混凝土结构中册习题答案第11章11．1 解：1、求支座截面出现塑性铰时的均布荷载q 1 首先计算支座截面极限抗弯承载力M uA ：C20混凝土查得f c =9.6N/mm 2, 316 A s =603mm 2KNmx h f A M h mm bf f A x y s uAb c ys 6.75)294465(300603)2(942006.9300603001=-⨯=-=<=⨯⨯==ξα 按弹性分析：,122ql M M uAA == kNm l M q uA 2.2566.75121222=⨯== m kN q /2.251=∴2、计算跨中极限抗弯承载力1u M ：216 As=402mm 2mm x 632006.9300402=⨯⨯=, kNm M u 3.522634653004021=⎪⎭⎫ ⎝⎛-⨯=总弯矩kNm M M M u uA 9.1273.526.751=+=+=总 由82l p M u =总 得 m kN l M p u /4.2869.1278822=⨯==总 3、若均布荷载同为p u ,按弹性计算的支座弯矩kNm M M Ae 3.859.1273232=⨯==总 则调幅系数114.03.856.753.85=-=-=Ae Au Ae M M M β11．2 解：A s1=A sA =644mm 2/m ， f y =210N/mm 2, h 0=120-20=100mm01.1410006.9210644h mm x b ξ<=⨯⨯=, m kNm M u /58.12)214100(210644=-⨯=m kNm M M u /2.252==总222/6.12142.25818m kN l M p n u =⨯⨯=⨯=总11．3 解：塑性铰线位置如图所示。

AB1 8/10@1008/10@100 ABaa al -B4al - 2)(31a l -⋅取出一块梯形板块为隔离体，对铰支座AB 取力矩平衡：()()()⎥⎥⎦⎤⎢⎢⎣⎡-⋅⋅-+-⋅⋅⎪⎭⎫ ⎝⎛-=-⋅422312)(2a l a a l a l a l p a l m u()()()()()()a l a l mp a l p a l a a l a l p a l a a l p m u u u u -+=∴+-=⎪⎭⎫ ⎝⎛+--⋅=⎥⎦⎤⎢⎣⎡-+-=224224388242第12章12．1 解：影响线 267.066.1808.064.445.0075.0645.014213===+====y y y y kND P P D kNP kN y P D yi i435.2221152.222.2211528.9108.9185.22225.2479.011515.29.09.015.2267.0808.0075.01max max min min min max max =⨯=⋅==-⨯+⨯==⨯=⨯⨯===+++=∑∑水平荷载系数12.0=α()kNT kN T k k 93.75.222115098.4098.48.91094.312.041max,=⨯==⨯+⨯=12．2 解：○1计算柱顶水平集中力k W ：柱顶标高处,0.1≈z μ 檐口处07.1≈z μ()()[]()kNW W W k k k 54.7645.007.112.01.23.1645.007.12.16.05.01.25.08.021=⨯⨯⨯-⨯=⨯⨯⨯⨯-+⨯+=+=○2 m kN B w q z s k /16.2645.00.18.001=⨯⨯⨯==μμm kN q k /35.1645.00.15.02-=⨯⨯⨯-=0.45 4.4 1.15 4.41.6y 1y 2 y 3y 4○3 剪力分配系数计算：;2.05.104.85.10369.05.192.7148.038.1413.2=-=====λB A n n()()930301096.21369.012.013868.2757.5008.0131148.012.013因只需相对值，故略去=⎪⎭⎫⎝⎛-+==+=⎪⎭⎫⎝⎛-+=BA C C;72.57196.25.191;24.411868.238.143333023⋅=⨯⋅=∆⋅=⨯⨯==∆c c B c c A c A E H E H u E H E H C I E H u()3396.9872.5724.4111HE H E u u c c B A =+=∆+∆417.096.9824.41==A η， 583.096.9872.57==B η ○4 计算约束反力A R 、R B ：371.011.8008.31369.012.0181369.012.013362.096.98028.31148.012.0181148.012.01334113411==⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛-+⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛-+===⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛-+⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛-+=BAC C()()←=⨯⨯==←=⨯⨯==kN HC q R kN HC q R B B A A 26.5371.05.1035.121.8362.05.1016.2112111∑=+=kN R 47.1326.521.8○5 剪力计算：A Bk W q 2k()()()()kNW R kNW R k B k A 25.1201.21583.054.747.13583.076.801.21417.054.747.13417.0=⨯=+⋅=+=⨯=+⋅=+∑∑ηηA 柱顶剪力 V A =8.76-8.21=0.55Kn (→)B 柱顶剪力 V B =12.25-5.26=7kN (→)○6 弯矩图：kNm M kNm H V H q M B A A 8.1475.1075.1035.12185.1245.1055.05.1016.22121221=⨯+⨯⨯==⨯+⨯⨯=+=底底12．3 解：从前题知 n A =0.148, n B =0.369, 318.0115.3==λ ○1 计算剪力分配系数：84.21369.01318.01353.21148.01318.0133030=⎪⎭⎫⎝⎛-+==⎪⎭⎫⎝⎛-+=BA C C38.55184.25.19138.36153.238.14133=⨯=⋅∆=⨯=⋅∆H E u H E u c B c A3376.91)38.5538.36(11HE H E u u c c B A =+=∆+∆ （相对值）4.076.9138.36==A η， 6.076.9138.55==B η ○2 计算约束反力R A 、R B ：M A =124.85kNM B=147.8kNm278.1055.1899.05.11369.01318.01318.015.1138.1185.1899.05.11148.01318.01318.015.1323323=⋅=⎪⎭⎫⎝⎛-+-⋅==⋅=⎪⎭⎫⎝⎛-+-⋅=B AC C()()→-=⨯-=⋅=←=⨯=⋅=kN C H M R kN C H M R B B A A 22.15278.11113185.15138.1112.1533231()∑←=-=kN R 63.022.1585.15○3 柱顶剪力：()()∑∑→-=⨯--=-=←=⨯-=-=kN R R V kN R R V B B B A A A 6.1563.06.022.156.1563.04.085.15ηη○4 弯矩图：12．4 解：f y =300 N/mm 2, F V =324 kN, Fh=78 kN, a=250 mm()23373031241830010782.14080030085.025010324mm A s =+=⨯⋅+-⨯⨯⨯⨯=小于最小配筋量612的面积，故按构造配筋12．5 由于解答不唯一，故从略。

第15章15．1 解：查得砌体抗压强度设计值f=1.5 N/mm 2,mm N M e 4.3210250101.836=⨯⨯==; 97.1062068000===h H β; 052.06204.32==h e ;73.01846.01121052.01211846.0110015.011112220=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎭⎫⎝⎛-++==⨯+=+=ϕαβϕkN N kN fA 25066.3326204905.173.0=>=⨯⨯⨯=ϕM 2AB54.6kNm 98.6kNm18.4kNm54.6kNm76.4kNm40.6kNm按轴压计算时 88.134906800==βϕϕ>=⨯+=776.088.130015.01120 承载力应会满足∴。

15．2 解：抗压强度设计值 f=1.19 N/mm 2, 翼墙间距s=6.8 m,层高H=3.5 m, 2H>s>H,∴计算高度 m H s H 42.35.32.08.64.0024.00=⨯+⨯=+=8.1919.01.142.3=⨯=β, 0015.0=α63.08.190015.01120=⨯+=ϕ 底层轴力N=118+3.36*3.5=129.76Kn76.12947.142190100019.163.00>=⨯⨯⨯=fA ϕ15．3 解：,180kN N l = ,500kN N = ()201752.024********m A =⨯⨯+=2/3.1mm N f =, mm mm a 2402153.1600100<==2053750250215mm b a A b l =⨯==;326.3537501752000>==l A A 0=∴ψ 2526.1126.335.01<=-+=γl l N kN fA <=⨯⨯⨯=64.74537503.1526.17.0ηγ局部受压不满足要求。

15．4 解：首先计算中和轴位置y()mm y 20224036005004905.024*********.074049022=⨯+⨯⨯⨯-+⨯⨯=惯性矩 I:()()210101010102323105386.3105019.0103583.0100234.110655.122402022404903600122404903600274053874049012740490mm I ⨯=⨯+⨯+⨯+⨯=⎪⎭⎫ ⎝⎛--+⨯-+⎪⎭⎫ ⎝⎛-⨯+⨯=2610109.12403600500490mm A ⨯=⨯+⨯=4903600538y=2022 500240114回转半径 mm A I i 17910109.1105386.3610=⨯⨯==折算厚度 mm i h T 6255.3==， 高厚比 12.1362582000===T h H β mm N M e 11410350104036=⨯⨯==, 182.0625114==T h e , 744.02.13002.01120=⨯+=ϕ403.01744.01121182.012112=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎭⎫⎝⎛-++=ϕ; 查得2/3.1mm N f =kN N kN fA 35058110109.13.1403.06=>=⨯⨯⨯=ϕ承载力满足要求。