课程名称︰钢筋混凝土学
课程性质︰土木工程学系必修
课程教师︰欧昱辰
开课学院：工学院
开课系所︰土木工程学系
考试日期（年月日）︰2018/10/4~2018/10/11
考试时限（分钟）：壹万零捌十分钟
试题 :
作业三(Homework three)
缴交日期(Due date):10/11
(1)若钢筋混凝土梁矩形断面为单筋梁，且配置单排钢筋，断面尺寸如下所示，混凝土规定
强度fc' = 280kgf/cm^2，钢筋降伏强度fy = 4200kgf/cm^2。
https://imgur.com/J36yPNb （本人制图）
(a)何谓平衡钢筋比？
(b)此断面之平衡钢筋用量Asb为何？
(c)依此规范规定如何定义纵向钢筋最大钢筋比？
(d)此断面之纵向钢筋最大钢筋量Asmax为何？
(e)规范限制最大钢筋用量的目的为何？
(f)此断面规范规定之最小钢筋量Asmin为何？
(g)规范限制最小钢筋用量的目的为何？
注：
(a)本题修改、合并自87年结构技师、99年与106年高考题。92年、98年、101年与104年普
考题。
(b)第(c)小题，目前规范没有直接定义最大钢筋比，而是定义εt有一最小值，此最小值
间接限制了最大可使用的钢筋量。若为单排钢筋，讲义有推导最大钢筋量的公式。
A singly reinforced concrete rectangular beam as shown above is designed
with one layer of longitudinal tension reinforcement. The concrete compress-
ive strength fc' = 280 kgf/cm^2. The reinforcement yield strength fy = 4200
kgf/cm^2.
(a)What is the balanced reinforcement ratio?(Definition)
(b)What is the balanced reinforcement area Asb of this beam?
(c)How is the maximum reinforcement ratio defined based on the code?
(d)What is the maximum reinforcement area Asmax for the longitudinal reinfo-
rcement of this beam?
(e)What is the perpose for the code to specify a maximum reinforcemant area?
(f)What is the minimum reinforcement area Asmin for the longitudinal reinfo-
rcement of this beam?
(g)What is the perpose for the code to specify a minimum reinforcement area?
Note:
(a)This question is modified from past nationa exams.
(b)For question (c), the code does not directly specify a maximum reinforce-
ment ratio. Instead, the code specifies a minimum value for εt, which indi-
rectly impose an upper limit on the longitudinal tension reinforcement. For
only one layer of longitudinal reinforcement, the lecture notes provide an
equation to calculate the maximum reinforcement ratio.
(2)下图为两端均为固定端之矩形断面(35cm宽 x 70cm深)RC梁，梁净跨度长L =7.5m，受
垂直均布载重：静载重wD = 8tf/m(含梁自重)与活载重wL = 7tf/m；梁跨中点弯矩采用
wL^2/24。拉力钢筋之有效深度，统一取63cm作计算。混凝土fc' = 280kgf/cm^2，梁主
筋fy = 4200kgf/cm^2。
(见上面连结图)
(a)计算跨度中央断面所需拉力筋总面积。
(b)采#8钢筋，每根断面积Ab = 5.067cm^2，采单排配置，请决定至少需配置的根数。需
检查净间距是否符合规范规定，假设保护层厚度为4cm，采#3箍筋，最大粗粒料直径为2cm。
(c)根据最终配置的拉力筋总面积As，检查极限状态εt是否满足规范对梁的规定。
(d)检查是否满足规范最小钢筋量规定。
(e)检核φMn >= Mu。
注：
(a)本题修改、合并自90、91与105年结构技师、97、104与105年高考题。107年普考题。
(b)载重组合采1.2D+1.6L。
A RC beam as shown above is fixed at two ends and has a rectangular section
of 35 cm x 70 cm(width x height). The beam has a span L = 7.5m and is subje-
cted to vertical uniformly distributed loads: a dead load of wD = 8tf/m (in-
cluding the self-weight of the beam)and a live load of wL = 7tf/m. Use wL^2
/24 to calculate the moment at the mid span. The effective depth of the
beam is assumed to be 63 cm. Concrete fc' = 280 kgf/cm^2 and beam longitud-
inal reinforcement fy = 4200 kgf/cm^2.
(a)Calculate the required area of longitudinal tension reinforcement at the
midspan of the beam.
(b)Use #8 reinforcing bars for longitudinal reinforcement (Ab = 5.067 cm^2).
Use one layer design. Determine the minimum number of the reinforcing bars
to satisfy(a). Need to check the clear spacing of the bars based on the
code. Assume the cover depth is 4 cm. Use#3 transverse reinforcement. The
maximum size of the aggregate is 2 cm.
(c)Based on the final design of longitudinal reinforcement area As, check
whether εt at the ultimate state satisfies the code requirement on beams.
(d)Check the minimum reinforcement area based on the code.
(e)Check φMn >= Mu.
Note:
(a)This question is modified from past national exams.
(b)Use 1.2D+1.6L for the loading combination.