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山东大学高起专《钢筋混凝土结构》三 参考答案

山东大学高起专《钢筋混凝土结构》三 参考答案
山东大学高起专《钢筋混凝土结构》三 参考答案

《钢筋混凝土结构》专科试卷三

一、填空题

1、 0.2为对应于残余应变为0.2% 的应力，称之为无明显屈服点钢筋的条件屈服强度。

2、经冷加工后的钢筋在强度提高的同时，伸长率显著降低，除冷拉钢筋仍具有明显的屈服点外，其余冷加工钢筋均无明显屈服点和屈服台阶。

3、结构可靠性是指结构在规定的时间内，在规定的条件下，完成预定的概率。

4、在荷载作用下，钢筋混凝土梁正截面受力和变形的发展过程可划分为三个阶段，截面抗裂验算是建立在第Ⅰ阶段的基础之上，构件使用阶段的变形和裂缝宽度验算是建立在第Ⅱ阶段的基础之上，而截面的承载力计算则是建立在第Ⅲ阶段的基础之上。

5、双筋矩形截面受弯构件的计算公式中，为防止受压区混凝土在受拉区纵向受

力钢筋屈服前压碎，应满足条件；为防止受压区纵向受力钢筋在构件

破坏时达不到抗压强度设计值，应满足条件。

6、弯剪扭（M，V，T）构件承载力计算，分别按受弯和受扭计算的纵筋截面面积相叠加；分别按受剪和受扭计算的箍筋截面面积相叠加。

7、偏心受压构件的破坏类型是大偏心受压和小偏心受压。

8、影响无腹筋梁斜截面受剪承载力的主要因素有剪跨比、混凝土强度

和纵筋配筋率。

二、选择题

1、对构件施加预应力的主要目的是（B ）

(A) 提高承载力

(B) 避免裂缝或减少裂缝（使用阶段），发挥高强材料作用

(C) 对构件进行检验

2、受扭构件中，抗扭纵筋应（ C ）

(A) 在四角放置(B) 在截面左右两侧放置(C) 沿截面周边对称放置

3、大偏心受拉构件的破坏特征与（C ）构件类似。

(A) 受剪(B) 大偏心受压(C) 小偏心受拉

4、矩形截面小偏心受压构件截面设计时A s可按最小配筋率及构造要求配置，这是为了（C ）

(A) 保证构件破坏时，A s的应力能达到屈服强度f y，以充分利用钢筋的抗拉作用

(B) 保证构件破坏时，不是从A s一侧先被压坏引起

(C) 节约钢材用量，因为构件破坏时A s应力 s一般达不到屈服强度。

5、适筋梁在逐渐加载过程中，当纵向受拉钢筋达到屈服以后（ C ）

(A) 该梁即达到最大承载力而破坏

(B) 该梁达到最大承载力，一直维持到受压混凝土达到极限强度而破坏

(C) 该梁承载力略有所增大，但很快受压区混凝土达到极限压应变，承载力急剧下降而破坏

三、简答题

1、什么叫钢筋的冷拔？冷拉和冷拔对钢筋的性能的改变有何不同？

答：冷拔是将钢筋用强力拔过比其直径小的硬质合金拔丝模。冷拉可以提高钢筋的抗拉强度但不能提高钢筋的抗压强度。冷拔可以同时提高钢筋的抗拉强度和抗压强度。

2、荷载设计值与荷载标准值有什么关系？

答：荷载代表值乘以荷载分项系数后的值，称为荷载设计值。设计过程中，只是在按承载力极限状态计算荷载效应组合设计值的公式中引用了荷载分项系数，因此，只有在按承载力极限状态设计时才需要考虑荷载分项系数和荷载设计值。在按正常使用极限状态设计中，当考虑荷载短期效应组合时，恒载和活载都用标准值；当考虑荷载长期效应组合时，恒载用标准值，活载用准永久值。

3、受弯构件正截面有哪几种破坏形式？各属什么性质的破坏？严格按规范设计的构件如发生正截面破坏，一般会是哪种破坏形式？

答：受弯构件正截面可发生“适筋梁”、“超筋梁”、“少筋梁”三种破坏形式。“适筋梁”破坏形式属“塑性破坏”、“超筋梁”和“少筋梁”破坏形式均属“脆性破坏”。严格按规范设计的构件如发生正截面破坏，一般应为“适筋梁”破坏。

四、计算题

1、矩形截面偏心受拉构件，b=300mm ，h=500mm ，承受轴向拉力设计值N=600KN,弯矩设计值42M kN m =?，采用C20级混凝土，HRB335级钢筋，计算构件的配筋。(已知: 22'29.6/, 1.10/,300/c t y y f N mm f N mm f f N mm ====，a=a ‘=40mm )

1、 解：材料强度：22'29.6/, 1.10/,300/c t y y f N mm f N mm f f N mm ====。 取0'40,460a a mm h mm ===。

603421070210600102

M h e mm a mm N ?===<-=? 故为小偏心受拉。

00'0.5'25040702800.5'2504070140e h a e mm

e h a e mm =-+=-+==--=--=

代入公式

3203'20'600102801333.3(')300420

60010140666.7(')300420s y s y Ne A mm f h a Ne A mm f h a ??===-???===-?

min max(0.45/,0.002)0.002t y f f ρ==，2min 300bh mm ρ=，满足最小配筋率要求，

2、已知梁截面尺寸为b =200mm ，h =400mm ，混凝土为C20级，29.6N/mm c f = 21.10N/mm t f =，采用HRB335级钢筋，受拉钢筋为3φ20，2941s A mm =，受压钢筋为2φ14，'2308s A mm =，0.550b ξ=，承受弯矩设计值90kN m M =?，试验算该截面是否安全。

解：（1）基本参数

C20级混凝土29.6c f N mm =，HRB335级钢筋2300,0.550y b f N mm ξ==，等效矩形图形系数 1.0α=。

截面有效高度04002510365h mm =--=。

（2）计算受压区高度x

''

300(941308)98.99.6200y s y s c f A f A x mm f b -?-===? （3）计算受弯承载力u M

按式（5-26b ）计算得

'''00()()2

u c y s x M f bx h f A h a =-+- 9.620098.9(36598.92)300308(36535)=???-+??- 90.4kN m M =?>

满足安全要求。

'

270a mm >= 0200.75b h mm ξ<=

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