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1、2-1解:作轴力图如下:130kN70kNF/3F/3F/3400kN340kNF(a)(b)400340F2702F/3kNF/3FF/lF3FF2FFl/2ll(c)(d)2F3FF2FFNF/l2FFF2FF/l(e)(f)2F2FF题7-1图2-3(a)q=10kN/m解：求反力,FA=FB=40kN;取一半(如图)分析:CA1.2mB121mFAFB2mD4m2mΣMiC=04041042×−××q=10kN/mF==36.4kNN12.2FCxC由结点D的平衡:A1FCy∑Fix=0,F1=F2×5/2=40.7kN,FAFN2D3
2、6.410×3FN1σ(1)=FA1/1=−6=35.4MPa,115010×FN2340.710×Dσ=F/A==31.7MPa(2)22−6115010×2-3(b)解：去掉(1)杆的约束(如图),q=3kN/mA用约束力F1代之,DE1m2由3∑MA=0,F1=−××331.5/1=−13.5kNB1C由结点C的平衡1m1m1m∑Fix=0,F3=F1×2=−19.1kNq=3kN/mADE∑Fiy=0,F2=−F3×1/213.5=kN3F1F1−13.510×σ===−15.9MPaC(1)−6A85010×1313.510×σ==2
3、2.5MPaF2(2)60010×−6F3F13−19.110×Cσ==−38.2MPa(3)−650010×2-412FF解:(a)FN=F=10kN,12bb0F1410×3σ===350MPamax−6δAmin(2010)410−××(a)16kN.9kN17kN(b)轴力图如图,∴,31710×Dσ==213MPaABCAB8010×−6243810×kN17σ==267MPa8BC−6(b)3010×32410×（b)σ==200MPaCD−612010×A50cmB50cmC（c）12kN,(c)31210×σ==40MPaBC−
4、63010×31810×σ==45MPaAB−640010×∴σ=σ=45MPamaxBC2-6F200钢铝300解:由7-3式,FlFl0.20.3N11N22∆=l+=−F(+)=−0.49−69−6EAEA20010××10010010××7010××20020010××1122解得,F=1932kN.2-7FlFl解:ρgAl2l22F+ρgAl22Fl3ρgAl∆=∆l=−(+)=−(+)BBCEAEAEA2EA方向向下2-11解:由平衡条件可求得CF×2AF===40kN,2BFAF1A1A1A10.5m33B3B3B3GF1mF×
5、1DEFB==20kN,1m1m3CABF=F=60kNCCDΔlADΔlBE3Fl4010××1AAD∆l===0.4mmAD9−6,EA1120010××50010×36010××0.5∆l==0.2mmCG9−610010××150010×32010××1.∆lBE=9−6=0.67mmG点的竖直位移1010××300010×0.67−0.4∆=CC′+∆l=0.4++0.2=0.69mmGCG32m2-12AFAB4mΔF2mHB2m2m3m12解:水压力:FH=×102××=240kN2F×2/3AB杆的内力HF==11.1kNAB4
6、3/5×FABσ=≤[σ]由强度条件7-7式A3故d≥FAB=11.110×=3610×−3m6[σπ]/41110××3.14/42-13解:AB杆的轴力:2.5FFF==2.5FNoA2sin30×C30°FBND由强度条件7-7式2m0.5mFNσ=≤[σ]A260.0317010×××π[σ]A4得F≤==48.0kN2.52.5所以容许荷载[F]=48kN.2-14解:最大轴力:FF=F+ρgA+ρgANmax1122由强度条件7-7式,3mFNmaxσ=≤[σ]A20.4ma式中A=a22所以332F+ρgAl1610×+1610×
7、×0.4×311a≥==0.573m63[σ]−ρ2gh0.0810×−2010××0.42-16解:变形协调的几何关系:aF23∆=∆l1l21AF2aF1物理关系ACDBFaFa212∆=l,∆=lΔlΔl212a1aEAEA得补充方程3F=2F12由平衡条件F+5F=2F124F6F解得F1=,F2=11114F6Fσ=,σ=所以1211A11A3-1解:作扭矩图如图:10Nm90Nm500Nm/mT2TT6T2T1001002001.5aaaa(b)(a)2T10T3T1004T（Nm）1kNm4kNm1kNm2kNm3kNm1.5kN
8、maaaa2aaa(c)(d)5321.52.51(kN.m)(kN.m)3-2TT213d/3解:由8-5式,8-7式和8-2式,3−3,Mρ210
