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| author | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
| commit | 7bc77cb1ed33745c720952c92b3b2747c5cbf2df (patch) | |
| tree | 449d555969bfd7befe906877abab098c6e63a0e8 /3886/CH3 | |
| parent | d1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff) | |
| download | Scilab-TBC-Uploads-7bc77cb1ed33745c720952c92b3b2747c5cbf2df.tar.gz Scilab-TBC-Uploads-7bc77cb1ed33745c720952c92b3b2747c5cbf2df.tar.bz2 Scilab-TBC-Uploads-7bc77cb1ed33745c720952c92b3b2747c5cbf2df.zip | |
Diffstat (limited to '3886/CH3')
56 files changed, 430 insertions, 0 deletions
diff --git a/3886/CH3/EX3.1/3_1.txt b/3886/CH3/EX3.1/3_1.txt new file mode 100644 index 000000000..0931c10a8 --- /dev/null +++ b/3886/CH3/EX3.1/3_1.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_1.sce', -1)
+MA=28301.270189 N-mm,Anticlockwise
\ No newline at end of file diff --git a/3886/CH3/EX3.1/Ex3_1.sce b/3886/CH3/EX3.1/Ex3_1.sce new file mode 100644 index 000000000..4ad0268ac --- /dev/null +++ b/3886/CH3/EX3.1/Ex3_1.sce @@ -0,0 +1,6 @@ +//Determine the moment
+//Refer fig. 3.5
+//Take clockwise moment as positive
+//Apply Varignon's Theorem
+MA=100*300*cosd(60)-100*500*sind(60) //N-mm
+printf("MA=%f N-mm,Anticlockwise",-MA)
diff --git a/3886/CH3/EX3.10/3_10.txt b/3886/CH3/EX3.10/3_10.txt new file mode 100644 index 000000000..140103f95 --- /dev/null +++ b/3886/CH3/EX3.10/3_10.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_10.sce', -1)
+tension T in cable BC is T=100.38 kN.
+RA=114.543 kN
+alpha=32.17 degree
\ No newline at end of file diff --git a/3886/CH3/EX3.10/Ex3_10.sce b/3886/CH3/EX3.10/Ex3_10.sce new file mode 100644 index 000000000..a25fea531 --- /dev/null +++ b/3886/CH3/EX3.10/Ex3_10.sce @@ -0,0 +1,10 @@ +//tension in the cable and reaction at a point
+//refer fig. 3.21 (a),(b)&(c)
+//Taking moment about A we get
+T=((25*12*cosd(30))+(10*6*cosd(30)))/(12*sind(15)) //kN
+//applying equilibrium conditions
+HA=T*cosd(15) //kN
+VA=10+25+T*sind(15) //kN
+RA=sqrt(HA^2+VA^2) //kN
+alpha=atand(VA/HA) //degree
+printf("tension T in cable BC is T=%.2f kN.\nRA=%.3f kN\nalpha=%.2f degree",T,RA,alpha)
diff --git a/3886/CH3/EX3.11/3_11.txt b/3886/CH3/EX3.11/3_11.txt new file mode 100644 index 000000000..53f1ee37e --- /dev/null +++ b/3886/CH3/EX3.11/3_11.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_11.sce', -1)
+The required force is F=298.3 N
\ No newline at end of file diff --git a/3886/CH3/EX3.11/Ex3_11.sce b/3886/CH3/EX3.11/Ex3_11.sce new file mode 100644 index 000000000..ba7b746cc --- /dev/null +++ b/3886/CH3/EX3.11/Ex3_11.sce @@ -0,0 +1,7 @@ +//Horizontal force required at bottom to avoid slipping
+//refer fig 3.22 (a)&(b)
+//Taking moment about A
+RB=(700*2*cotd(60)+100*1.5*cotd(60))/3 //N
+F=RB //N
+printf("The required force is F=%.1f N",F)
+
diff --git a/3886/CH3/EX3.12/3_12.txt b/3886/CH3/EX3.12/3_12.txt new file mode 100644 index 000000000..4b8daeb6d --- /dev/null +++ b/3886/CH3/EX3.12/3_12.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_12.sce', -1)
+Required force is F=447.45 N
\ No newline at end of file diff --git a/3886/CH3/EX3.12/Ex3_12.sce b/3886/CH3/EX3.12/Ex3_12.sce new file mode 100644 index 000000000..53212d39e --- /dev/null +++ b/3886/CH3/EX3.12/Ex3_12.sce @@ -0,0 +1,10 @@ +//Horizontal force required at certain height to avoid slipping
+//Refer fig. 3.23
+//applying equilibrium conditions we get
+//F=RB...(1)
+//-RB*3+700*2*cotd(60)+100*1.5*cotd(60)+F=0...(2)
+//Solving (1)&(2) we get
+F=(700*2+100*1.5)*cotd(60)/2 //N
+printf("Required force is F=%.2f N",F)
+
+
diff --git a/3886/CH3/EX3.13/3_13.txt b/3886/CH3/EX3.13/3_13.txt new file mode 100644 index 000000000..2ab7a6aed --- /dev/null +++ b/3886/CH3/EX3.13/3_13.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_13.sce', -1)
+The reactions are:-
+RD=1466.7 N
+HC=1154.7 N
\ No newline at end of file diff --git a/3886/CH3/EX3.13/Ex3_13.sce b/3886/CH3/EX3.13/Ex3_13.sce new file mode 100644 index 000000000..e539d1129 --- /dev/null +++ b/3886/CH3/EX3.13/Ex3_13.sce @@ -0,0 +1,17 @@ +//Reactions at supports C and D
+//Refer fig.3.24(a),(b)&(c)
+//applying equilibrium conditions for roller
+R2=2000/cosd(30) //N
+//consider equilibrium of bar CD
+//Taking moment about C
+RD=((800*2.5*cosd(30))+(R2*2))/(5*cosd(30)) //N
+VC=800+R2*cosd(30)-RD //N
+HC=R2*sind(30) //N
+printf("The reactions are:-\nRD=%.1f N\nHC=%.1f N",RD,HC)
+
+
+
+
+
+
+
diff --git a/3886/CH3/EX3.14/3_14.txt b/3886/CH3/EX3.14/3_14.txt new file mode 100644 index 000000000..e21cc4dd6 --- /dev/null +++ b/3886/CH3/EX3.14/3_14.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_14.sce', -1)
+Required values are:-
+T=51.962 kN
+R1=23.660 kN
+R2=6.34 kN
\ No newline at end of file diff --git a/3886/CH3/EX3.14/Ex3_14.sce b/3886/CH3/EX3.14/Ex3_14.sce new file mode 100644 index 000000000..87699a2ea --- /dev/null +++ b/3886/CH3/EX3.14/Ex3_14.sce @@ -0,0 +1,10 @@ +//Tension in the cable and reaction at axles
+//Refer fig.3.25 (a)&(b)
+//assume T as tension in the rope parallel to track
+//applying equilibrium conditions
+T=60*sind(60) //kN
+//applying moment equilibrium condition about upper axle reaction point we get
+R1=(-T*600+60*800*sind(60)+60*600*cosd(60))/1200 //kN
+R2=60*cosd(60)-R1 //kN
+printf("Required values are:-\nT=%.3f kN\nR1=%.3f kN\nR2=%.2f kN",T,R1,R2)
+
diff --git a/3886/CH3/EX3.15/3_15.txt b/3886/CH3/EX3.15/3_15.txt new file mode 100644 index 000000000..ab230155a --- /dev/null +++ b/3886/CH3/EX3.15/3_15.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_15.sce', -1)
+Minimum weight of hollow cylinder is W=0.75 kN
\ No newline at end of file diff --git a/3886/CH3/EX3.15/Ex3_15.sce b/3886/CH3/EX3.15/Ex3_15.sce new file mode 100644 index 000000000..245f221d0 --- /dev/null +++ b/3886/CH3/EX3.15/Ex3_15.sce @@ -0,0 +1,14 @@ +//Minimum weight of hollow cylinder
+//Refer fig.3.26 (a),(b)&(c)
+O1O2=400+600 //mm
+O2D=1600-400-600 //mm
+alpha=acosd(O2D/O1O2) //degree
+//considering equilibrium of spheres
+//Taking moment about O2
+R1=(600)/(1000*sind(alpha)) //kN
+R2=R1 //kN
+R3=1+3 //kN
+//Assume minimum weight W.During tipping there will be no reaction at point B
+//Taking moment about A
+W=(0.75*1000*sind(53.13))/(800) //kN
+printf("Minimum weight of hollow cylinder is W=%.2f kN",W)
diff --git a/3886/CH3/EX3.16/3_16.txt b/3886/CH3/EX3.16/3_16.txt new file mode 100644 index 000000000..24d0518e7 --- /dev/null +++ b/3886/CH3/EX3.16/3_16.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_16.sce', -1)
+The tension is:-
+T=374.0 N
\ No newline at end of file diff --git a/3886/CH3/EX3.16/Ex3_16.sce b/3886/CH3/EX3.16/Ex3_16.sce new file mode 100644 index 000000000..12502f76f --- /dev/null +++ b/3886/CH3/EX3.16/Ex3_16.sce @@ -0,0 +1,10 @@ +//Tension in horizontal rope
+//Refer fig. 3.27 (a),(b)&(c)
+//Considering equilibrium of the entire system
+RB=500/2 //N
+RA=RB //N (symmetry)
+R1=500/(2*cosd(60)) //N
+R2=R1 //N
+//Taking moment about C
+T=((500*0.866)+(250*1.2*0.5))/(1.8*sind(60)) //N
+printf("The tension is:-\nT=%.1f N",T)
diff --git a/3886/CH3/EX3.17/3_17.txt b/3886/CH3/EX3.17/3_17.txt new file mode 100644 index 000000000..815be1e70 --- /dev/null +++ b/3886/CH3/EX3.17/3_17.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_17.sce', -1)
+Required values:-
+VA=52.32 kN
+HA=10.00 kN
+MA=99.64 kN-m
\ No newline at end of file diff --git a/3886/CH3/EX3.17/Ex3_17.sce b/3886/CH3/EX3.17/Ex3_17.sce new file mode 100644 index 000000000..f232296ec --- /dev/null +++ b/3886/CH3/EX3.17/Ex3_17.sce @@ -0,0 +1,9 @@ +//Reactions developed in cantilever beam
+//Refer fig. 3.44 (a)&(b)
+//assumptions are made as shown in fig. 3.44 (a)&(b)
+//applying equilibrium conditions
+VA=15+(10*2)+(20*sind(60)) //kN
+HA=20*cosd(60) //kN
+//Taking moment about A
+MA=10*2*1+20*2*sind(60)+15*3 //kN-m
+printf("Required values:-\nVA=%.2f kN\nHA=%.2f kN\nMA=%.2f kN-m",VA,HA,MA)
diff --git a/3886/CH3/EX3.18/3_18.txt b/3886/CH3/EX3.18/3_18.txt new file mode 100644 index 000000000..adbcaa93b --- /dev/null +++ b/3886/CH3/EX3.18/3_18.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_18.sce', -1)
+Required values:-
+VA=105.00 kN
+HA=0.00 kN
+MA=180.00 kN-m
\ No newline at end of file diff --git a/3886/CH3/EX3.18/Ex3_18.sce b/3886/CH3/EX3.18/Ex3_18.sce new file mode 100644 index 000000000..cd08724c7 --- /dev/null +++ b/3886/CH3/EX3.18/Ex3_18.sce @@ -0,0 +1,10 @@ +//Reactions developed in cantilever beam
+//Refer fig. 3.45 (a)&(b)
+//Make assumptions as shown in fig. 3.45(a) and(b)
+//applying equilibrium conditions
+VA=60+45*2/2 //kN
+HA=0 //kN
+//Taking moment about A
+MA=((45*2*2)/(3*2))+(60*2.5) //kN-m
+printf("Required values:-\nVA=%.2f kN\nHA=%.2f kN\nMA=%.2f kN-m",VA,HA,MA)
+
diff --git a/3886/CH3/EX3.19/3_19.txt b/3886/CH3/EX3.19/3_19.txt new file mode 100644 index 000000000..b5cd7e401 --- /dev/null +++ b/3886/CH3/EX3.19/3_19.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_19.sce', -1)
+The reactions are:-
+RA=44.44 kN
+RB=115.56 kN
\ No newline at end of file diff --git a/3886/CH3/EX3.19/Ex3_19.sce b/3886/CH3/EX3.19/Ex3_19.sce new file mode 100644 index 000000000..938f7e1b7 --- /dev/null +++ b/3886/CH3/EX3.19/Ex3_19.sce @@ -0,0 +1,7 @@ +//Reactions developed in simply supported beam
+//Refer fig. 3.46 (a)&(b)
+//make assumptions as shown in fig. 3.46 (a)&(b)
+//Taking moment about B
+RA=((20*4*2)+((4*40*4)/(3*2)))/(6) //kN
+RB=80+80-RA //kN
+printf("The reactions are:-\nRA=%.2f kN\nRB=%.2f kN",RA,RB)
diff --git a/3886/CH3/EX3.2/3_2.txt b/3886/CH3/EX3.2/3_2.txt new file mode 100644 index 000000000..a132c6a87 --- /dev/null +++ b/3886/CH3/EX3.2/3_2.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_2.sce', -1)
+y-intercept will be y=2 m
\ No newline at end of file diff --git a/3886/CH3/EX3.2/Ex3_2.sce b/3886/CH3/EX3.2/Ex3_2.sce new file mode 100644 index 000000000..de00c41c7 --- /dev/null +++ b/3886/CH3/EX3.2/Ex3_2.sce @@ -0,0 +1,7 @@ +//Finding y-intercept
+//Apply law of Transmissibility and resolve 5000 N force at B
+Fx=5000*4/5 //N
+Fy=5000*3/5 //N
+//Apply Varignon's Theorem
+y=8000/4000 //m
+printf("y-intercept will be y=%.0d m",y)
diff --git a/3886/CH3/EX3.20/3_20.txt b/3886/CH3/EX3.20/3_20.txt new file mode 100644 index 000000000..ee913bc7b --- /dev/null +++ b/3886/CH3/EX3.20/3_20.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_20.sce', -1)
+The reactions developed are:-
+HA=27.13 kN
+VA=12.77 kN
+RB=18.87 kN
\ No newline at end of file diff --git a/3886/CH3/EX3.20/Ex3_20.sce b/3886/CH3/EX3.20/Ex3_20.sce new file mode 100644 index 000000000..048e9e8ce --- /dev/null +++ b/3886/CH3/EX3.20/Ex3_20.sce @@ -0,0 +1,9 @@ +//Reactions developed at A and B
+//Refer fig. 3.47 (a) and (b)
+//Make proper assumptions from this fig.
+//applying equilibrium conditions
+HA=15*cosd(30)+20*cosd(45) //kN
+//Taking moment about A
+RB=(10*4+15*6*sind(30)+20*10*sind(45))/12 //kN
+VA=-RB+10+15*sind(30)+20*sind(45) //kN
+printf("The reactions developed are:-\nHA=%.2f kN\nVA=%.2f kN\nRB=%.2f kN",HA,VA,RB)
diff --git a/3886/CH3/EX3.21/3_21.txt b/3886/CH3/EX3.21/3_21.txt new file mode 100644 index 000000000..361bccb4c --- /dev/null +++ b/3886/CH3/EX3.21/3_21.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_21.sce', -1)
+The reactions are:-
+RA=87.02 kN
+RB=100.45 kN
+As shown in fig. 3.48
\ No newline at end of file diff --git a/3886/CH3/EX3.21/Ex3_21.sce b/3886/CH3/EX3.21/Ex3_21.sce new file mode 100644 index 000000000..75eab1950 --- /dev/null +++ b/3886/CH3/EX3.21/Ex3_21.sce @@ -0,0 +1,9 @@ +//determine magnitude and direction of support reactions
+//Refer fig. 3.48 (a)&(b)
+//Taking moment about A
+RB=((60*1*sind(60))+(80*3*sind(75))+(50*5.5*sind(60)))/(6*sind(60)) //kN (At 60 degree to the horizontal)
+HA=(-60*cosd(60))+(80*cosd(75))-(50*cosd(60))+(100.45*cosd(60)) //kN
+VA=(-100.45*sind(60))+(60*sind(60))+(80*sind(75))+(50*sind(60)) //kN
+RA=sqrt((HA^2)+(VA^2)) //kN
+alphaA=atand(VA/HA) //Degree
+printf("The reactions are:-\nRA=%.2f kN \nRB=%.2f kN \nAs shown in fig. 3.48",RA,RB)
diff --git a/3886/CH3/EX3.22/3_22.txt b/3886/CH3/EX3.22/3_22.txt new file mode 100644 index 000000000..c36db50db --- /dev/null +++ b/3886/CH3/EX3.22/3_22.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_22.sce', -1)
+The reactions are:-
+RA=91.65 kN
+HB=42.43 kN
+VB=90.78 kN
+As shown in fig.3.49
\ No newline at end of file diff --git a/3886/CH3/EX3.22/Ex3_22.sce b/3886/CH3/EX3.22/Ex3_22.sce new file mode 100644 index 000000000..ebd0d264f --- /dev/null +++ b/3886/CH3/EX3.22/Ex3_22.sce @@ -0,0 +1,8 @@ +//determine support reactions
+//Refer fig. 3.49
+//Taking moment about B
+RA=(20*7+30*4*5+60*2*sind(45))/9 //kN
+HB=60*cosd(45) //kN
+VB=20+30*4+60*sind(45)-RA //kN
+printf("The reactions are:-\nRA=%.2f kN \nHB=%.2f kN \nVB=%.2f kN \nAs shown in fig.3.49",RA,HB,VB)
+
diff --git a/3886/CH3/EX3.23/3_23.txt b/3886/CH3/EX3.23/3_23.txt new file mode 100644 index 000000000..ddeb18ea2 --- /dev/null +++ b/3886/CH3/EX3.23/3_23.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_23.sce', -1)
+The support reactions are:-
+RA=42.60 kN
+RB=89.40 kN
+As shown in fig. 3.50
\ No newline at end of file diff --git a/3886/CH3/EX3.23/Ex3_23.sce b/3886/CH3/EX3.23/Ex3_23.sce new file mode 100644 index 000000000..9664daed5 --- /dev/null +++ b/3886/CH3/EX3.23/Ex3_23.sce @@ -0,0 +1,7 @@ +//determine support reactions
+//Refer fig. 3.50
+//Taking moment about A
+RB=(30*1+24*3*(2+1.5)+(1.5*40/2)*(5+1.5/3))/5 //kN
+RA=30+72+30-RB //kN
+printf("The support reactions are:-\nRA=%.2f kN \nRB=%.2f kN \nAs shown in fig. 3.50",RA,RB)
+
diff --git a/3886/CH3/EX3.24/3_24.txt b/3886/CH3/EX3.24/3_24.txt new file mode 100644 index 000000000..092a184cf --- /dev/null +++ b/3886/CH3/EX3.24/3_24.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_24.sce', -1)
+Required values are:-
+RB=71.01 kN
+HA=21.21 kN
+VA=9.80 kN downwards
\ No newline at end of file diff --git a/3886/CH3/EX3.24/Ex3_24.sce b/3886/CH3/EX3.24/Ex3_24.sce new file mode 100644 index 000000000..b2c6e39ec --- /dev/null +++ b/3886/CH3/EX3.24/Ex3_24.sce @@ -0,0 +1,8 @@ +//determine support reactions
+//Refer fig. 3.51
+//Taking moment about A
+RB=(40+30*5*sind(45)+20*2*7)/6 //kN
+HA=30*cosd(45) //kN
+VA=30*sind(45)-RB+40 //kN (downwards)
+printf("Required values are:-\nRB=%.2f kN\nHA=%.2f kN\nVA=%.2f kN downwards",RB,HA,-VA)
+
diff --git a/3886/CH3/EX3.25/3_25.txt b/3886/CH3/EX3.25/3_25.txt new file mode 100644 index 000000000..0ca325774 --- /dev/null +++ b/3886/CH3/EX3.25/3_25.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_25.sce', -1)
+Required values are:-
+RA=26.00 kN
+RB=34.00 kN
\ No newline at end of file diff --git a/3886/CH3/EX3.25/Ex3_25.sce b/3886/CH3/EX3.25/Ex3_25.sce new file mode 100644 index 000000000..1a96052a4 --- /dev/null +++ b/3886/CH3/EX3.25/Ex3_25.sce @@ -0,0 +1,9 @@ +//determine support reactions
+//Refer fig. 3.52
+//Taking moment about B
+RA=((10/2)*(1/3+5)+(2*10/2)*(5-2/3)+10*3*1.5+3*10*1/2)/5 //kN
+RB=-26+5+10+30+15 //kN
+printf("Required values are:-\nRA=%.2f kN\nRB=%.2f kN",RA,RB)
+
+
+
diff --git a/3886/CH3/EX3.26/3_26.txt b/3886/CH3/EX3.26/3_26.txt new file mode 100644 index 000000000..8b2b6b301 --- /dev/null +++ b/3886/CH3/EX3.26/3_26.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_26.sce', -1)
+The required distance is x=5 m
\ No newline at end of file diff --git a/3886/CH3/EX3.26/Ex3_26.sce b/3886/CH3/EX3.26/Ex3_26.sce new file mode 100644 index 000000000..475823bf9 --- /dev/null +++ b/3886/CH3/EX3.26/Ex3_26.sce @@ -0,0 +1,14 @@ +//determine distance of support C from end A
+//Refer fig. 3.53
+//assume that the support at C is at a distance of x metres from end A
+//applying equilibrium conditions
+RC=(30+120+50)/2 //kN
+RD=RC //kN (Given)
+//Taking moment about A
+x=(1000+1200-100*12)/200 //m
+printf("The required distance is x=%.0f m",x)
+
+
+
+
+
diff --git a/3886/CH3/EX3.28/3_28.txt b/3886/CH3/EX3.28/3_28.txt new file mode 100644 index 000000000..e9ebddfa9 --- /dev/null +++ b/3886/CH3/EX3.28/3_28.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_28.sce', -1)
+Required reactions are:-
+RA=12.86 kN
+RB=8.57 kN,
+RD=16.07 kN
\ No newline at end of file diff --git a/3886/CH3/EX3.28/Ex3_28.sce b/3886/CH3/EX3.28/Ex3_28.sce new file mode 100644 index 000000000..e96a19817 --- /dev/null +++ b/3886/CH3/EX3.28/Ex3_28.sce @@ -0,0 +1,15 @@ +//determine reactions at A,B and D
+//Refer fig. 3.55
+//Taking moment about C
+RD=(3*5*2.5+(5*9*2*5)/(2*3))/7 //kN
+RC=15+22.5-RD //kN
+//Taking moment about A
+RB=(2*RC/5) //kN
+RA=RC-RB //kN
+printf("Required reactions are:-\nRA=%.2f kN\nRB=%.2f kN,\nRD=%.2f kN",RA,RB,RD)
+
+
+
+
+
+
diff --git a/3886/CH3/EX3.29/3_29.txt b/3886/CH3/EX3.29/3_29.txt new file mode 100644 index 000000000..24f8e9d25 --- /dev/null +++ b/3886/CH3/EX3.29/3_29.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_29.sce', -1)
+Required Values are:-
+HA=28.28 kN
+VA=9.43 kN downward
+RD=40.91 kN
+HC=10.00 kN
+VC=34.12 kN
\ No newline at end of file diff --git a/3886/CH3/EX3.29/Ex3_29.sce b/3886/CH3/EX3.29/Ex3_29.sce new file mode 100644 index 000000000..c81d38777 --- /dev/null +++ b/3886/CH3/EX3.29/Ex3_29.sce @@ -0,0 +1,18 @@ +//determine reactions at A,C and D
+//Refer fig. 3.56
+//Taking moment about A
+RE=(20*3+40*4*sind(45))/3 //kN
+HA=40*cosd(45) //kN
+VA=20+40*sind(45)-RE //kN (Downwards)
+//Taking moment about C
+RD=((20*sind(60))-(10)+(57.71*2))/3 //kN
+HC=20*cosd(60) //kN
+VC=20*sind(60)+RE-RD //kN
+printf("Required Values are:-\nHA=%.2f kN\nVA=%.2f kN downward\nRD=%.2f kN\nHC=%.2f kN\nVC=%.2f kN",HA,-VA,RD,HC,VC)
+
+
+
+
+
+
+
diff --git a/3886/CH3/EX3.3/3_3.txt b/3886/CH3/EX3.3/3_3.txt new file mode 100644 index 000000000..e7c451717 --- /dev/null +++ b/3886/CH3/EX3.3/3_3.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_3.sce', -1)
+R=68.06 kN as shown in fig. 3.12(a)
diff --git a/3886/CH3/EX3.3/Ex3_3.sce b/3886/CH3/EX3.3/Ex3_3.sce new file mode 100644 index 000000000..9115560e8 --- /dev/null +++ b/3886/CH3/EX3.3/Ex3_3.sce @@ -0,0 +1,11 @@ +//Determine Resultant
+//horizontal direction is assumed as x-axis and vertical as y-axis
+Rx=-20*cosd(60) //kN (towards left)
+Ry=-20-30-20*sind(60) //kN (downwards)
+R=sqrt(Rx^2+Ry^2) //kN
+alpha=atand(Ry/Rx) //degree (as shown in fig. 3.12(b))
+//Taking moment about A
+MA=20*1.5+30*3+20*6*sind(60) //kN-m
+//x-intercept of the resultant is
+x=MA/Ry //m (shown in fig.)
+printf("R=%.2f kN as shown in fig. 3.12(a)",R)
diff --git a/3886/CH3/EX3.4/3_4.txt b/3886/CH3/EX3.4/3_4.txt new file mode 100644 index 000000000..d2a876941 --- /dev/null +++ b/3886/CH3/EX3.4/3_4.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_4.sce', -1)
+R=108 kN is the resultant as shown in fig. 3.13 (a)
\ No newline at end of file diff --git a/3886/CH3/EX3.4/Ex3_4.sce b/3886/CH3/EX3.4/Ex3_4.sce new file mode 100644 index 000000000..18472dba7 --- /dev/null +++ b/3886/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,11 @@ +//Find resultant
+Rx=60-100*cosd(60)-120*cosd(30) //kN (towards left)
+Ry=-80+100*sind(60)-120*sind(30) //kN (downwards)
+R=sqrt(Rx^2+Ry^2) //kN
+alpha=atand(Ry/Rx) //degree (shown in fig. 3.13(b))
+MA=(80*100*cosd(60)+60*100*sind(60)+120*100*sind(30)) //kN-mm
+//intercept on x-axis is
+x=MA/Ry //mm (as shown in fig. 3.13(a))
+printf("R=%.0f kN is the resultant as shown in fig. 3.13 (a)",R)
+
+
diff --git a/3886/CH3/EX3.5/3_5.txt b/3886/CH3/EX3.5/3_5.txt new file mode 100644 index 000000000..7f9684413 --- /dev/null +++ b/3886/CH3/EX3.5/3_5.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_5.sce', -1)
+The resultant of the system is R=4.656 kN as shown in fig. 3.14(b)
\ No newline at end of file diff --git a/3886/CH3/EX3.5/Ex3_5.sce b/3886/CH3/EX3.5/Ex3_5.sce new file mode 100644 index 000000000..d74c2e7a4 --- /dev/null +++ b/3886/CH3/EX3.5/Ex3_5.sce @@ -0,0 +1,17 @@ +//Find the resultant
+//refer fig. 3.14 (a) and (b)
+theta1=atand(10/10)
+theta2=atand(30/40)
+theta3=atand(10/20)
+Rx=2*cosd(theta1)+5*cosd(theta2)-1.5*cosd(theta3) //kN
+Ry=2*sind(theta1)-5*sind(theta2)-1.5*sind(theta3) //kN
+R=sqrt(Rx^2+Ry^2) //kN
+alpha=atand(-Ry/Rx) //Degree
+//Moment of forces about O is
+MO=2*30*cosd(45)+5*50*sind(theta2)+1.5*10*sind(theta3) //kN-mm
+//distance d of resultant R from O is given as
+d=MO/R
+printf("The resultant of the system is R=%.3f kN as shown in fig. 3.14(b)",R)
+
+
+
diff --git a/3886/CH3/EX3.6/3_6.txt b/3886/CH3/EX3.6/3_6.txt new file mode 100644 index 000000000..6ce70e17b --- /dev/null +++ b/3886/CH3/EX3.6/3_6.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_6.sce', -1)
+The Resultant is R=2671.2 N as shown in fig.3.15
\ No newline at end of file diff --git a/3886/CH3/EX3.6/Ex3_6.sce b/3886/CH3/EX3.6/Ex3_6.sce new file mode 100644 index 000000000..d5ab057b3 --- /dev/null +++ b/3886/CH3/EX3.6/Ex3_6.sce @@ -0,0 +1,11 @@ +//Determine magnitude,direction and point of application
+//refer fig.3.15(a)&(b)
+Rx=500*cosd(60)-700 //N (towards left)
+Ry=-500*sind(60)-1000-1200 //N (Downwards)
+R=sqrt((Rx^2)+(Ry^2)) //N
+alpha=atand(-Ry/Rx) //degree
+//taking moment about O
+MO=-500*300*sind(60)-1000*150+1200*150*cosd(60)-700*300*sind(60)
+//let point of application of resultant be at a distance of x from point O along the horizontal then
+x=MO/Ry //mm
+printf("The Resultant is R=%.1f N as shown in fig.3.15",R)
diff --git a/3886/CH3/EX3.7/3_7.txt b/3886/CH3/EX3.7/3_7.txt new file mode 100644 index 000000000..f8815e4c2 --- /dev/null +++ b/3886/CH3/EX3.7/3_7.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_7.sce', -1)
+x=3.467 m lies in the middle third of base.Hence dam is safe
\ No newline at end of file diff --git a/3886/CH3/EX3.7/Ex3_7.sce b/3886/CH3/EX3.7/Ex3_7.sce new file mode 100644 index 000000000..2a1405899 --- /dev/null +++ b/3886/CH3/EX3.7/Ex3_7.sce @@ -0,0 +1,10 @@ +//Safety of dam
+//refer fig. 3.16
+Rx=300 //kN (towards right)
+Ry=100-1200-400 //kN (Downwards)
+//taking moment about O
+MO=300*3-100*1+1200*2+400*5
+//assume that the resultant cut the base at a distance of x from O
+x=MO/Ry //m
+printf("x=%.3f m lies in the middle third of base.Hence dam is safe",-x)
+
diff --git a/3886/CH3/EX3.8/3_8.txt b/3886/CH3/EX3.8/3_8.txt new file mode 100644 index 000000000..78f01d7bb --- /dev/null +++ b/3886/CH3/EX3.8/3_8.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_8.sce', -1)
+The resultant is R=580.3 N as shown in fig.3.17 (a)
\ No newline at end of file diff --git a/3886/CH3/EX3.8/Ex3_8.sce b/3886/CH3/EX3.8/Ex3_8.sce new file mode 100644 index 000000000..ea4e58868 --- /dev/null +++ b/3886/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,13 @@ +//determine the resultant
+//refer fig. 3.17 (a) and (b)
+Rx=-400*cosd(45)-150*cosd(30) //N (towards left)
+Ry=200+400*sind(45)-150*sind(30)
+R=sqrt(Rx^2+Ry^2) //N
+alpha=atand(Ry/Rx) //degree
+//assume that the resultant intersects arm AB at a distance of x from A
+//taking moment about A
+MA=-400*3*sind(45)-400*0.6*cosd(45)+50+150*6*sind(30)+150*1*cosd(30) //N-m (anticlockwise)
+x=MA/Ry //m
+printf("The resultant is R=%.1f N as shown in fig.3.17 (a)",R)
+
+
diff --git a/3886/CH3/EX3.9/3_9.txt b/3886/CH3/EX3.9/3_9.txt new file mode 100644 index 000000000..a90e43daf --- /dev/null +++ b/3886/CH3/EX3.9/3_9.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\3. Resultant and equilibrium of system of coplanar non-concurrent forces\Ex3_9.sce', -1)
+Equilibriant is equal and opposite to resultant.
+R=116.52 kN
+alpha=76.82 degree
+x=1.517 m
+As shown in fig.3.18 (a)
\ No newline at end of file diff --git a/3886/CH3/EX3.9/Ex3_9.sce b/3886/CH3/EX3.9/Ex3_9.sce new file mode 100644 index 000000000..7d2158b17 --- /dev/null +++ b/3886/CH3/EX3.9/Ex3_9.sce @@ -0,0 +1,13 @@ +//determine equilibriant
+//two 40 kN forces have no moment about the pulley centre hence can be considered acting at pulley centre
+//Accordingly
+Rx=20*cosd(45)-30*cosd(60)-50*cosd(30)+40*cosd(20)-40*sind(30) //kN (towards left)
+Ry=-20*sind(45)-20+20-30*sind(60)-50*sind(30)-40*sind(20)-40*cosd(30) //kN (Downwards)
+R=sqrt(Rx^2+Ry^2) //kN
+alpha=atand(Ry/Rx) //degree
+//Taking moment about A
+MA=20*4-20*4+30*6*sind(60)+50*2*sind(30)-50*2*cosd(30)+40*3*cosd(20)-40*3*sind(30)
+//assume that the resultant intersects AB at a distance x from A,then
+x=MA/Ry //m
+printf("Equilibriant is equal and opposite to resultant.\nR=%.2f kN\nalpha=%.2f degree\nx=%.3f m\nAs shown in fig.3.18 (a)",R,alpha,-x)
+
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