1.500 m X Y m m m m m m m m m m m m N/ N/ ( ) qa N/ N/ 2 2
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2 1.500 m X Y m m m m m m m m m m m m N/ N/ ( ) qa N/ N/ X(m) Y(m) (kn/m 2 )
3 kn N/m y P = m xp = m b p = m σ ck = 21 N/mm 2 σ ca = 7 N/mm 2 τ ca = 0.7 N/mm 2 σ sa = 180 N/mm 2 τ a 2.1 N/mm 2 X P P y p 0.26 N/mm 2 t(mm) U(mm) No X Y Xn+1 - Xn / m2 γc kn, ) Σ i ( i 2/3 i)} i Σ i ( i 1/3 i)} i i
4 (kn) (m) (knm) δ= () j= () α= 55 () kn kn a kn KA=2 Pa/(γs h 2 ) = KH=Ka cos(δ ) = KV=Ka sin(δ ) = 0.102
5 PH=1/2 h^2 γs KH My=PH Y PV=1/2 h^2 γs KV Mx=PV X QwH=Qw h KH My=QwH Y QwV=Qw h KV Mx=QwV X m Σ kn Σ kn M kNm e/2 / m m OK Nµ+CB Fs H NO d B/2 e m m q kn/m 2 q kn/m 2 qa kn/m OK
6 Df1= m γ1= kn/m 3 Df2= m γ2= kn/m kn/m 2 φ= 30.0 c= 0.0 kn/m 2 B= m L= m V= kn HB= 9.63 kn e= m Q = / Ae ( α K C Nc c + K q Nq q + 1/2 γ 2 β Be Nr r ) = 1/ ( / ) = 51 kn/m 2 a : a= 3 C : kn/m2 q : (kn/m2) q γ1 Df1 γ2 Df (kn/m2) Ae : (m2) Ae Be L (m2) Be : (m) Be B - 2 eb (m) α,β : α= 1.00 β= 1.00 K : K = * Df'/ Be Nc,Nq,Nr : tanθ = B/ Nc = 9.25 Nq = 6.34 Nr = 1.92 λ µ-1/3 c ') λ ' q ') ' γ ') µ ' B B0 B
7
8 Hk Nµ+CB P k Fs H OK
9 kn kn = P= kn y P = m ( ) b p = m 1m P u = kn/m knm/m kn/m kn knm/m kn/m
10 ( B2 = 200 mm b = 1000 mm i = 60 mm d = 140 mm A s = 285 mm 2 U = 120 mm b σ c x/3 x=kd M C h A s d z=jd T=σ s A s E A S S n = = 15 np = n = Ec b d k = 2 k ( np) + 2np np = j = 1 = M= 4.39E+06 Nmm S= 7.74E+03 N 2.2 N/mm N/mm 2 OK N/mm 2 σsa= N/mm 2 OK N/mm 2 τca= 0.70 N/mm 2 OK τca= 2.10 OK d = 140 mm M= 0.00E+00 Nmm A s = 285 mm 2 S= 0.00E+00 N U = 120 mm K = j = N/mm N/mm 2 OK 0.0 N/mm 2 σsa= N/mm 2 OK N/mm 2 τca= 0.70 N/mm 2 OK N/mm τa= 2.10 OK
11 l lh2/2 l 0.100m H3 = 0.20 m H2 = 0.20 m B= 1.40 m HS = 0.20 m l m =B5 = 1.20 m l s = 1.10 m W1=H3γc(hp-H3)γs kn/m2 W2=H2γc(hp-H2)γs 4.90 (W 2 W 1 )l W s 3 W 1 B kn/m2 q 1 = kn/m 2 q 2 = kn/m 2 q 3 = kn/m 2 q 4 = kn/m 2 ls/2q1q4w1w kn/m l m 2 /6 2 q1 W1 q3 W knm/m
12 T = mm b = mm i = mm d = mm A s = mm 2 U = mm E S = = 15 A n S np = n = Ec b d ( np) + np = k j = = k = 2 np M= 4.90E+06 Nmm S= 7.76E+03 N 2.5 N/mm 2 σca= 7.0 N/mm 2 OK N/mm 2 σca= N/mm 2 OK d = 140 mm N/mm 2 τca= 0.70 N/mm 2 OK N/mm τca= 2.10 OK
13 OK OK NO OK OK OK OK OK OK OK OK OK OK OK OK OK #DIV/0! 0.00 #DIV/0! #DIV/0! #DIV/0! #DIV/0! OK #DIV/0! #DIV/0! #DIV/0! 0.00 #DIV/0! 0.00 #DIV/0! 0.00 OK #DIV/0! OK #DIV/0! OK OK OK NO NO OK OK OK OK OK OK OK OK OK OK OK OK #DIV/0! 0.00 #DIV/0! #DIV/0! #DIV/0! #DIV/0! OK #DIV/0! #DIV/0! #DIV/0! 0.00 #DIV/0! 0.00 #DIV/0! 0.00 OK #DIV/0! OK #DIV/0! OK
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20 3 Copyright (2007) by P.W.R.I. All rights reserved. No part of this book may be reproduced by any means, nor transmitted, nor translated into a machine language without the written permission of the
(1)基礎の設計に関する基本事項
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無筋擁壁設計システム Ver4.2 適用基準 土地改良事業計画設計基準 設計 農道 (H7/3) 土地改良事業計画設計基準 設計 水路工 (H26/3) 日本道路協会 道路土工 擁壁工指針 (H24/7) 土木学会 大型ブロック積み擁壁設計 (H6/6) 宅地防災マニュアルの解説 第二次改訂版 (H9/2) 出力例 ブロック積み擁壁の計算書 ( 安定計算および部材断面計算 ) 開発 販売元 ( 株
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1- 擁壁断面の形状 寸法及び荷重の計算 ( 常時 ) フェンス荷重 1 kn/m 1,100 0 上載荷重 10 m kn/ 3, (1) 自重 地表面と水平面とのなす角度 α=0.00 壁背面と鉛直面とのなす角度 θ=.73 擁壁
構造計算例鉄筋コンクリート造擁壁の構造計算例 1 常時 1-1 設計条件 (1) 擁壁の型式及び高さ型式 : 片持梁式鉄筋コンクリート造 L 型擁壁擁壁の高さ :H'=3.00m 擁壁の全高 :H =3.50m () 外力土圧の作用面は縦壁背面とする 上載荷重 : q=10kn/ mフェンス荷重 ( 水平力 ) : 1kN/ m (3) 背面土土質の種類 : 関東ローム土の単位体積重量 :γs=16.0/
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Super Build/宅造擁壁 出力例1
宅造擁壁構造計算書 使用プログラム : uper Build/ 宅造擁壁 Ver.1.60 工事名 : 日付 : 設計者名 : 宅地防災マニュアル事例集 015/01/7 UNION YTEM INC. Ⅶ-1 建設地 : L 型擁壁の設計例 壁体背面を荷重面としてとる場合 *** uper Build/ 宅造擁壁 *** 160-999999 [ 宅地防災マニュアル Ⅶ-1] 015/01/7 00:00
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6 x x 6.1 t P P = P t P = I P P P 1 0 1 0,, 0 1 0 1 cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ x θ x θ P x P x, P ) = t P x)p ) = t x t P P ) = t x = x, ) 6.1) x = Figure 6.1 Px = x, P=, θ = θ P
1. 1 A : l l : (1) l m (m 3) (2) m (3) n (n 3) (4) A α, β γ α β + γ = 2 m l lm n nα nα = lm. α = lm n. m lm 2β 2β = lm β = lm 2. γ l 2. 3
1. 1 A : l l : (1) l m (m 3) (2) m (3) n (n 3) (4) A 2 1 2 1 2 3 α, β γ α β + γ = 2 m l lm n nα nα = lm. α = lm n. m lm 2β 2β = lm β = lm 2. γ l 2. 3 4 P, Q R n = {(x 1, x 2,, x n ) ; x 1, x 2,, x n R}
t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ
4 5 ( 5 3 9 4 0 5 ( 4 6 7 7 ( 0 8 3 9 ( 8 t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ S θ > 0 θ < 0 ( P S(, 0 θ > 0 ( 60 θ
1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1
1 I 1.1 ± e = = - =1.602 10 19 C C MKA [m], [Kg] [s] [A] 1C 1A 1 MKA 1C 1C +q q +q q 1 1.1 r 1,2 q 1, q 2 r 12 2 q 1, q 2 2 F 12 = k q 1q 2 r 12 2 (1.1) k 2 k 2 ( r 1 r 2 ) ( r 2 r 1 ) q 1 q 2 (q 1 q 2
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道路土工擁壁工指針 (H24) に準拠 重力式擁壁の安定計算 ( 盛土土圧対応 ) 正規版 Ver 基本データの入力 2 地形データの入力 3 計算実行 Ver /01/18 Civil Tech 洋洋 本ソフトの概要 機能 道路土工 擁壁工指針 ( 平成 24 年度
道路土工擁壁工指針 (H24) に準拠 重力式擁壁の安定計算 ( 盛土土圧対応 ) 正規版 Ver.1.10 1 基本データの入力 2 地形データの入力 3 計算実行 Ver 1.10 2019/01/18 Civil Tech 洋洋 本ソフトの概要 機能 道路土工 擁壁工指針 ( 平成 24 年度版 ) に準拠して 重力式擁壁の安定計算を行ないます 滑動 転倒 地盤支持力の安定検討を行うことができます
Microsoft Word - 部材規格追記 doc
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集水桝の構造計算(固定版編)V1-正規版.xls
集水桝の構造計算 集水桝 3.0.5 3.15 横断方向断面の計算 1. 計算条件 11. 集水桝の寸法 内空幅 B = 3.000 (m) 内空奥行き L =.500 (m) 内空高さ H = 3.150 (m) 側壁厚 T = 0.300 (m) 底版厚 Tb = 0.400 (m) 1. 土質条件 土の単位体積重量 γs = 18.000 (kn/m 3 ) 土の内部摩擦角 φ = 30.000
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Report JS0.5 J Simplicity February 4, 2012 1 J Simplicity HOME http://www.jsimplicity.com/ Preface 2 Report 2 Contents I 5 1 6 1.1..................................... 6 1.2 1 1:................ 7 1.3
x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y)
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