diff options
Diffstat (limited to '3886/CH4')
| -rw-r--r-- | 3886/CH4/EX4.1/4_1.txt | 9 | ||||
| -rw-r--r-- | 3886/CH4/EX4.1/Ex4_1.sce | 23 | ||||
| -rw-r--r-- | 3886/CH4/EX4.10/4_10.txt | 9 | ||||
| -rw-r--r-- | 3886/CH4/EX4.10/Ex4_10.sce | 33 | ||||
| -rw-r--r-- | 3886/CH4/EX4.11/4_11.txt | 10 | ||||
| -rw-r--r-- | 3886/CH4/EX4.11/Ex4_11.sce | 30 | ||||
| -rw-r--r-- | 3886/CH4/EX4.12/4_12.txt | 10 | ||||
| -rw-r--r-- | 3886/CH4/EX4.12/Ex4_12.sce | 38 | ||||
| -rw-r--r-- | 3886/CH4/EX4.2/4_2.txt | 11 | ||||
| -rw-r--r-- | 3886/CH4/EX4.2/Ex4_2.sce | 23 | ||||
| -rw-r--r-- | 3886/CH4/EX4.3/4_3.txt | 15 | ||||
| -rw-r--r-- | 3886/CH4/EX4.3/Ex4_3.sce | 32 | ||||
| -rw-r--r-- | 3886/CH4/EX4.4/4_4.txt | 17 | ||||
| -rw-r--r-- | 3886/CH4/EX4.4/Ex4_4.sce | 29 | ||||
| -rw-r--r-- | 3886/CH4/EX4.5/4_5.txt | 18 | ||||
| -rw-r--r-- | 3886/CH4/EX4.5/Ex4_5.sce | 28 | ||||
| -rw-r--r-- | 3886/CH4/EX4.6/4_6.txt | 6 | ||||
| -rw-r--r-- | 3886/CH4/EX4.6/Ex4_6.sce | 11 | ||||
| -rw-r--r-- | 3886/CH4/EX4.7/4_7.txt | 7 | ||||
| -rw-r--r-- | 3886/CH4/EX4.7/Ex4_7.sce | 30 | ||||
| -rw-r--r-- | 3886/CH4/EX4.8/4_8.txt | 6 | ||||
| -rw-r--r-- | 3886/CH4/EX4.8/Ex4_8.sce | 15 | ||||
| -rw-r--r-- | 3886/CH4/EX4.9/4_9.txt | 10 | ||||
| -rw-r--r-- | 3886/CH4/EX4.9/Ex4_9.sce | 21 |
24 files changed, 441 insertions, 0 deletions
diff --git a/3886/CH4/EX4.1/4_1.txt b/3886/CH4/EX4.1/4_1.txt new file mode 100644 index 000000000..5ffb41d6b --- /dev/null +++ b/3886/CH4/EX4.1/4_1.txt @@ -0,0 +1,9 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_1.sce', -1)
+The required forces in members are:-
+AB=120.00 kN (Tension)
+BC=56.57 kN (Tension)
+CD=40.00 kN (Compression)
+DE=40.00 kN (Compression)
+BE=113.14 kN (Compression)
+BD=40.00 kN (Tension)
\ No newline at end of file diff --git a/3886/CH4/EX4.1/Ex4_1.sce b/3886/CH4/EX4.1/Ex4_1.sce new file mode 100644 index 000000000..eea95d2a1 --- /dev/null +++ b/3886/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,23 @@ +//finding forces in members of truss
+//Refer fig. 4.8
+//Step 1
+theta=atand(3/3) //Degree
+//Step 2
+//Assume Notations as in fig. 4.8
+//Step 3
+//applying equilibrium conditions
+FCB=40/sind(45) //kN
+FCD=FCB*cosd(45) //kN
+//Step 4
+//Mark and analyse at joint C
+//Step 5
+//Analyse joint D
+FDB=40 //kN
+FDE=40 //kN
+//step 6
+//Analysis of joint B
+FBE=(40+56.57*sind(45))/sind(45) //kN
+FBA=FBE*cosd(45)+56.57*cosd(45) //kN
+//step 7
+//Tabulating answers
+printf("The required forces in members are:-\nAB=%.2f kN (Tension)\nBC=%.2f kN (Tension)\nCD=%.2f kN (Compression)\nDE=%.2f kN (Compression)\nBE=%.2f kN (Compression)\nBD=%.2f kN (Tension)",FBA,FCB,FCD,FDE,FBE,FDB)
diff --git a/3886/CH4/EX4.10/4_10.txt b/3886/CH4/EX4.10/4_10.txt new file mode 100644 index 000000000..ad3bcaa84 --- /dev/null +++ b/3886/CH4/EX4.10/4_10.txt @@ -0,0 +1,9 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_10.sce', -1)
+The required forces are:-
+Force in member AB=120 kN
+Force in member BC=56 kN
+Force in member CD=-39 kN
+Force in member DE=-40 kN
+Force in member EB=-113 kN
+Force in member BD=40 kN
\ No newline at end of file diff --git a/3886/CH4/EX4.10/Ex4_10.sce b/3886/CH4/EX4.10/Ex4_10.sce new file mode 100644 index 000000000..24552c8fe --- /dev/null +++ b/3886/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,33 @@ +//Finding unknown forces
+//Refer fig. 4.8
+//Let us assume joint E as origin,EC as x-axis,EA as y-direction
+//accordingly the co-ordinates are
+//A(0,3),B(3,3),C(6,0),D(3,0),E(0,0)
+YD=-40 //kN
+YC=-40 //kN
+//Using co-ordinates lengths are found out to be
+LAB=3 //m
+LBC=3*sqrt(2) //m
+LCD=3 //m
+LDE=3 //m
+LBD=3 //m
+LBE=3*sqrt(2) //m
+//Consider joint C
+//applying equilibrium conditions
+tCB=40/3
+tCD=-40/3
+FCB=tCB*LBC //kN
+FCD=-13.333*LCD //kN
+//Consider joint D
+//applying equilibrium conditions
+tDE=tCD
+FDE=tCD*LCD //kN
+tDB=40/3
+FDB=tDB*LBD
+//Consider joint B
+//applying equilibrium conditions
+tBE=-(13.333+13.333)
+FBE=tBE*LBE //kN
+tBA=40
+FBA=tBA*LAB
+printf("The required forces are:-\nForce in member AB=%.2d kN\nForce in member BC=%.2d kN\nForce in member CD=%.2d kN\nForce in member DE=%.2d kN\nForce in member EB=%.2d kN\nForce in member BD=%.2d kN",FBA,FCB,FCD,FDE,FBE,FDB)
diff --git a/3886/CH4/EX4.11/4_11.txt b/3886/CH4/EX4.11/4_11.txt new file mode 100644 index 000000000..7ced3b5be --- /dev/null +++ b/3886/CH4/EX4.11/4_11.txt @@ -0,0 +1,10 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_11.sce', -1)
+The forces in different members are:-
+AB=-68 kN
+BC=-45 kN
+CD=-104 kN
+DE=52 kN
+EA=64 kN
+EB=22 kN
+EC=46 kN
\ No newline at end of file diff --git a/3886/CH4/EX4.11/Ex4_11.sce b/3886/CH4/EX4.11/Ex4_11.sce new file mode 100644 index 000000000..0f916a380 --- /dev/null +++ b/3886/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,30 @@ +//Analyse the truss by method of tension coefficient to determine the forces
+//Refer fig.4.19
+//Consider entire structure
+//Taking moment about A
+YD=(40*2+50*6+30*4*sind(60)+60*4)/8
+XA=-30 //kN
+YA=40+50+60-90.49 //kN
+//Take A as origin and determine co-ordinates of all other point
+//Consider equilibrium of individual joints
+//Joint A
+tAB=-(59.51/3.464)
+FAB=tAB*4 //kN
+tAE=64.36/4
+FAE=tAE*4 //kN
+//Joint B
+tBE=-11.547+17.18
+FBE=tBE*4 //kN
+tBC=0.5*(-17.18-5.637)
+FBC=tBC*4 //kN
+//Joint C
+tCD=-(14.434+37.818)/2
+FCD=4*tCD //kN
+tCD=4 //kN
+tCE=-14.434-tCD
+FCE=11.692*4 //kN
+//Joint D
+tDE=-0.5*(-26.126)
+FDE=tDE*4 //kN
+printf("The forces in different members are:-\nAB=%.2d kN\nBC=%.2d kN\nCD=%.2d kN\nDE=%.2d kN\nEA=%.2d kN\nEB=%.2d kN\nEC=%.2d kN",FAB,FBC,FCD,FDE,FAE,FBE,FCE)
+
diff --git a/3886/CH4/EX4.12/4_12.txt b/3886/CH4/EX4.12/4_12.txt new file mode 100644 index 000000000..1a68cffd4 --- /dev/null +++ b/3886/CH4/EX4.12/4_12.txt @@ -0,0 +1,10 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_12.sce', -1)
+The forces in different members are:-
+AB=20 kN
+BC=60 kN
+CD=84 kN
+DE=-60 kN
+EA=-40 kN
+EC=-60 kN
+EB=-28 kN
\ No newline at end of file diff --git a/3886/CH4/EX4.12/Ex4_12.sce b/3886/CH4/EX4.12/Ex4_12.sce new file mode 100644 index 000000000..ab6e62458 --- /dev/null +++ b/3886/CH4/EX4.12/Ex4_12.sce @@ -0,0 +1,38 @@ +//Analyse the truss
+//Refer fig. 4.20
+//Consider equilibrium of entire truss
+//taking moment about A
+YE=(60*8-40*4)/4 //kN
+XA=40 //kN
+YA=60-80 //kN
+//Take A as origin and determine co-ordinates of various point
+//Lengths in m are
+AB=4
+CE=4
+AE=4
+ED=4
+BE=4*sqrt(2)
+CD=4*sqrt(2)
+//Consider equilibrium of joints individually
+//Joint A
+tAB=5
+FAB=tAB*AB //kN
+tAE=-10
+FAE=tAE*AE //kN
+//Joint B
+tBE=-5
+FBE=tBE*BE //kN
+tBC=10-tBE
+BC=4
+FBC=tBC*BC //kN
+//Joint C
+tCD=15
+CD=4*sqrt(2)
+FCD=15*4*sqrt(2) //kN The answer provided in the textbook is wrong
+tCE=-15
+FCE=tCE*CE //kN
+//Joint D
+tDE=-15
+DE=4
+FDE=tDE*DE //kN
+printf("The forces in different members are:-\nAB=%.2d kN\nBC=%.2d kN\nCD=%.2d kN\nDE=%.2d kN\nEA=%.2d kN\nEC=%.2d kN\nEB=%.2d kN",FAB,FBC,FCD,FDE,FAE,FCE,FBE)
diff --git a/3886/CH4/EX4.2/4_2.txt b/3886/CH4/EX4.2/4_2.txt new file mode 100644 index 000000000..1100421fd --- /dev/null +++ b/3886/CH4/EX4.2/4_2.txt @@ -0,0 +1,11 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_2.sce', -1)
+RD=77.50 kN
+RA=72.50 kN
+FAB=83.72 kN (Compression)
+FAE=41.86 kN (Tension)
+FDC=89.49 kN (Compression)
+FDE=44.74 kN (Tension)
+FBE=37.53 kN (Tension)
+FBC=60.62 kN (Compression)
+FCE=31.75 kN (Tension)
\ No newline at end of file diff --git a/3886/CH4/EX4.2/Ex4_2.sce b/3886/CH4/EX4.2/Ex4_2.sce new file mode 100644 index 000000000..519bc8e53 --- /dev/null +++ b/3886/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,23 @@ +//finding forces in members of truss
+//Refer fig. 4.9
+//Find support reactions
+//applying equilibrium conditions
+//Taking moment about A
+RD=(40*1+60*2+50*3)/4 //kN
+RA=150-RD //kN
+//Consider equilibrium of joint A
+FAB=RA/sind(60) //kN (Compression)
+FAE=FAB*cosd(60) //kN (Tension)
+//Joint D
+FDC=RD/sind(60) //kN (Compression)
+FDE=FDC*cosd(60) //kN (Tension)
+//Joint B (Refer Fig. 4.9 (d)
+FBE=((FAB*sind(60))-40)/sind(60) //kN (Tension)
+FBC=FAB*cosd(60)+FBE*cosd(60) //kN (Compression)
+//Joint C (Refer fig. 4.9 (e))
+FCE=(FDC*sind(60)-50)/sind(60) //kN (Tension)
+//Refer fig. 4.9 (e),(f)
+printf("RD=%.2f kN\nRA=%.2f kN\nFAB=%.2f kN (Compression)\nFAE=%.2f kN (Tension)\nFDC=%.2f kN (Compression)\nFDE=%.2f kN (Tension)\nFBE=%.2f kN (Tension)\nFBC=%.2f kN (Compression)\nFCE=%.2f kN (Tension)",RD,RA,FAB,FAE,FDC,FDE,FBE,FBC,FCE)
+
+
+
diff --git a/3886/CH4/EX4.3/4_3.txt b/3886/CH4/EX4.3/4_3.txt new file mode 100644 index 000000000..4912b7aa7 --- /dev/null +++ b/3886/CH4/EX4.3/4_3.txt @@ -0,0 +1,15 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_3.sce', -1)
+FED=25.00 kN (Tension)
+FEF=15.00 kN (Compression)
+RC=15.00 kN
+VA=20.00 kN
+HA=15.00 kN
+FAB=20.00 kN (Compression)
+FAF=15.00 kN (Compression)
+FCB=25.00 kN (Compression)
+FCD=20.00 kN (Tension)
+FBF=0.00 kN
+FBD=15.00 kN (Tension)
+FFD=0.00 kN
+FBF=0.00 kN
\ No newline at end of file diff --git a/3886/CH4/EX4.3/Ex4_3.sce b/3886/CH4/EX4.3/Ex4_3.sce new file mode 100644 index 000000000..0d0d41764 --- /dev/null +++ b/3886/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,32 @@ +//Analysing the truss
+//Refer fig. 4.10(a)
+//inclined members make an angle theta with the horizontal
+theta=atand(4/3) //Degree
+//Joint E
+//Refer fig. 4.10 (c)
+//applying equilibrium conditions
+FED=20/sind(theta) //kN (Tension)
+FEF=25*cosd(theta) //kN (Compression)
+//Refer fig 4.10 (b)
+//Taking moment about A
+RC=20*6/8 //kN
+VA=20 //kN
+HA=RC //kN
+//Joint A
+//Refer fig.4.10 (d)
+//applying equilibrium conditions
+FAB=VA //kN (Compression)
+FAF=HA //kN (Compression)
+//Joint C
+//Refer fig. 4.10 (E)
+FCB=RC/cosd(theta) //kN (Compression)
+FCD=FCB*sind(theta) //kN (Tension)
+//Joint B
+//Refer fig. 4.10 (f)
+FBF=(FBC*sind(theta)-FAB)/sind(theta) //kN
+FBD=0+25*cosd(theta) //kN (Tension)
+//Joint F
+//Refer Fig. 4.10(g)
+FFD=0
+FBF=0
+printf("FED=%.2f kN (Tension)\nFEF=%.2f kN (Compression)\nRC=%.2f kN\nVA=%.2f kN\nHA=%.2f kN\nFAB=%.2f kN (Compression)\nFAF=%.2f kN (Compression)\nFCB=%.2f kN (Compression)\nFCD=%.2f kN (Tension)\nFBF=%.2f kN\nFBD=%.2f kN (Tension)\nFFD=%.2f kN\nFBF=%.2f kN",FED,FEF,RC,VA,HA,FAB,FAF,FCB,FCD,FBF,FBD,FFD,FBF)
diff --git a/3886/CH4/EX4.4/4_4.txt b/3886/CH4/EX4.4/4_4.txt new file mode 100644 index 000000000..b432b7c96 --- /dev/null +++ b/3886/CH4/EX4.4/4_4.txt @@ -0,0 +1,17 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_4.sce', -1)
+Required values are:-
+FHG=25.00 kN (Compression)
+FHF=15.00 kN (Tension)
+RG=42.00 kN
+VA=10.00 kN (Downwards)
+HA=0.00 kN
+FAC=18.03 kN (Compression)
+FAB=15.00 kN (Tension)
+FBC=0.00 kN
+FCE=18.03 kN (Compression)
+FDE=0.00 kN
+FDF=15.00 kN (Tension)
+FEF=0.00 kN
+FEG=18.03 kN (Compression)
+FAG=12.00 kN (Compression)
\ No newline at end of file diff --git a/3886/CH4/EX4.4/Ex4_4.sce b/3886/CH4/EX4.4/Ex4_4.sce new file mode 100644 index 000000000..92223721f --- /dev/null +++ b/3886/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,29 @@ +//Finding forces in all members
+//Refer fig. 4.11(a)
+theta1=atand(4/6) //Degree
+theta2=atand(8/6) //Degree
+theta3=atand(4/3) //Degree
+//Joint H
+FHG=20/sind(53.13) //kN (Compression)
+FHF=25*cosd(53.13) //kN (Tension)
+//Taking moment about A
+RG=(20*9+12*6)/6 //kN
+VA=32-42 //kN (Downwards)
+HA=0
+//Joint A
+//applying equilibrium conditions
+FAC=10/sind(33.69) //kN (Compression)
+FAB=18.03*cosd(33.69) //kN (Tension)
+//Joint B
+FBC=0
+FCE=FAC //kN (Compression)
+//Joint D
+FDE=0
+FDF=FBD //kN (Tension)
+//Joint E
+FEF=0
+FEG=FCE //kN (Compression)
+//Joint F
+FAG=12 //kN (Compression)
+printf("Required values are:-\nFHG=%.2f kN (Compression)\nFHF=%.2f kN (Tension)\nRG=%.2f kN\nVA=%.2f kN (Downwards)\nHA=%.2f kN\nFAC=%.2f kN (Compression)\nFAB=%.2f kN (Tension)\nFBC=%.2f kN\nFCE=%.2f kN (Compression)\nFDE=%.2f kN\nFDF=%.2f kN (Tension)\nFEF=%.2f kN\nFEG=%.2f kN (Compression)\nFAG=%.2f kN (Compression)",FHG,FHF,RG,-VA,HA,FAC,FAB,FBC,FCE,FDE,FDF,FEF,FEG,FAG)
+
diff --git a/3886/CH4/EX4.5/4_5.txt b/3886/CH4/EX4.5/4_5.txt new file mode 100644 index 000000000..8b98d9c1f --- /dev/null +++ b/3886/CH4/EX4.5/4_5.txt @@ -0,0 +1,18 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_5.sce', -1)
+
+Required Forces are:-
+FGF=23.09 kN (Tension)
+FGE=11.55 kN (Compression)
+FFE=23.09 kN (Compression)
+FFD=13.09 kN (Tension)
+RE=58.17 kN
+VA=31.83 kN
+HA=10.00 kN
+FAB=36.75 kN (Compression)
+FAC=8.37 (Tension)
+FBC=9.43 kN (Compression)
+FBD=13.65 kN (Compression)
+FCD=9.43 kN (Tension)
+FCE=1.06 kN (Compression)
+FDE=44.07 kN (Compression)
\ No newline at end of file diff --git a/3886/CH4/EX4.5/Ex4_5.sce b/3886/CH4/EX4.5/Ex4_5.sce new file mode 100644 index 000000000..bca4a3ebf --- /dev/null +++ b/3886/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,28 @@ +//Analyse truss
+//Refer fig. 4.12 (a)
+//All triangles are equilateral
+//applying equilibrium conditions At
+//Joint G
+FGF=20/sind(60) //kN (Tension)
+FGE=FGF*cosd(60) //kN (Compression)
+//Joint F
+FFE=FGF //kN (Compression)
+FFD=FGF*cosd(60)+FFE*cosd(60)-10 //kN (Tension)
+//Consider eqiulibrium of entire truss
+RE=(-10*3*sind(60)+40*3*cosd(60)+30*(3+3*cosd(60))+20*9)/6 //kN
+VA=(40+30+20)-58.17 //kN
+HA=10 //kN
+//Joint A
+FAB=31.83/sind(60) //kN (Compression)
+FAC=36.75*cosd(60)-10 //kN (Tension)
+//Joint B
+FBC=(40-FAB*sind(60))/sind(60) //kN (Compression)
+FBD=36.75*cosd(60)-9.44*cosd(60) //kN (Compression)
+//Joint C
+FCD=FBC //kN (Tension)
+FCE=9.44*cosd(60)+9.44*cosd(60)-8.38 //kN (Compression)
+//Joint D
+FDE=(30+FCD*sind(60))/sind(60) //kN (Compression)
+printf("\nRequired Forces are:-\nFGF=%.2f kN (Tension)\nFGE=%.2f kN (Compression)\nFFE=%.2f kN (Compression)\nFFD=%.2f kN (Tension)\nRE=%.2f kN\nVA=%.2f kN\nHA=%.2f kN\nFAB=%.2f kN (Compression)\nFAC=%.2f (Tension)\nFBC=%.2f kN (Compression)\nFBD=%.2f kN (Compression)\nFCD=%.2f kN (Tension)\nFCE=%.2f kN (Compression)\nFDE=%.2f kN (Compression)",FGF,FGE,FFE,FFD,RE,VA,HA,FAB,FAC,FBC,FBD,FCD,FCE,FDE)
+
+
diff --git a/3886/CH4/EX4.6/4_6.txt b/3886/CH4/EX4.6/4_6.txt new file mode 100644 index 000000000..8f49c82ed --- /dev/null +++ b/3886/CH4/EX4.6/4_6.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_6.sce', -1)
+The required values are:-
+FFH=69 kN (Compressive)
+FGH=05 kN (Compressive)
+FGI=72 kN (Tension)
\ No newline at end of file diff --git a/3886/CH4/EX4.6/Ex4_6.sce b/3886/CH4/EX4.6/Ex4_6.sce new file mode 100644 index 000000000..273c55e46 --- /dev/null +++ b/3886/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,11 @@ +//Determine The Forces in the member
+//Using symmetry
+RA=70/2 //kN
+RB=RA //kN
+//Consider fig.4.13 (b)
+//Taking moment about G
+FFH=(35*12-10*10-10*6-10*2)/(4*sind(60)) //kN (Compression)
+FGH=(35-10-10-10)/sind(60) //kN (Compression)
+FGI=69.28+5.77*cosd(60) //kN (Tension)
+printf("The required values are:-\nFFH=%.2d kN (Compressive)\nFGH=%.2d kN (Compressive)\nFGI=%.2d kN (Tension)",FFH,FGH,FGI)
+
diff --git a/3886/CH4/EX4.7/4_7.txt b/3886/CH4/EX4.7/4_7.txt new file mode 100644 index 000000000..3f5b52fdd --- /dev/null +++ b/3886/CH4/EX4.7/4_7.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_7.sce', -1)
+The required forces are:-
+Member Force
+U3U4= 456 kN (Compression)
+L3L4= 412 kN (Tension)
+U4L3= 62 kN (Tension)
\ No newline at end of file diff --git a/3886/CH4/EX4.7/Ex4_7.sce b/3886/CH4/EX4.7/Ex4_7.sce new file mode 100644 index 000000000..3d3848461 --- /dev/null +++ b/3886/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,30 @@ +//finding magnitude and nature of forces
+//refer fig. 4.14(a)
+//considering equilibrium if entire truss
+//taking moment about L0
+R2=(200*6+200*12+150*18+100*24+100*30)/36 //kN
+R1=200+200+150+100+100-R2 //kN
+//consider equilibrium of right hand side of section (1)-(1)
+theta1=atand(1/6) //degree
+theta2=atand(6/8) //degree
+//taking moment about U4
+FL3L4=(-100*6+325*12)/8 //kN (tension)
+//taking moment about L3
+FU3U4=456.2 //kN (compression)
+FU4L3=(456.2*cosd(9.46)-412.5)/sind(36.87) //kN (tension)
+printf("The required forces are:-\nMember Force\nU3U4= %.2d kN (Compression)\nL3L4= %.2d kN (Tension)\nU4L3= %.2d kN (Tension)",FU3U4,FL3L4,FU4L3)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3886/CH4/EX4.8/4_8.txt b/3886/CH4/EX4.8/4_8.txt new file mode 100644 index 000000000..b9bca3950 --- /dev/null +++ b/3886/CH4/EX4.8/4_8.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_8.sce', -1)
+The required forces are:-
+F1=109 kN (Compression)
+F2=51 kN (Tension)
+F3=69 kN (Tension)
\ No newline at end of file diff --git a/3886/CH4/EX4.8/Ex4_8.sce b/3886/CH4/EX4.8/Ex4_8.sce new file mode 100644 index 000000000..47b6ff003 --- /dev/null +++ b/3886/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,15 @@ +//Finding forces in members
+//Refer fig. 4.15 (a)
+RA=7*20/2 //kN
+RB=RA //kN (symmetry)
+//CE is perpendicular on AB
+CE=5.196 //m
+DE=3 //m
+theta=atand(5.196/3) //degree
+//The fact that 20 kN loads are equidistant can be used to find out horizontal distances of loads from A
+//Consider equilibrium of left hand side portion of section (1)-(1)
+//taking moment about A
+F2=(20*2.25+20*4.5+20*6.75)/(6*sind(60)) //kN (Tension)
+F1=(70-20-20-20+51.96*sind(60))/sind(30) //kN (compression)
+F3=-51.96*cosd(60)+110*cosd(30) //kN (Tension)
+printf("The required forces are:-\nF1=%.2d kN (Compression)\nF2=%.2d kN (Tension)\nF3=%.2d kN (Tension)",F1,F2,F3)
diff --git a/3886/CH4/EX4.9/4_9.txt b/3886/CH4/EX4.9/4_9.txt new file mode 100644 index 000000000..a09c3d4f2 --- /dev/null +++ b/3886/CH4/EX4.9/4_9.txt @@ -0,0 +1,10 @@ +
+--> exec('E:\My program EM\4. Analysis of pin-jointed plane frames\Ex4_9.sce', -1)
+
+The required forces are:-
+FAB=92 kN (Compression)
+FBC=71 kN (Compression)
+FBF=21 kN (Compression)
+FAF=40 kN (Tension)
+FFC=40 kN (Tension)
+FAE=30 kN (Tension)
\ No newline at end of file diff --git a/3886/CH4/EX4.9/Ex4_9.sce b/3886/CH4/EX4.9/Ex4_9.sce new file mode 100644 index 000000000..59a686b2d --- /dev/null +++ b/3886/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,21 @@ +//Finding required forces
+//Refer fig. 4.16 (a)
+//Symmetry gives
+RE=(15+30+30+30+15)/2 //kN
+RA=RE //kN
+FAE=30 //kN
+//After construction as shown in ref. fig
+//Taking moment about C
+FAE=(60*5-15*5-30*2.5)/5 //kN (Tension)
+//assumptions are made as shown in fig. 4.16 (b)
+//Apply equilibrium conditions and solving equations
+FFC=15/0.366 //kN (Tension)
+FBC=(0.866*40.98+15)/0.707 //kN (Compression)
+//Lets analyse Joint B
+//Applying equilibrium conditions
+FBF=30*cosd(45) //kN (Compression)
+FAB=71.41+21.21 //kN (Compression)
+//Lets analyse Joint A
+//Applying equilibrium conditions
+FAF=(92.62*sind(45)-45)/sind(30) //kN (Tension)
+printf("\nThe required forces are:-\nFAB=%.2d kN (Compression)\nFBC=%.2d kN (Compression)\nFBF=%.2d kN (Compression)\nFAF=%.2d kN (Tension)\nFFC=%.2d kN (Tension)\nFAE=%.2d kN (Tension)",FAB,FBC,FBF,FAF,FFC,FAE)
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