diff options
Diffstat (limited to '1205/CH4')
| -rw-r--r-- | 1205/CH4/EX4.1/S_4_1.sce | 21 | ||||
| -rw-r--r-- | 1205/CH4/EX4.10/S_4_10.sce | 20 | ||||
| -rw-r--r-- | 1205/CH4/EX4.2/S_4_2.sce | 13 | ||||
| -rw-r--r-- | 1205/CH4/EX4.3/S_4_3.sce | 19 | ||||
| -rw-r--r-- | 1205/CH4/EX4.4/S_4_4.sce | 21 | ||||
| -rw-r--r-- | 1205/CH4/EX4.5/S_4_5.sce | 25 | ||||
| -rw-r--r-- | 1205/CH4/EX4.6/S_4_6.sce | 32 | ||||
| -rw-r--r-- | 1205/CH4/EX4.7/S_4_7.sce | 31 | ||||
| -rw-r--r-- | 1205/CH4/EX4.8/S_4_8.sce | 20 | ||||
| -rw-r--r-- | 1205/CH4/EX4.9/S_4_9.sce | 24 |
10 files changed, 226 insertions, 0 deletions
diff --git a/1205/CH4/EX4.1/S_4_1.sce b/1205/CH4/EX4.1/S_4_1.sce new file mode 100644 index 000000000..b9fb4f379 --- /dev/null +++ b/1205/CH4/EX4.1/S_4_1.sce @@ -0,0 +1,21 @@ +clc;
+//Determination of B
+//At equillibrium +sum(M_A)=0
+//B*1.5m-(9.81kN)(2 m)-(23.5 kN)(6 m)=0, B assumed to be in +ve X direction
+B=(9.81*2+23.5*6)/1.5//kN
+printf("B=%.2f kN \n +ve sign shows reaction is directed as assumed ",B);
+//Determination of Ax
+//Sum Fx=0
+//Ax+B=0
+Ax=-B;//kN
+printf("Ax=%.2f kN\n",Ax);
+//Determination of Ay
+//Sum Fy=0
+//Ay-9.81 kN-23.5kN=0
+Ay=9.81+23.5;//kN
+printf("Ay=%.2f kN\n",Ay);
+A=[Ax,Ay];//kN Adding component
+A=norm(A);//Magnitude of force A
+theta=atan(Ay/Ax);//radians
+theta=theta*180/%pi;//degrees, conversion into degrees
+printf("Reaction at A is A=%.2f kN making angle %.2f degrees with + ve x axis ",A,theta);
diff --git a/1205/CH4/EX4.10/S_4_10.sce b/1205/CH4/EX4.10/S_4_10.sce new file mode 100644 index 000000000..dbeafedea --- /dev/null +++ b/1205/CH4/EX4.10/S_4_10.sce @@ -0,0 +1,20 @@ +clc;
+AC=[3.6,3.6];//m, Position vector of AC
+AE=[1.8,3.6,0];//m, Position vector of AE
+W=[0,-2000];//N, load
+AD=[3.6,3.6,-1.8];//m, Position vector of AD
+lambda=AD/norm(AD)//m, Unit vector along AD
+ACW=AC(1)*W(2)-AC(2)*W(1);//N.m K Cross product of AC and w
+lACW=-lambda(3)*ACW;//N.m,dot product of lambada and ACW
+lAET=-lACW;//N.m, by SumM_AD=0,triple product of lambada, AE and T .....1
+lAE=[lambda(2)*AE(3)-lambda(3)*AE(2),lambda(3)*AE(1)-lambda(1)*AE(3),lambda(1)*AE(2)-lambda(2)*AE(1)];//Cross product of lambada and AE
+unit=lAE/norm(lAE);// Unit vector of cross product.....2
+T=lAET/(lAE(1)+lAE(2)+lAE(3));//N, by 1 amd 2
+Tmin=unit*T;//N, Vector of minimum tension
+printf("Minimum Value of tension vector T=(%.1f N)i +(%.1f N)j +(%.1f N)k is %.0f N \n",Tmin(1),Tmin(2),Tmin(3),norm(Tmin));
+
+//Location of G
+//EG and Tmin are having same direction, so their component should be in proportion
+x=-1.8/Tmin(3)*Tmin(1)+1.8;//m, X co-ordinate of G
+y=-1.8/Tmin(3)*Tmin(2)+3.6;//m, Y co-ordinate of G
+printf("Co-ordinates of G are x=%.0f m and y= %.1f m",x,y);
diff --git a/1205/CH4/EX4.2/S_4_2.sce b/1205/CH4/EX4.2/S_4_2.sce new file mode 100644 index 000000000..fd84c26bf --- /dev/null +++ b/1205/CH4/EX4.2/S_4_2.sce @@ -0,0 +1,13 @@ +clc;
+
+//At equillibrium equations are +-> sum Fx=0, +sum(M_A)=0, +sum(M_B)=0
+//Sum Fx=0 gives
+Bx=0;//kN
+printf("Bx=%.0f kN \n",Bx);
+//+sum(M_A)=0 gives -(70kN)(0.9m)+By(2.7m)-(27kN)(3.3m)-(27kN)(3.9m)=0, B assumed to be in +ve Y direction
+By=(70*0.9+27*3.3+27*3.9)/2.7//kN
+printf("By=%.2f kN +ve sign shows reaction is directed as assumed \n",By);
+
+//+sum(M_B)=0 gives -A(2.7m)+(70kN)(1.8m)-(27kN)(0.6m)-(27kN)(1.2m)=0, A assumed to be in +ve Y direction
+A=(70*1.8-27*0.6-27*1.2)/2.7//kN
+printf("A=%.2f kN +ve sign shows reaction is directed as assumed \n",A);
diff --git a/1205/CH4/EX4.3/S_4_3.sce b/1205/CH4/EX4.3/S_4_3.sce new file mode 100644 index 000000000..a99731244 --- /dev/null +++ b/1205/CH4/EX4.3/S_4_3.sce @@ -0,0 +1,19 @@ +clc;
+//Take x axis parallel to track and Y axis perpendicular to track
+W=25;//kN
+// Resolving weight
+Wx=W*cos(25*%pi/180);//kN
+Wy=-W*sin(25*%pi/180);//kN
+//At equillibrium equations are +-> sum Fx=0, +sum(M_A)=0, +sum(M_B)=0
+
+//+sum(M_A)=0 gives -(10.5kN)(625 mm)-(22.65 kN)(150 mm)+ R2(1250 mm)=0, R2 assumed to be in +ve Y direction
+R2=(10.5*625+22.65*150)/1250;//kN
+printf("R2=%.0f kN +ve sign shows reaction is directed as assumed \n",R2);
+
+//+sum(M_B)=0 gives (10.5kN)(625 mm)-(22.65 kN)(150 mm)+ R1(1250 mm)=0, R1 assumed to be in +ve Y direction
+R1=(10.5*625-22.65*150)/1250;//kN
+printf("R1=%.1f kN +ve sign shows reaction is directed as assumed \n",R1);
+
+//Sum Fx=0 gives, 22.65 N-T=0
+T=22.65;//kN
+printf("T=%.2f kN +ve sign shows reaction is directed as assumed \n",T);
diff --git a/1205/CH4/EX4.4/S_4_4.sce b/1205/CH4/EX4.4/S_4_4.sce new file mode 100644 index 000000000..2a3f25751 --- /dev/null +++ b/1205/CH4/EX4.4/S_4_4.sce @@ -0,0 +1,21 @@ +clc;
+
+//At equillibrium equations are +-> sum Fx=0, +sum(M_A)=0, +sum(M_B)=0
+
+//Sum Fx=0 gives,
+Ax=600*cos(10*%pi/180)-375*cos(20*%pi/180);//N
+printf("Ax=%.2f kN \n",Ax);
+
+
+//Sum Fy=0 gives, Ay-1600 N -(375 N)sin(20 degree)-(600 N)sin(10 degree)=0
+Ay=600*sin(10*%pi/180)+375*sin(20*%pi/180)+1600;//N
+printf("Ay=%.2f kN \n",Ay);
+
+A=[Ax,Ay];//N Reaction force at A
+A=norm(A);//N Magnitude of A
+theta=atan(Ay/Ax);//Radian , Angle made by A with +ve X axis
+theta=theta*180/%pi;//Degrees
+printf("A=%.0f kN Theta=%.1f\n",A,theta);
+//+sum(M_A)=0 gives M_A-(375 N)cos(20 degree)(6 m)+(600 N)cos(10 degree)(6 m)=0,
+M_A=-600*cos(10*%pi/180)*6+375*cos(20*%pi/180)*6;//N.m
+printf("M_A=%.0f kN +ve sign shows reaction is directed as assumed \n",M_A);
diff --git a/1205/CH4/EX4.5/S_4_5.sce b/1205/CH4/EX4.5/S_4_5.sce new file mode 100644 index 000000000..c7243aeec --- /dev/null +++ b/1205/CH4/EX4.5/S_4_5.sce @@ -0,0 +1,25 @@ +clc;
+
+//At equillibrium +sum(Mo)=0,
+//s=r*theta;
+//F=k*s=k*r*theta;
+k=45;//N/mm
+r=75;//mm
+W=1800;//N
+l=200;//mm
+
+
+//+sum(Mo)=0 W*l*sin(theta)-r(k*r*theta)=0,
+//sin(theta)=k*r^2*theta/(W*l)
+
+// trial and error
+printf("Probable answers by trial and error method are \n");
+for i=0:0.1:%pi/2 // from 0 to 90 degrees
+
+difference=(sin(i)-k*r^2*(i)/(W*l));
+if difference<0.01 then // Approximation
+ theta=i;
+ theta=theta*180/%pi;//Degrees , conversion into degrees
+printf("Theta=%.2f degrees\n",theta);
+end
+end
diff --git a/1205/CH4/EX4.6/S_4_6.sce b/1205/CH4/EX4.6/S_4_6.sce new file mode 100644 index 000000000..2ec437e50 --- /dev/null +++ b/1205/CH4/EX4.6/S_4_6.sce @@ -0,0 +1,32 @@ +clc;
+
+m=10;//kg mass of joist
+g=9.81;//m/s^2 gravitational acceleration
+W=m*g;//N
+AB=4;//m
+// Three force body
+BF=AB*cos(45*%pi/180);//m
+AF=BF;//m
+
+AE=1/2*AF;//m
+EF=AE;//m
+CD=AE;//m
+BD=CD/tan((45+25)*%pi/180);//m
+DF=BF-BD;//m
+CE=DF;//m
+alpha=atan(CE/AE);//radians
+alpha=alpha*180/%pi;//degrees
+
+//From geometry
+
+G=90-alpha;//degrees
+B=alpha-(90-(45+25));//degrees
+C=180-(G+B);//Degrees
+
+//Force triangle
+//T/sin(G)=R/sin(C)=W/sin(B)..... sine law
+
+T=W/sin(B*%pi/180)*sin(G*%pi/180);//N
+R=W/sin(B*%pi/180)*sin(C*%pi/180);//N
+printf("Tension in cable T= %.1f N\n Reaction At A is R= %.1f N with angle alpha= %.1f degrees with +ve X axis",T,R,alpha);
+
diff --git a/1205/CH4/EX4.7/S_4_7.sce b/1205/CH4/EX4.7/S_4_7.sce new file mode 100644 index 000000000..93142916c --- /dev/null +++ b/1205/CH4/EX4.7/S_4_7.sce @@ -0,0 +1,31 @@ +clc;
+
+m1=80;//kg mass of man
+m2=20;//kg, mass of ladder
+m=m1+m2;//kg
+g=9.81;//m/s^2 gravitational acceleration
+W=-m*g;//N, j
+
+//Equillibrium equations
+//At equillibrium +sum(F)=0, gives
+// Ay j+ Az k+ By k+ Bz k + W j + C k=0
+// i.e. (Ay + By +W)j+(Az+Bz+C)k=0
+//At equillibrium +sum(M_A)=0, sum (r*F)=0
+//1.2i*(By j + Bz k)+ (0.9 i -0.6 k)*(W j)+(0.6 i+3 j-1.2 k)*(C k)=0
+//By rules of vector product it can be simply written as
+//1.2Byk-1.2Bzj+0.9Wk+0.6Wi-0.6Cj+3Ci=0
+//i.e (3C+0.6W)i -(1.2Bz+0.6C)j +(1.2By+0.9W)k=0
+// Equating coeeficients of i, jand k to zero
+
+C=-0.6*W/3;//N
+Bz=-0.6*C/1.2;//N
+By=-0.9*W/1.2;//N
+
+printf(" Reaction At B is B= (%.0f) N j +(%.1f N)k\n",By,Bz);
+printf(" Reaction At C is C= (%.2f) N j\n",C);
+Ay=-W-By;//N
+Az=-C-Bz;//N
+
+
+printf(" Reaction At A is A= (%.0f) N j +(%.1f N)k \n",Ay,Az);
+
diff --git a/1205/CH4/EX4.8/S_4_8.sce b/1205/CH4/EX4.8/S_4_8.sce new file mode 100644 index 000000000..7aed83b71 --- /dev/null +++ b/1205/CH4/EX4.8/S_4_8.sce @@ -0,0 +1,20 @@ +clc;
+W=-1200;//N,j Weight
+BD=[-2.4,1.2,-2.4];//m, Vector BD
+EC=[-1.8,0.9,0.6];//m, Vector EC
+//T_BD=norm(T_BD)*BD/norm(BD);// m, vector of tension in BD
+//T_EC=norm(T_EC)*EC/norm(EC);// m, vector of tension in EC
+// Applying equillibrium conditions we get
+// Sum_F=0, and Sum(M_A)=0 and setting co-efficient equal to zero
+A=[0.8,0.771;1.6,-0.514];//MAtrix of co-efficient
+b=[-1440;0];//matrix b
+x=linsolve(A,b);// solution matrix
+T_BD=x(1);// N,Tension in BD
+T_EC=x(2);//N, Tension in EC
+printf("T_BD= (%.0f N) and T_EC= (%.0f N) \n",x(1),x(2));
+
+Ax=2/3*T_BD+6/7*T_EC;//N, x component of reaction at A
+Ay=-(1/3*T_BD+3/7*T_EC+W);//N, Y component of rection at A
+Az=2/3*T_BD-2/7*T_EC;//N, z component of reaction at A
+
+printf("Reaction at A is A=(%.0f N)i +(%.0f N)j +(%.1f N)k \n",Ax,Ay,Az);
diff --git a/1205/CH4/EX4.9/S_4_9.sce b/1205/CH4/EX4.9/S_4_9.sce new file mode 100644 index 000000000..80d72ae05 --- /dev/null +++ b/1205/CH4/EX4.9/S_4_9.sce @@ -0,0 +1,24 @@ +clc;
+theta=30;//degree,
+theta=theta*%pi/180;//rad, Conversion
+e_AH=[0,sin(theta),cos(theta)];//Unit vector along AH
+r_HC=[-50,250,0];
+r_DC=[300,0,0];
+r_FC=[350,20,0];
+
+//Applying theory of Eqyuillibrium equations and equating coefficient to zero we get following equtions for cross product
+//by sum(Mc)=0
+// Coefficient of i
+T=100*10^3/216.5;//N,Tension wire AH
+// Coefficient of j
+Dz=-43.3*461.1/300;//N
+//coefficient of k
+Dy=(140*10^3+25*461.9)/300;//N
+printf("Tension wire AH is %.0f N\n",T);
+printf("Reaction at D is D=((%.0f N)j +(%.1f N)k \n",Dy,Dz);
+
+//Applying sumF=0
+Cx=0;//N, Xcomponent of C
+Cy=-461.9*0.5-505.1+400;//N, Y component of C
+Cz=-461.9*0.866-66.67;//N, Zcomponent of c
+printf("Reaction at C is C=(%.0f N)i +(%.0f N)j +(%.0f N)k \n",Cx,Cy,Cz);
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