土木人

一
規範:土木401-86a
B點彎矩`M_u=1.4*(3+2.4*0.6*0.4)*2^2/2+1.7*10*2=44.0128tf-m`
m=`(4200)/(0.85*350)`=14.1176
`R_n=(44.0128*10^5)/(0.9*40*52^2)=45.214`
`rho=1/m*(1-sqrt(1-(2*45.214*14.1176)/(4200)))=0.01174`
`A_s=rho*b*d=24.414cm^2`
土木401-93
B點彎矩`M_u=1.2*(3+2.4*0.6*0.4)*2^2/2+1.6*10*2=40.5824tf-m`
m=14.1176
`R_n=(40.5824*10^5)/(0.9*40*52^2)=41.69`
`rho=1/m*(1-sqrt(1-(2*41.69*14.1176)/(4200)))=0.01074`
`A_s=rho*b*d=22.34cm^2`

二
`P_n=208.059tf`
`M_n=6.234tf-m`

三
土木401-86a
`Z=f_s*(d_c*A)^(1/3)<={(31000(室內)),(26000(室外))]`
`f_s`=0.6*4200=2520
`d_c=4+1.27+3.23//2=6.885cm`
`A=(2b*d_s)/n=2*40*8.807/(4.59)=153.51`
`Z=2520*(6.885*153.51)^(1/3)=25669.28<26000`(OK!)
401-93
S=min`(38*2800//f_y-2.5C_c,30*2800//f_y)`=8.121cm

以上供参

401-93規範錯了
是s<Smin(38*2800/fs-2.5Cc,30*2800/fs)
才對

第三題
`f_s=2/3*f_y`
S=min`(38*(2800)/(f_s)-2.5C_c,30*(2800)/(f_s))=20.788cm`


第四題

s=min`(8d_b,24d_(bs),d/4,30cm)`=19.25cm

設設計靜載重不含RC梁自重
則係數化載重`w_u=2tf//m+1.2*2.4tf//m^3*0.6*0.9=3.555tf//m`
有疑問的是
`V_u=3.555*(4.5-0.77)=13.261tf<phi*V_c//2=0.75*0.53*sqrt(350)*60*77//2=17.178tf`
按照規範是不用配剪力筋?

laimaddux31 寫:第三題
`f_s=2/3*f_y`
S=min`(38*(2800)/(f_s)-2.5C_c,30*(2800)/(f_s))=20.788cm`


第四題

s=min`(8d_b,24d_(bs),d/4,30cm)`=19.25cm

設設計靜載重不含RC梁自重
則係數化載重`w_u=2tf//m+1.2*2.4tf//m^3*0.6*0.9=3.555tf//m`
有疑問的是
`V_u=3.555*(4.5-0.77)=13.261tf<phi*V_c//2=0.75*0.53*sqrt(350)*60*77//2=17.178tf`
按照規範是不用配剪力筋?

ㄜ...這是耐震設計 不是一般剪力筋的算法

主要公式是 Ve=(Mp1+Mp2)/L+w_u*L^2/2

其他細節請參考規範的耐震設計章節

第四題
依據耐震設計規範
`M_(pl)=258.783tf-m,M_(pr)=178.296tf-m`
`V_e=(M_(pl)+M_(pr))/(L_n)+-(w_u*L_n)/(2)=48.564+-9={(57.564tf),(39.564tf)]`
`(M_(pl)+M_(pr))/(L_n)=48.564>(V_e)/2=28.782tf,:.V_c=0`
`S=min(8d_b,24d_(bs),d/4,30,(A_v*f_y*d)/(V_e),(A_v*f_y)/(3.5*b_w))=9.513cm`

第四題.
個人拙解..幫樓上L大補充

B1 = 0.8
fy = 4200 kgf/cm^2

(1) Mpr1

As' = 9*8.14 = 73.26 cm^2 → p'=0.0159
As = 6*8.14 = 48.84 cm^2 → p=0.0106
Pcy = 0.85*B1*f'c/1.25fy*6120/(6120-fy)*d'/d + p' = 0.0403
p < pcy 壓力筋不降伏、拉力筋降伏
fs' = 6.12*(x-d')/x = 6.12 - 79.56/x
Cs = 73.26*[fs'-0.85*fc'] = 426.556 - 5828.566/x (tf)
T = 48.84*1.25*4.2 = 256.41 (tf)
Cc = 0.85*fc'*B1*x*b = 14.28x (tf)

Cc+Cs=T 14.28*x^2 + 170.146*x - 5828.556 = 0 → x = 15.11 cm
check fs' = 854.61 kgf/cm^2 < fy (O.K)

Mpr1 = Cc(d-a/2) + Cs(d-d') = 179.223 (tf-m)

(2) Mpr2

As' = 6*8.14 = 48.84 cm^2 → p=0.0106
As = 9*8.14 = 73.26 cm^2 → p'=0.0159
Pcy = 0.85*B1*f'c/1.25fy*6120/(6120-fy)*d'/d + p' = 0.035
p < pcy 亦為壓力筋不降伏、拉力筋降伏
fs' = 6.12*(x-d')/x = 6.12 - 79.56/x
Cs = 48.84*[fs'-0.85*fc'] = 284.371 - 3885.71/x (tf)
T = 73.26*1.25*4.2 = 384.615 (tf)
Cc = 0.85*fc'*B1*x*b = 14.28x (tf)

Cc+Cs=T 14.28*x^2 - 100.244*x - 3885.71 = 0 → x = 20.37 cm
check fs' = 2214 kgf/cm^2 < fy (O.K)

Mpr2 = Cc(d-a/2) + Cs(d-d') = 260.19 (tf-m)

(3) Ve

(假設wu已包含梁自重，取wu=2 tf/m)
Ve = (Mpr1+Mpr2)/Ln + wu*Ln/2 = 57.825 (tf)
(Mpr1+Mpr2)/Ln = 48.82 > Ve/2 = 28.9
故取Vc=0

(3) S

強度折減因子 φ=0.75
φ*(Av*fyt*d/S) > Ve
0.75*2*1.27*2.8*77/S > 57.825
S < 7.103 cm
Smax = min(8db,24dh,d/4,30cm) = 19.25 cm
S < Smax (O.K)

Ans. S < 7.103 cm

r96521230 寫:第四題.
個人拙解..幫樓上L大補充

B1 = 0.8
fy = 4200 kgf/cm^2

(1) Mpr1

As' = 9*8.14 = 73.26 cm^2 → p'=0.0159
As = 6*8.14 = 48.84 cm^2 → p=0.0106
Pcy = 0.85*B1*f'c/1.25fy*6120/(6120-fy)*d'/d + p' = 0.0403
p < pcy 壓力筋不降伏、拉力筋降伏
fs' = 6.12*(x-d')/x = 6.12 - 79.56/x
Cs = 73.26*[fs'-0.85*fc'] = 426.556 - 5828.566/x (tf)
T = 48.84*1.25*4.2 = 256.41 (tf)
Cc = 0.85*fc'*B1*x*b = 14.28x (tf)

Cc+Cs=T 14.28*x^2 + 170.146*x - 5828.556 = 0 → x = 15.11 cm
check fs' = 854.61 kgf/cm^2 < fy (O.K)

Mpr1 = Cc(d-a/2) + Cs(d-d') = 179.223 (tf-m)

(2) Mpr2

As' = 6*8.14 = 48.84 cm^2 → p=0.0106
As = 9*8.14 = 73.26 cm^2 → p'=0.0159
Pcy = 0.85*B1*f'c/1.25fy*6120/(6120-fy)*d'/d + p' = 0.035
p < pcy 亦為壓力筋不降伏、拉力筋降伏
fs' = 6.12*(x-d')/x = 6.12 - 79.56/x
Cs = 48.84*[fs'-0.85*fc'] = 284.371 - 3885.71/x (tf)
T = 73.26*1.25*4.2 = 384.615 (tf)
Cc = 0.85*fc'*B1*x*b = 14.28x (tf)

Cc+Cs=T 14.28*x^2 - 100.244*x - 3885.71 = 0 → x = 20.37 cm
check fs' = 2214 kgf/cm^2 < fy (O.K)

Mpr2 = Cc(d-a/2) + Cs(d-d') = 260.19 (tf-m)

(3) Ve

(假設wu已包含梁自重，取wu=2 tf/m)
Ve = (Mpr1+Mpr2)/Ln + wu*Ln/2 = 57.825 (tf)
(Mpr1+Mpr2)/Ln = 48.82 > Ve/2 = 28.9
故取Vc=0

(3) S

強度折減因子 φ=0.75
φ*(Av*fyt*d/S) > Ve
0.75*2*1.27*2.8*77/S > 57.825
S < 7.103 cm
Smax = min(8db,24dh,d/4,30cm) = 19.25 cm
S < Smax (O.K)

Ans. S < 7.103 cm

第四題我看92結技試題解析實力于耀的解法，他是設定拉力筋發揮1. 25As fy，而忽略壓力筋的貢獻喔

請問L大,第一題B點的彎矩計算,為甚麼可以忽略左側(B點)的影響?可否略為說明一下,謝謝.

chc 寫:請問L大,第一題B點的彎矩計算,為甚麼可以忽略左側(B點)的影響?可否略為說明一下,謝謝.

當把集中力放在最右側時
B點的彎矩最大
取右邊算彎矩
才會等於B點彎矩`M_u=1.2*(3+2.4*0.6*0.4)*2^2/2+1.6*10*2=40.5824tf-m`

kingofmen 寫:

chc 寫:請問L大,第一題B點的彎矩計算,為甚麼可以忽略左側(B點)的影響?可否略為說明一下,謝謝.

當把集中力放在最右側時
B點的彎矩最大
取右邊算彎矩
才會等於B點彎矩`M_u=1.2*(3+2.4*0.6*0.4)*2^2/2+1.6*10*2=40.5824tf-m`

.

我也是找到彎矩最大值是集中力放在最右側時,
這題讓我寫了40多分鐘才算完,