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高分请大家帮帮忙：步进电机单片机控制程序C翻译成汇编
这是生成的汇编代码
和C语言一一对应的,看的差不多了就把C语言的语句删掉编译才能通过.
126: ?C_STARTUP: LJMP STARTUP1
127:
128: RSEG ?C_C51STARTUP
129:
130: STARTUP1:
131:
132: IF IDATALEN 0
C:0x0000 020003 LJMP STARTUP1(C:0003)
133: MOV R0,#IDATALEN - 1
C:0x0003 787F MOV R0,#0x7F
134: CLR A
C:0x0005 E4 CLR A
135: IDATALOOP: MOV @R0,A
C:0x0006 F6 MOV @R0,A
136: DJNZ R0,IDATALOOP
C:0x0007 D8FD DJNZ R0,IDATALOOP(C:0006)
185: MOV SP,#?STACK-1
186:
187: ; This code is required if you use L51_BANK.A51 with Banking Mode 4
188: ;h Code Banking
189: ; q Select Bank 0 for L51_BANK.A51 Mode 4
190: #if 0
191: ; i Initialize bank mechanism to code bank 0 when using L51_BANK.A51 with Banking Mode 4.
192: EXTRN CODE (?B_SWITCH0)
193: CALL ?B_SWITCH0 ; init bank mechanism to code bank 0
194: #endif
195: ;/h
C:0x0009 75814B MOV SP(0x81),#0x4B
196: LJMP ?C_START
C:0x000C 02004A LJMP C:004A
C:0x000F 020118 LJMP main(C:0118)
C:0x0012 E4 CLR A
C:0x0013 93 MOVC A,@A+DPTR
C:0x0014 A3 INC DPTR
C:0x0015 F8 MOV R0,A
C:0x0016 E4 CLR A
C:0x0017 93 MOVC A,@A+DPTR
C:0x0018 A3 INC DPTR
C:0x0019 4003 JC C:001E
C:0x001B F6 MOV @R0,A
C:0x001C 8001 SJMP C:001F
C:0x001E F2 MOVX @R0,A
C:0x001F 08 INC R0
C:0x0020 DFF4 DJNZ R7,C:0016
C:0x0022 8029 SJMP C:004D
C:0x0024 E4 CLR A
C:0x0025 93 MOVC A,@A+DPTR
C:0x0026 A3 INC DPTR
C:0x0027 F8 MOV R0,A
C:0x0028 5407 ANL A,#0x07
C:0x002A 240C ADD A,#0x0C
C:0x002C C8 XCH A,R0
C:0x002D C3 CLR C
C:0x002E 33 RLC A
C:0x002F C4 SWAP A
C:0x0030 540F ANL A,#0x0F
C:0x0032 4420 ORL A,#0x20
C:0x0034 C8 XCH A,R0
C:0x0035 83 MOVC A,@A+PC
C:0x0036 4004 JC C:003C
C:0x0038 F4 CPL A
C:0x0039 56 ANL A,@R0
C:0x003A 8001 SJMP C:003D
C:0x003C 46 ORL A,@R0
C:0x003D F6 MOV @R0,A
C:0x003E DFE4 DJNZ R7,C:0024
C:0x0040 800B SJMP C:004D
C:0x0042 0102 AJMP C:0002
C:0x0044 04 INC A
C:0x0045 08 INC R0
C:0x0046 102040 JBC 0x24.0,C:0089
C:0x0049 8090 SJMP C:FFDB
C:0x004B 00 NOP
C:0x004C 8FE4 MOV 0xE4,R7
C:0x004E 7E01 MOV R6,#0x01
C:0x0050 93 MOVC A,@A+DPTR
C:0x0051 60BC JZ C:000F
C:0x0053 A3 INC DPTR
C:0x0054 FF MOV R7,A
C:0x0055 543F ANL A,#0x3F
C:0x0057 30E509 JNB 0xE0.5,C:0063
C:0x005A 541F ANL A,#0x1F
C:0x005C FE MOV R6,A
C:0x005D E4 CLR A
C:0x005E 93 MOVC A,@A+DPTR
C:0x005F A3 INC DPTR
C:0x0060 6001 JZ C:0063
C:0x0062 0E INC R6
C:0x0063 CF XCH A,R7
C:0x0064 54C0 ANL A,#0xC0
C:0x0066 25E0 ADD A,ACC(0xE0)
C:0x0068 60A8 JZ C:0012
C:0x006A 40B8 JC C:0024
C:0x006C E4 CLR A
C:0x006D 93 MOVC A,@A+DPTR
C:0x006E A3 INC DPTR
C:0x006F FA MOV R2,A
C:0x0070 E4 CLR A
C:0x0071 93 MOVC A,@A+DPTR
C:0x0072 A3 INC DPTR
C:0x0073 F8 MOV R0,A
C:0x0074 E4 CLR A
C:0x0075 93 MOVC A,@A+DPTR
C:0x0076 A3 INC DPTR
C:0x0077 C8 XCH A,R0
C:0x0078 C582 XCH A,DPL(0x82)
C:0x007A C8 XCH A,R0
C:0x007B CA XCH A,R2
C:0x007C C583 XCH A,DPH(0x83)
C:0x007E CA XCH A,R2
C:0x007F F0 MOVX @DPTR,A
C:0x0080 A3 INC DPTR
C:0x0081 C8 XCH A,R0
C:0x0082 C582 XCH A,DPL(0x82)
C:0x0084 C8 XCH A,R0
C:0x0085 CA XCH A,R2
C:0x0086 C583 XCH A,DPH(0x83)
C:0x0088 CA XCH A,R2
C:0x0089 DFE9 DJNZ R7,C:0074
C:0x008B DEE7 DJNZ R6,C:0074
C:0x008D 80BE SJMP C:004D
C:0x008F 012F AJMP C:002F
C:0x0091 00 NOP
C:0x0092 0A INC R2
C:0x0093 423F ORL 0x3F,A
C:0x0095 06 INC @R0
C:0x0096 5B ANL A,R3
C:0x0097 4F ORL A,R7
C:0x0098 66 XRL A,@R0
C:0x0099 6D XRL A,R5
C:0x009A 7D07 MOV R5,#0x07
C:0x009C 7F6F MOV R7,#0x6F
C:0x009E 08 INC R0
C:0x009F 2111 AJMP C:0111
C:0x00A1 99 SUBB A,R1
C:0x00A2 88CC MOV 0xCC,R0
C:0x00A4 4466 ORL A,#0x66
C:0x00A6 22 RET
C:0x00A7 33 RLC A
C:0x00A8 0D INC R5
C:0x00A9 35EE ADDC A,0xEE
C:0x00AB DEBE DJNZ R6,C:006B
C:0x00AD 7EED MOV R6,#0xED
C:0x00AF DDBD DJNZ R5,C:006E
C:0x00B1 7DEB MOV R5,#0xEB
C:0x00B3 DBBB DJNZ R3,C:0070
C:0x00B5 7BE7 MOV R3,#0xE7
C:0x00B7 04 INC A
C:0x00B8 300000 JNB f0(0x20.0),C:00BB
C:0x00BB 07 INC @R1
C:0x00BC D004 POP 0x04
C:0x00BE 2B ADD A,R3
C:0x00BF 00 NOP
C:0x00C0 00 NOP
C:0x00C1 00 NOP
C:0x00C2 00 NOP
C:0x00C3 C181 AJMP C:0681
C:0x00C5 00 NOP
103: void delay1(unsigned int t)
104: {unsigned int i,k,j;
105:
106: for(k=0;kt;k++)
C:0x00C6 E4 CLR A
C:0x00C7 FD MOV R5,A
C:0x00C8 FC MOV R4,A
C:0x00C9 C3 CLR C
C:0x00CA ED MOV A,R5
C:0x00CB 9F SUBB A,R7
C:0x00CC EC MOV A,R4
C:0x00CD 9E SUBB A,R6
C:0x00CE 5021 JNC C:00F1
107: for(i=0;i=50;i++)
C:0x00D0 E4 CLR A
C:0x00D1 FB MOV R3,A
C:0x00D2 FA MOV R2,A
108: for(j=0;j=10;j++);
C:0x00D3 900000 MOV DPTR,#C_STARTUP(0x0000)
C:0x00D6 A3 INC DPTR
C:0x00D7 E582 MOV A,DPL(0x82)
C:0x00D9 640B XRL A,#0x0B
C:0x00DB 4583 ORL A,DPH(0x83)
C:0x00DD 70F7 JNZ C:00D6
C:0x00DF 0B INC R3
C:0x00E0 BB0001 CJNE R3,#0x00,C:00E4
C:0x00E3 0A INC R2
C:0x00E4 EB MOV A,R3
C:0x00E5 6433 XRL A,#0x33
C:0x00E7 4A ORL A,R2
C:0x00E8 70E9 JNZ C:00D3
C:0x00EA 0D INC R5
C:0x00EB BD0001 CJNE R5,#0x00,C:00EF
C:0x00EE 0C INC R4
C:0x00EF 80D8 SJMP C:00C9
109: }
110:
111:
112:
113: const unsigned char keycode[]=
114: {0xee,0xde,0xbe,0x7e,
115: 0xed,0xdd,0xbd,0x7d,0xeb,0xdb,0xbb,0x7b,0xe7,
116: };
117:
118:
119: long timecount=2000,count=0;
120: bit f0,flag99999=1;
C:0x00F1 22 RET
77: void a_step(uchar d)
78: {
79: if (d0x01)
C:0x00F2 EF MOV A,R7
C:0x00F3 30E00D JNB 0xE0.0,C:0103
80: {
81: if (np==0)
C:0x00F6 E534 MOV A,np(0x34)
C:0x00F8 7005 JNZ C:00FF
82: np=7;
C:0x00FA 753407 MOV np(0x34),#0x07
C:0x00FD 8010 SJMP C:010F
83: else np--;
C:0x00FF 1534 DEC np(0x34)
84: }
85: else
C:0x0101 800C SJMP C:010F
86: {
87: if (np==7)
C:0x0103 E534 MOV A,np(0x34)
C:0x0105 B40705 CJNE A,#0x07,C:010D
88: np=0;
C:0x0108 E4 CLR A
C:0x0109 F534 MOV np(0x34),A
C:0x010B 8002 SJMP C:010F
89: else np++;
90:
C:0x010D 0534 INC np(0x34)
91: }
92:
93: P3=motortb[np];
94:
C:0x010F 7421 MOV A,#motortb(0x21)
C:0x0111 2534 ADD A,np(0x34)
C:0x0113 F8 MOV R0,A
C:0x0114 E6 MOV A,@R0
C:0x0115 F5B0 MOV P3(0xB0),A
95: }
C:0x0117 22 RET
121: void main(void)
122: { unsigned char temp,key,shudu=39;
C:0x0118 750A27 MOV 0x0A,#0x27
123: TMOD=0x02;
C:0x011B 758902 MOV TMOD(0x89),#0x02
124: TH0=0x00;
C:0x011E E4 CLR A
C:0x011F F58C MOV TH0(0x8C),A
125: TL0=TH0;// 未完，
C:0x0121 858C8A MOV TL0(0x8A),TH0(0x8C)
126: PT0=1;
C:0x0124 D2B9 SETB PT0(0xB8.1)
127: ET0=1;
C:0x0126 D2A9 SETB ET0(0xA8.1)
128: TR0=1;
C:0x0128 D28C SETB TR0(0x88.4)
129: EA=1;
C:0x012A D2AF SETB EA(0xA8.7)
130: while(1)
131: { P1=0x0F;
C:0x012C 75900F MOV P1(0x90),#0x0F
132: if((P10x0f)!=0x0F);
C:0x012F E590 MOV A,P1(0x90)
C:0x0131 540F ANL A,#0x0F
C:0x0133 640F XRL A,#0x0F
133: }}
C:0x0135 80F5 SJMP C:012C
29: void delay(uchar t)
30: {
31: uchar i;
32: uint j;
33: for (i=0;it;i++)
C:0x0137 E4 CLR A
C:0x0138 FE MOV R6,A
C:0x0139 EE MOV A,R6
C:0x013A C3 CLR C
C:0x013B 9F SUBB A,R7
C:0x013C 5011 JNC C:014F
34: for (j=0;j900;j++);
C:0x013E E4 CLR A
C:0x013F FD MOV R5,A
C:0x0140 FC MOV R4,A
C:0x0141 0D INC R5
C:0x0142 BD0001 CJNE R5,#0x00,C:0146
C:0x0145 0C INC R4
C:0x0146 BC03F8 CJNE R4,#0x03,C:0141
C:0x0149 BD84F5 CJNE R5,#0x84,C:0141
C:0x014C 0E INC R6
C:0x014D 80EA SJMP C:0139
35: }
36:
37: //
38: void select(uchar LED)
39: {
40: switch(LED)
41: {
42: case 0:P2_0=0;break;
43: case 1:P2_1=0;break;
44: default: P2=0xff;
45: }
46: }
47:
C:0x014F 22 RET
48: void dispone(uchar LED,uchar number) //number 0--9
49: {
50:
51: select(LED);
C:0x0150 120177 LCALL select(C:0177)
52: P0=LEDDATA[number];
53:
C:0x0153 7442 MOV A,#LEDDATA(0x42)
C:0x0155 2D ADD A,R5
C:0x0156 F8 MOV R0,A
C:0x0157 E6 MOV A,@R0
C:0x0158 F580 MOV P0(0x80),A
54: delayus(200);
C:0x015A 7FC8 MOV R7,#0xC8
C:0x015C 7E00 MOV R6,#0x00
C:0x015E 120188 LCALL delayus(C:0188)
55: P2=0xff;
C:0x0161 75A0FF MOV PPAGE_SFR(0xA0),#0xFF
56: }
57:
C:0x0164 22 RET
58: void disp()
59: {
60: uchar i;
61:
62: for(i=0;i2;i++)
C:0x0165 E4 CLR A
C:0x0166 FB MOV R3,A
63: {
64:
65: dispone(i,LEDBUFFER[i]);
66:
C:0x0167 AF03 MOV R7,0x03
C:0x0169 7429 MOV A,#LEDBUFFER(0x29)
C:0x016B 2B ADD A,R3
C:0x016C F8 MOV R0,A
C:0x016D E6 MOV A,@R0
C:0x016E FD MOV R5,A
C:0x016F 120150 LCALL dispone(C:0150)
67: }
C:0x0172 0B INC R3
C:0x0173 BB02F1 CJNE R3,#0x02,C:0167
68: }
C:0x0176 22 RET
38: void select(uchar LED)
39: {
40: switch(LED)
C:0x0177 EF MOV A,R7
C:0x0178 14 DEC A
C:0x0179 6006 JZ C:0181
C:0x017B 04 INC A
C:0x017C 7006 JNZ C:0184
41: {
42: case 0:P2_0=0;break;
C:0x017E C2A0 CLR P2_0(0xA0.0)
C:0x0180 22 RET
43: case 1:P2_1=0;break;
C:0x0181 C2A1 CLR P2_1(0xA0.1)
C:0x0183 22 RET
44: default: P2=0xff;
C:0x0184 75A0FF MOV PPAGE_SFR(0xA0),#0xFF
45: }
C:0x0187 22 RET
21: void delayus(uint us)
22: {
23: while(us--);
C:0x0188 EF MOV A,R7
C:0x0189 1F DEC R7
C:0x018A AC06 MOV R4,0x06
C:0x018C 7001 JNZ C:018F
C:0x018E 1E DEC R6
C:0x018F 4C ORL A,R4
C:0x0190 70F6 JNZ delayus(C:0188)
24: }
C:0x0192 22 RET
KTR是外国品牌吗?
是的
德国KTR 品牌简介
KUPPLUNGSTECHNIK GMBH自从进入中国市场以来,凭借其优良的品质、丰富的产品种类以及热情的服务,在中国联轴器市场取得了极大的成功,尤其是其ROTEX系列联轴器,在中国的CNC以及工程机械行业占领了较大的市场份额。
KTR连轴器广泛应用于工程机械、机床、冶金、石油化工设备及各种通用机械等，几乎所有需要动力传递的机械设备中都要用到KTR的产品。由于其的性能和优良的品质，KTR的产品已为世界各地的设备厂商所采用。
KTR联轴器特点：
有钢质轴套,扭向弹性,免维护,吸收振动;
轴向插入式安装,失效保护;
良好的动态特性;设计紧凑,惯性小;
成品孔径公差按照ISO标准为H7,键槽宽公差标准DIN 6885/1为JS9.
KTR联轴器弹性体的正常工作温度为-40-+100℃,允许的zui高瞬时温度为120℃.弹性体的肖氏硬度通常为92 Shore A,若需传递更高扭矩,可选用硬度为95/98 Shore A和64D-F的弹性体.弹性体耐磨,抗油,抗臭氧,抗老化,其耐水解性适合热带气候地区.由于具有的内部缓冲,能保护传动不受过载的影响.
德国KTR公司主要产品有：KTR联轴器、KTR曲面齿联轴器、KTR尼龙曲面齿联轴器、KTR特种曲面齿联轴器、KTR扭力限制器、KTR涨紧套、KTR力矩转速检测仪
＂基础工程＂计算题，求解答计算，急急急急急！
你题目没有给出竖向力大小。
按750kN计算时，抗冲切承载力验算可以通过。
按800kN计算时，抗冲切承载力验算不能通过。
阶梯基础计算
项目名称_____________日 期_____________
设 计 者_____________校 对 者_____________
一、设计依据
《建筑地基基础设计规范》 (GB50007-2011)①
《混凝土结构设计规范》 (GB50010-2010)②
二、示意图
三、计算信息
构件编号: JC-1 计算类型: 验算截面尺寸
1. 几何参数
台阶数 n=1
矩形柱宽 bc=400mm 矩形柱高 hc=600mm
基础高度 h1=500mm
一阶长度 b1=800mm b2=800mm 一阶宽度 a1=1000mm a2=1000mm
2. 材料信息
基础混凝土等级: C25 ft_b=1.27N/mm2 fc_b=11.9N/mm2
柱混凝土等级: C30 ft_c=1.43N/mm2 fc_c=14.3N/mm2
钢筋级别: HRB400 fy=360N/mm2
3. 计算信息
结构重要性系数: γo=1.0
基础埋深: dh=1.500m
纵筋合力点至近边距离: as=40mm
基础及其上覆土的平均容重: γ=20.000kN/m3
最小配筋率: ρmin=0.150%
Fgk=750.000kN Fqk=0.000kN
Mgxk=212.500kN*m Mqxk=0.000kN*m
Mgyk=0.000kN*m Mqyk=0.000kN*m
Vgxk=0.000kN Vqxk=0.000kN
Vgyk=0.000kN Vqyk=0.000kN
永久荷载分项系数rg=1.20
可变荷载分项系数rq=1.40
Fk=Fgk+Fqk=750.000+(0.000)=750.000kN
Mxk=Mgxk+Fgk*(A2-A1)/2+Mqxk+Fqk*(A2-A1)/2
=212.500+750.000*(1.300-1.300)/2+(0.000)+0.000*(1.300-1.300)/2
=212.500kN*m
Myk=Mgyk+Fgk*(B2-B1)/2+Mqyk+Fqk*(B2-B1)/2
=0.000+750.000*(1.000-1.000)/2+(0.000)+0.000*(1.000-1.000)/2
=0.000kN*m
Vxk=Vgxk+Vqxk=0.000+(0.000)=0.000kN
Vyk=Vgyk+Vqyk=0.000+(0.000)=0.000kN
F1=rg*Fgk+rq*Fqk=1.20*(750.000)+1.40*(0.000)=900.000kN
Mx1=rg*(Mgxk+Fgk*(A2-A1)/2)+rq*(Mqxk+Fqk*(A2-A1)/2)
=1.20*(212.500+750.000*(1.300-1.300)/2)+1.40*(0.000+0.000*(1.300-1.300)/2)
=255.000kN*m
My1=rg*(Mgyk+Fgk*(B2-B1)/2)+rq*(Mqyk+Fqk*(B2-B1)/2)
=1.20*(0.000+750.000*(1.000-1.000)/2)+1.40*(0.000+0.000*(1.000-1.000)/2)
=0.000kN*m
Vx1=rg*Vgxk+rq*Vqxk=1.20*(0.000)+1.40*(0.000)=0.000kN
Vy1=rg*Vgyk+rq*Vqyk=1.20*(0.000)+1.40*(0.000)=0.000kN
F2=1.35*Fk=1.35*750.000=1012.500kN
Mx2=1.35*Mxk=1.35*212.500=286.875kN*m
My2=1.35*Myk=1.35*(0.000)=0.000kN*m
Vx2=1.35*Vxk=1.35*(0.000)=0.000kN
Vy2=1.35*Vyk=1.35*(0.000)=0.000kN
F=max(|F1|,|F2|)=max(|900.000|,|1012.500|)=1012.500kN
Mx=max(|Mx1|,|Mx2|)=max(|255.000|,|286.875|)=286.875kN*m
My=max(|My1|,|My2|)=max(|0.000|,|0.000|)=0.000kN*m
Vx=max(|Vx1|,|Vx2|)=max(|0.000|,|0.000|)=0.000kN
Vy=max(|Vy1|,|Vy2|)=max(|0.000|,|0.000|)=0.000kN
5. 修正后的地基承载力特征值
fa=2000.000kPa
四、计算参数
1. 基础总长 Bx=b1+b2+bc=0.800+0.800+0.400=2.000m
2. 基础总宽 By=a1+a2+hc=1.000+1.000+0.600=2.600m
A1=a1+hc/2=1.000+0.600/2=1.300m A2=a2+hc/2=1.000+0.600/2=1.300m
B1=b1+bc/2=0.800+0.400/2=1.000m B2=b2+bc/2=0.800+0.400/2=1.000m
3. 基础总高 H=h1=0.500=0.500m
4. 底板配筋计算高度 ho=h1-as=0.500-0.040=0.460m
5. 基础底面积 A=Bx*By=2.000*2.600=5.200m2
6. Gk=γ*Bx*By*dh=20.000*2.000*2.600*1.500=156.000kN
G=1.35*Gk=1.35*156.000=210.600kN
五、计算作用在基础底部弯矩值
Mdxk=Mxk-Vyk*H=212.500-0.000*0.500=212.500kN*m
Mdyk=Myk+Vxk*H=0.000+0.000*0.500=0.000kN*m
Mdx=Mx-Vy*H=286.875-0.000*0.500=286.875kN*m
Mdy=My+Vx*H=0.000+0.000*0.500=0.000kN*m
六、验算地基承载力
1. 验算轴心荷载作用下地基承载力
pk=(Fk+Gk)/A=(750.000+156.000)/5.200=174.231kPa 【①5.2.1-2】
因γo*pk=1.0*174.231=174.231kPa≤fa=2000.000kPa
轴心荷载作用下地基承载力满足要求
2. 验算偏心荷载作用下的地基承载力
因 Mdyk=0 Pkmax_x=Pkmin_x=(Fk+Gk)/A=(750.000+156.000)/5.200=174.231kPa
eyk=Mdxk/(Fk+Gk)=212.500/(750.000+156.000)=0.235m
因 |eyk| ≤By/6=0.433m y方向小偏心
Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx)
=(750.000+156.000)/5.200+6*|212.500|/(2.6002*2.000)
=268.536kPa
Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx)
=(750.000+156.000)/5.200-6*|212.500|/(2.6002*2.000)
=79.926kPa
3. 确定基础底面反力设计值
Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk
=(174.231-174.231)+(268.536-174.231)+174.231
=268.536kPa
γo*Pkmax=1.0*268.536=268.536kPa≤1.2*fa=1.2*2000.000=2400.000kPa
偏心荷载作用下地基承载力满足要求
七、基础冲切验算
1. 计算基础底面反力设计值
1.1 计算x方向基础底面反力设计值
ex=Mdy/(F+G)=0.000/(1012.500+210.600)=0.000m
因 ex≤ Bx/6.0=0.333m x方向小偏心
Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By)
=(1012.500+210.600)/5.200+6*|0.000|/(2.0002*2.600)
=235.212kPa
Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By)
=(1012.500+210.600)/5.200-6*|0.000|/(2.0002*2.600)
=235.212kPa
1.2 计算y方向基础底面反力设计值
ey=Mdx/(F+G)=286.875/(1012.500+210.600)=0.235m
因 ey ≤By/6=0.433 y方向小偏心
Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx)
=(1012.500+210.600)/5.200+6*|286.875|/(2.6002*2.000)
=362.523kPa
Pmin_y=(F+G)/A-6*|Mdx|/(By2*Bx)
=(1012.500+210.600)/5.200-6*|286.875|/(2.6002*2.000)
=107.900kPa
1.3 因 Mdx≠0 并且 Mdy=0
Pmax=Pmax_y=362.523kPa
Pmin=Pmin_y=107.900kPa
1.4 计算地基净反力极值
Pjmax=Pmax-G/A=362.523-210.600/5.200=322.023kPa
2. 验算柱边冲切
YH=h1=0.500m, YB=bc=0.400m, YL=hc=0.600m
YB1=B1=1.000m, YB2=B2=1.000m, YL1=A1=1.300m, YL2=A2=1.300m
YHo=YH-as=0.460m
2.1 因 (YH≤800) βhp=1.0
2.2 x方向柱对基础的冲切验算
x冲切位置斜截面上边长 bt=YB=0.400m
x冲切位置斜截面下边长 bb=YB+2*YHo=1.320m
x冲切不利位置 bm=(bt+bb)/2=(0.400+1.320)/2=0.860m
x冲切面积 Alx=max((YL1-YL/2-ho)*(YB1+YB2)+(YB1-YB/2-ho)2/2-(YB2-YB/2-ho)2/2,(YL2-YL/2-ho)*(YB1+YB2)+(YB2-YB/2-ho)2/2-(YB2-YB/2-ho)2/2
=max((1.300-0.600/2-0.460)*(1.000+1.000)+(1.000-0.400/2-0.460)2/2-(1.000-0.400/2-0.460)2/2,(1.300-0.600/2-0.460)*(1.000+1.000)+(1.000-0.400/2-0.460)2/2-(1.000-0.400/2-0.460)2/2)
=max(1.138,1.138)
=1.080m2
x冲切截面上的地基净反力设计值 Flx=Alx*Pjmax=1.080*322.023=347.785kN
γo*Flx=1.0*347.785=347.78kN
γo*Flx≤0.7*βhp*ft_b*bm*YHo (6.5.5-1)
=0.7*1.000*1.27*860*460
=351.69kN
x方向柱对基础的冲切满足规范要求
2.3 y方向柱对基础的冲切验算
y冲切位置斜截面上边长 at=YL=0.600m
y冲切位置斜截面下边长 ab=YL+2*YHo=1.520m
y冲切面积 Aly=max((YB1-YB/2-ho)*(YL+2*ho)+(YB1-YB/2-ho)2,(YB2-YB/2-ho)*(YL+2*ho)+(YB2-YB/2-ho)2)
=max((1.000-0.400/2-0.460)*(0.600+0.460)+(1.000-0.400/2-0.460)2,(1.000-0.400/2-0.460)*(0.600+0.460)+(1.000-0.400/2-0.460)2)
=max(0.632,0.632)
=0.632m2
y冲切截面上的地基净反力设计值 Fly=Aly*Pjmax=0.632*322.023=203.647kN
γo*Fly=1.0*203.647=203.65kN
γo*Fly≤0.7*βhp*ft_b*am*YHo (6.5.5-1)
=0.7*1.000*1.27*1060*460
=433.48kN
y方向柱对基础的冲切满足规范要求
八、基础受剪承载力验算
1. 计算剪力
Az=A1+A2
=1000+1000
=2000mm
Bz=B1+B2
=800+800
=1600mm
A'=Bz*(max(A1,A2)-0.5*hc)
=1600.0*(max(1000.0,1000.0)-0.5*400.0)
=0.80m2
Vs=A'*pk=0.8*174.2=139.4kN
基础底面短边尺寸大于柱宽加两倍基础有效高度，不需验算受剪承载力！
九、柱下基础的局部受压验算
因为基础的混凝土强度等级小于柱的混凝土强度等级，验算柱下扩展基础顶面的局部受压承载力。
1. Fl=F=1012.500kN
混凝土等级为C25, fcc=0.85*fc=0.85*11.9=10.115N/mm2, ω=1.0
2. 计算混凝土局部受压面积
YB1=b1+bc/2=0.800+0.400/2=1.000m2
YB2=b2+bc/2=0.800+0.400/2=1.000m2
YA1=a1+hc/2=1.000+0.600/2=1.300m2
YA2=a2+hc/2=1.000+0.600/2=1.300m2
Al=bc*hc=0.400*0.600=0.240m2
Ab=(bc+2*min(bc,hc))*(hc+2*min(bc,hc,))
=(0.400+2*min(0.400, 0.600))*(0.600+2*min(0.400, 0.600))
=1.680m2
3. 计算混凝土局部受压时的强度提高系数 【②7.8.1-2】
β1=sqrt(Ab/Al)=sqrt(1.680/0.240)=2.646
4. 因 γo*Fl=1.0*1012.500=1012.50kN
γo*Fl≤ω*β1*fcc*Al
=1.000*2.646*10.115*240000.000/1000
=6422.83kN
柱下基础局部受压承载力满足规范要求
十、基础受弯计算
因Mdx=0 Mdy≠0 并且 ex≤Bx/6=0.333m x方向单向受压且小偏心
a=(Bx-bc)/2=(2.000-0.400)/2=0.800m
P=(Bx-a)*(Pmax-Pmin)/Bx+Pmin
=(2.000-0.800)*(362.523-107.900)/2.000+107.900
=260.674kPa
MI_1=1/12*a2*((2*By+hc)*(Pmax+P-2*G/A)+(Pmax-P)*By)
=1/12*0.8002*((2*2.600+0.600)*(362.523+260.674-2*210.600/5.200)+(362.523-260.674)*2.600)
=181.84kN*m
MII_1=1/48*(By-hc)2*(2*Bx+bc)*(Pmax+Pmin-2*G/A)
=1/48*(2.600-0.600)2*(2*2.000+0.400)*(362.523+107.900-2*210.600/5.200)
=142.79kN*m
十一、计算配筋
10.1 计算Asx
Asx_1=γo*MI_1/(0.9*(H-as)*fy)
=1.0*181.84*106/(0.9*(500.000-40.000)*360)
=1220.1mm2
Asx1=Asx_1=1220.1mm2
Asx=Asx1/By=1220.1/2.600=469mm2/m
Asx=max(Asx, ρmin*H*1000)
=max(469, 0.150%*500*1000)
=750mm2/m
选择钢筋12@150, 实配面积为754mm2/m。
10.2 计算Asy
Asy_1=γo*MII_1/(0.9*(H-as)*fy)
=1.0*142.79*106/(0.9*(500.000-40.000)*360)
=958.1mm2
Asy1=Asy_1=958.1mm2
Asy=Asy1/Bx=958.1/2.000=479mm2/m
Asy=max(Asy, ρmin*H*1000)
=max(479, 0.150%*500*1000)
=750mm2/m
选择钢筋12@150, 实配面积为754mm2/m。
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