1=100cm2,许用应力 1=7MPa,钢杆横截面面积 A2=6cm2,许用应力 2=160MPa,试求许可载荷F。FBAC30钢木21解:FF3 1NFF2 2N1N11AF13AF1311AF)kN(4.402N22AF22AF2222AF)kN(48)kN(4.40F变截面杆,已知A1=8cm2,A2=4cm2,E=200GPa,求杆的总伸长l。刘题刘题2.18 FN图+40kNA120040kN60kN20kN200A220kN0kN2 1NF0kN4 2NFniiiiiAElFl1N解:222N2111N1AElFAElF1.0025.0mm0
3、75.0例例1 图示结构,杆1为35号钢,s=315MPa,b=530MPa,A1=3cm2,杆2为铸铁,抗拉强度极限bt=180MPa,抗拉强度极限bc=600MPa,安全因数均为n=3。求:(1)按1杆的强度确定许可载荷;(2)在上述的许可载荷下确定2杆的面积。FABC121.2m0.9m351FFN0sin1FFN解:0,xF0,cos12NNFF34 2FFN,0yFFFN2FN1AFABC121.2m0.9m1杆:1N11AF135AF5 31AF53001053)N(109.183)kN(9.182杆:2N22AF234AFc 34c2FA 2003109.1843)mm(1262
4、)kN(9.18F许可载荷ns)MPa(1053315nbcc)MPa(2003600例例2 图示结构,1杆为35号钢,s=315MPa,b=530MPa,A1=3cm2,2杆为铸铁,抗拉强度极限bt=180MPa,抗压强度极限bc=600MPa,安全因数均为n=3。求:(1)按1杆的强度确定许可载荷;(2)在上述的许可载荷下确定2杆的面积。E2 aFaACBD2aaH12252NFF033522NaFaF解:0,EM0,521NaFaF25 N1FF,0AME2 aFaACBD2aaH12FN1EFDHFN2整体三角形DHE1杆:1N11AF125AF5 21AF53001052)N(102
5、.283)kN(2.282杆:2N22AF225AFc 25c2FA 2002102.2853)mm(1572)kN(2.28F许可载荷ns)MPa(1053315nbcc)MPa(2003600图示结构,杆1和杆2为圆截面钢杆,=160MPa,d1=30mm,d2=20mm,确定许可载荷。单题单题2-16FABC301245错误解题步骤:1)设1杆达到最大值,判断2杆是否破坏;2)2杆破坏,由2杆确定结构许可载荷;3)设2杆达到最大值,求出1杆的内力 4)求许可载荷。此方法不可取!单题单题2-16图示结构,杆1和杆2为圆截面钢杆,=160MPa,d1=30mm,d2=20mm,确定许可载荷。
6、FABC301245正确的解题步骤:(1)求出各杆内力;(2)由各杆强度求出载荷大小;(3)取最小值为许可载荷。FABC301245FF518.0N2FF732.0N11N11AF1732.0AFkN5.154F2N22AF2518.0AFkN2.97FkN2.97 F图示结构,各杆均为圆截面钢杆,1、2、3杆的直径为 d1=30mm,4、5、6、7、8杆的直径为 d2=20mm,9杆的直径为d3=40mm,=160MPa,确定许可载荷。正确解题步骤:(1)求出各杆内力;(2)由各杆强度求出载荷大小;(3)取最小值为许可载荷。错误解题步骤:(1)设1杆达到最大值,判断2杆是否破坏;(2)2杆破
7、坏,设2杆达到最大值,判断3杆是否破坏;(3)如此循环 ABCDF123456789例例34m4m4m孙题孙题2-22(P48)解:整体 求支反力RARBkN220BARRA节点 kN7.366ACNC节点 kN293CDNARAFACCFACFCD图示结构,各杆均为两根等边角钢组成,=170MPa,试选择杆AC和CD的角钢型号。ABCD220kN220kN4m4m4m2ACACACAN)(10781702107.36623mm2 cm 86.10780ACA的角钢,选AC杆:CD杆:2CDCDCDAN 2ACACNA 2CDCDNA)(86217021029323mm2cm 797.8675
8、A的角钢,选图示屋架结构,杆AE和EG均为两个75mm8mm的等边角钢组成,q=20kN/m,试求杆AE和EG横截面上的应力。孙题孙题2-4 解:整体 求支反力RARBkN4.177BARRqABCDEGF9m4.37m4.37m1m1.2mqACDE左半部分 RA0cMNEGkN6.357EGNENEGNEANEDkN8.366EAN)cm(006.23503.1122AE节点 AEAEAEAN等直杆,已知杆的横截面面积A和材料的弹性模量E,试作轴力图,并求杆端点D的位移。孙题孙题2-7 FN图+FFF+2FF2F3l3l3lABCD孙题孙题6-4 aaaFACEB12 图示结构,AB为刚性
9、杆,1杆和2杆的长度相等,横截面面积相等,材料相同,A=1000mm2,F=50kN。试求两杆的轴力和应力。aaaFACEB12aaaFACEBN1N2aaaFACEBN1N2,0Am 03221aFaNaN平衡方程:122 ll 11EAlNl l1FACEB12l2 22EAlNl 变形协调方程:物理方程:孙题孙题6-5 图示结构,AB为刚性杆,1杆和2杆为钢杆,=170MPa,A1=400mm2,l1=1m,A2=200mm2,l2=1.8m,试校核杆的强度。q=30kN/m2mACEB121mDq=30kN/m2mACBF11mF2,0AM 05.13312N1NqFF(1)平衡方程:
10、解:设1杆受压、2杆受拉l2q=30kN/m2mACEB121mDl1123 ll(2)变形协调方程:111N1EAlFl(3)物理方程:222NEAlF(4)补充方程:111N3EAlF2N1N2.1 FFkN32N2FkN5.38N1FMPa961MPa1602 22N22EAlFl q=30kN/m2mACBF11mF2,0AM 05.13312N1NqFF(1)平衡方程:解:设1杆和2杆均受拉l2q=30kN/m2mACEB121mDl1123 ll(2)变形协调方程:111N1EAlFl(3)物理方程:222NEAlF(4)补充方程:111N3 EAlF2N1N2.1 FFkN32N
11、2FkN5.38 N1FMPa961MPa1602 22N22EAlFl 刘题刘题2.43 P(65)FaaAClB123aaFACBFN1FN2FN3 图示结构,AC梁为刚杆,杆1,2,3的长度相等,横截面面积相等,材料相同,试求三杆的轴力。,0AM 0232aNaN平衡方程:,0yF 0321FNNNaaFACBFN1FN2FN33122llll1l2 1N1EAlFl 变形协调方程:物理方程:AClB123l3 2N2EAlFl 3N3EAlFl 651NFF 312NFF 61 3NFF刘题刘题2.43 另解另解 FaaAClB123aaFACBFN1FN2FN3,0AM 0232aN
12、aN平衡方程:,0yF 0321FNNNaaFACBFN1FN2FN33132)(2llll 1N1EAlFl 变形协调方程:物理方程:l1l2AB123l3 2N2EAlFl 3N3EAlFl 651NFF 312NFF 613NFF孙题孙题6-9 ABC60kN40kN1.2m2.4m1.2mD 图示阶梯状杆,下端与支座距离=1mm,上、下段杆的横截面面积分别为600mm2和300mm2,材料的弹性模量E=210GPa,试作杆的轴力图。解:解:ABC60kN40kN1.2m2.4m1.2mDABC60kN40kNDRBRAABC60kN40kNDRBRA04060BARRCBDCADABl
13、lllCBCBCBDCDCDCADADADEAlFEAlFEAlFNNNABlCBCBCBDCDCDCADADADEAlFEAlFEAlFNNNAADRFN60NADCRF4060NACBRF(1)(3)100AR(2)(4)又因为:ABC60kN40kNDRBRACBCBADCDCAADADAEAlREAlREAlR)100()40()(85 kNAR代入(1)得)(15 kNBRkN85NADFkN25NDCFkN15NCBF解得代入(4)得:ABC60kN40kN1.2m2.4m1.2mD+852515单位:单位:kN已知:杆子面积A=200 mm2,长l=2m,=1mm,受外力P=60
14、kN,材料的弹性模量E=200GPa,试画出杆子的轴力图。例例4单题单题2-32F30123C图示结构,由杆1,2,3组成,1=80MPa,2=60MPa,3=120MPa,A1=A2=2A3,E1=160GPa,E2=100GPa,E3=200GPa,F=160kN,l1=1m。试确定各杆横截面面积。F30FN1FN3FN2,0yF 030cos2N1NFF(1)平衡方程 030sin2N3NFFF解:F30123C,0 xFl130123CC l2l33030tan30sin123lll(2)变形协调方程(3)物理方程:1111N1AElFl 222N22AElFl 333N33AElFl
15、 l130123F30FN1FN3FN2CC l2l33030sin30cos30tan1213llll另设:单题单题2-26图示结构,杆1,2,3的横截面面积相同,求C点的位移。E=200GPa,A=100mm2,F=20kN,l=1m。FBC12453DFBCFN145DFN2FN3 1NF02NFkN103NF由平衡方程,得:FBC12453D 0.5mm1N31EAlFll 02l1lFBC123D 0.5mm1N31EAlFll 02l1l 已知:F=80kN,b=80mm,=10mm,d=16mm,=120MPa,许用挤压应力为bs=340MPa,许用应力=160MPa,试校核接头
16、的强度。单题单题2-39 2-39(P59)(P59)FFbd解:1、铆钉的剪切强度kN)(204SFFFFbFbF/4F/4F/4F/4d42SSdFAF MPa51.99d2、挤压强度dFbbs161010203SbFF MPa125bs4/FkN)(203、板子的拉伸强度FbF/4F/4F/4F/4F/43F/4FxFN3、板子的拉伸强度FbF/4F/4F/4F/4F/43F/4FxFN)(11dbF10)1680(10803MPa125 1122)2(22dbF10)16280(4108033MPa125 所以铆钉和板子均安全。一长为30cm的钢杆,横截面面积A=10cm2,E=210
17、GPa,试求:(1)AC、CD、DB各段的应力和变形;AB杆的总变形。北科大题北科大题1-6 10020kN20kN20kNABCD100100EAlFlACN)mm(01.0解:AC段kN)(20NFMPa)(20ANFCD段0NF00CDlEAlFlDBN)mm(01.0kN)(20NFMPa)(20ANFDB段ABl01.0001.0)mm(02.0试求杆BC段的变形,A=20mm2,l=1m,E=210GPa,例例 9kN5kN7kN3kN l l lABCD9kN5kNFNkN)(4NFEAlFlBCN2010210100010433)mm(952.0EAlFlBCN北科大题北科大题
18、1-15 斜杆BC由两角钢组成。=140MPa,Q=25kN,确定杆BC的型号。1.5mQACBD1.5m1.5mQABD1.5mFN1.5mQABD1.5mFN kN7.70NF平衡方程解:,0AMAF2N)mm(5.2522 A 2NF选45453等边角钢两根)mm(9.2652AFAyFAx北科大题北科大题1-19图示结构,AB为钢杆,横截面面积 A1=6cm2,=140MPa;BC为木杆,横截面面积 A2=300cm2,许用压应力 C=3.5MPa,试求最大许可载荷F。FBAC钢木3m1解:FF431NFF45 2N1N11AF143AF3 41AF)kN(1122N22AF245AF
19、C5 42AFC)kN(84)kN(84 F4m2FBFN1FN2两钢杆如图,已知横截面面积A1=1cm2,A2=2cm2,E=210GPa,线膨胀系数=12.510-61/C,当温度升30 C时,试求两杆内的最大应力。北科大题北科大题1-23 A1200200A2A2100A1(a)(b)lTl解:(a)22N211N1EAlFEAlFF510429.11875.0)kN(125.13FMPa)(131ANFFFFF0l解:(b)0llT 1NEAlF0ETAF1N1NAlTFMPa)(8.78两钢杆如图,已知横截面面积A1=1cm2,A2=2cm2,E=210GPa,线膨胀系数=12.510-61/C,当温度升30 C时,试求两杆内的最大应力。北科大题北科大题1-23 FN图+40kNA120040kN60kN20kN200A220kN0kN2 1NF0kN4 2NFniiiiiAElFl1N解:222N2111N1AElFAElF1.0025.0mm075.0