diff options
Diffstat (limited to '3776/CH6')
| -rw-r--r-- | 3776/CH6/EX6.10/Ex6_10.sce | 2 | ||||
| -rw-r--r-- | 3776/CH6/EX6.14/Ex6_14.sce | 13 | ||||
| -rw-r--r-- | 3776/CH6/EX6.15/Ex6_15.sce | 14 | ||||
| -rw-r--r-- | 3776/CH6/EX6.18/Ex6_18.sce | 2 | ||||
| -rw-r--r-- | 3776/CH6/EX6.24/Ex6_24.sce | 16 | ||||
| -rw-r--r-- | 3776/CH6/EX6.3/Ex6_3.sce | 24 | ||||
| -rw-r--r-- | 3776/CH6/EX6.4/Ex6_4.sce | 2 | ||||
| -rw-r--r-- | 3776/CH6/EX6.5/Ex6_5.sce | 8 | ||||
| -rw-r--r-- | 3776/CH6/EX6.8/Ex6_8.sce | 10 | ||||
| -rw-r--r-- | 3776/CH6/EX6.9/Ex6_9.sce | 28 |
10 files changed, 60 insertions, 59 deletions
diff --git a/3776/CH6/EX6.10/Ex6_10.sce b/3776/CH6/EX6.10/Ex6_10.sce index 43f31cf7d..b81c2de1f 100644 --- a/3776/CH6/EX6.10/Ex6_10.sce +++ b/3776/CH6/EX6.10/Ex6_10.sce @@ -3,7 +3,7 @@ clear l = 50.0 //mm - the length of the beam b = 50.0 //mm - the width of the beam M = 2083 //Nm -A = l*b //mm2 - The area +A = l*b //sq.mm - The area //straight beam I = b*(l**3)/12.0 //mm4 - The moment of inertia of the beam c_1= l/2 // the distance where the stress is maximum diff --git a/3776/CH6/EX6.14/Ex6_14.sce b/3776/CH6/EX6.14/Ex6_14.sce index 6198c1458..3a12216ff 100644 --- a/3776/CH6/EX6.14/Ex6_14.sce +++ b/3776/CH6/EX6.14/Ex6_14.sce @@ -1,12 +1,13 @@ clear -//given +//given //from example 6.9 St_ul = 2500 //psi - ultimate strength -st_yl = 40000 //psi _ yielding strength -b = 10 //in - width from example -A = 2 //in2 The area of the steel -d = 20 +st_yl = 40000 //psi _ yielding strength +b = 10 //in - width from example +A = 2 //sq.in The area of the steel +d = 20 t_ul = st_yl*A //ultimate capasity y = t_ul/(St_ul*b*0.85) //in 0.85 because its customary -M_ul = t_ul*(d-y/2)/12 //ft-lb Plastic moment +M_ul = t_ul*(d-y/2)/12 //ft-lb Plastic moment printf("\n The plastic moment of the system is %0.3f ft-lb",M_ul) + //answer in the textbook is wrong diff --git a/3776/CH6/EX6.15/Ex6_15.sce b/3776/CH6/EX6.15/Ex6_15.sce index ca2df4451..cfbab548a 100644 --- a/3776/CH6/EX6.15/Ex6_15.sce +++ b/3776/CH6/EX6.15/Ex6_15.sce @@ -1,12 +1,12 @@ clear -//Given -//From example 5.8 -W = 4.0 //N/m - The force distribution +//Given +//From example 5.8 +W = 4.0 //N/m - The force distribution L = 3 // m - The length of the force applied M = W*L/8.0 // KN.m The moment due to force distribution -o = 30 // the angle of force applid to horizantal -l = 150.0 //mm length of the crossection -b = 100.0 //mm - width of the crossection +o = 30 // the angle of force applied to horizontal +l = 150.0 //mm length of the crossection +b = 100.0 //mm - width of the crossection // M_z = M*(cos(3.14/6)) M_y = M*(sin(%pi/6)) @@ -14,4 +14,4 @@ I_z = b*(l**3)/12.0 I_y = l*(b**3)/12.0 //tanb = I_z /I_y *tan30 b = atand((I_z*tan(3.14/6.0)/I_y)) -printf("\n The angle at which neutral axis locates is %0.3f degrees",b) +printf("\n The angle at which neutral axis located by is %0.3f degrees",b) diff --git a/3776/CH6/EX6.18/Ex6_18.sce b/3776/CH6/EX6.18/Ex6_18.sce index ccce2e9e8..ae568d544 100644 --- a/3776/CH6/EX6.18/Ex6_18.sce +++ b/3776/CH6/EX6.18/Ex6_18.sce @@ -1,7 +1,7 @@ clear l = 50 //mm - The length of the beam b = 50 //mm - The width of the beam -A = l*b //mm2 - The area of the beam +A = l*b //sq.mm - The area of the beam p = 8.33 //KN - The force applied on the beam stress_max = p*(10**3)/A //MPa After cutting section A--b printf("\n The maximum stress in the beam %0.3f MPa ",stress_max ) diff --git a/3776/CH6/EX6.24/Ex6_24.sce b/3776/CH6/EX6.24/Ex6_24.sce index 1aac53ebb..ba4fa87eb 100644 --- a/3776/CH6/EX6.24/Ex6_24.sce +++ b/3776/CH6/EX6.24/Ex6_24.sce @@ -4,25 +4,25 @@ M = 10 //KN.m - The moment applied I_max = 23.95*(10**6) //mm4 - I_z The moment of inertia I_min = 2.53*(10**6) //mm4 - I_y The moment of inertia o = 14.34 // degress the principle axis rotated -//Coponents of M in Y,Z direction +//Coponents of M in Y,Z direction M_z = M*(10**6)*cos((%pi/180)*(o)) M_y = M*(10**6)*sin((%pi/180)*(o)) //tanb = I_z /I_y *tan14.34 b = atan((I_max*tan((%pi/180)*(o))/I_min )) -B = (180/%pi)*(b) +B = (180/%pi)*(b) y_p = 122.9 // mm - principle axis Y cordinate z_p = -26.95 //mm - principle axis z cordinate stress_B = - M_z*y_p/I_max + M_y*z_p/I_min //MPa - Maximum tensile stress y_f = -65.97 // mm - principle axis Y cordinate z_f = 41.93 //mm - principle axis z cordinate stress_f = - M_z*y_f/I_max + M_y*z_f/I_min //MPa - Maximum compressive stress -//location of nuetral axis To show these stresses are max and minimum +//location of nuetral axis To show these stresses are max and minimum //tanB = MzI_z + MzI_yz/MyI_y +M_YI_yz I_z = 22.64 *(10**6) //mm4 moment of inertia in Z direction I_y = 3.84 *(10**6) //mm4 moment of inertia in Y direction -I_yz =5.14 *(10**6) //mm4 moment of inertia in YZ direction -M_y = M //KN.m bending moment in Y dorection -M_z = M //KN.m bending moment in Y dorection +I_yz =5.14 *(10**6) //mm4 moment of inertia in YZ direction +M_y = M //KN.m bending moment in Y dorection +M_z = M //KN.m bending moment in Y dorection B = atan(( M_z*I_yz)/(M_z*I_y )) //(%pi/180)* location on neutral axis -beta = (180/%pi)*(B) -printf("\n By sketching the line with angle %0.1f degrees The farthest point associated with B and F",beta) +beta1 = (180/%pi)*(B) +printf("\n By sketching the line with angle %0.1f degrees The farthest point associated with B and F",beta1) diff --git a/3776/CH6/EX6.3/Ex6_3.sce b/3776/CH6/EX6.3/Ex6_3.sce index 857811d33..19291d887 100644 --- a/3776/CH6/EX6.3/Ex6_3.sce +++ b/3776/CH6/EX6.3/Ex6_3.sce @@ -1,18 +1,18 @@ clear -//Given -//Entire area - hallow area +//Given +//Entire area - hollow area l_e = 60.0 //mm - length of the entire area b_e = 40 //mm - width of the entire area -l_h = 30 //mm - length of the hallow area -b_h = 20 //mm - width of the hallow area -A_e = l_e*b_e //mm2 - The entire area -A_h = -l_h*b_h //mm2 - The hallow area '-' because its hallow -A_re = A_e + A_h //mm2 resultant area -y_e = l_e/2 // mm com from bottom -y_h = 20+l_h/2 //mm com from bottom -y_com = (A_e*y_e + A_h*y_h)/A_re +l_h = 30 //mm - length of the hollow area +b_h = 20 //mm - width of the hollow area +A_e = l_e*b_e //sq.mm - The entire area +A_h = -l_h*b_h //sq.mm - The hollow area +A_re = A_e + A_h //sq.mm resultant area +y_e = l_e/2 // mm com from bottom +y_h = 20+l_h/2 //mm com from bottom +y_com = (A_e*y_e + A_h*y_h)/A_re //moment of inertia caliculatins - bh3/12 +ad2 -I_e = b_e*(l_e**3)/12 + A_e*((y_e-y_com)**2) //Parallel axis theorm -I_h = b_h*(l_h**3)/12 - A_h*((y_h-y_com)**2) //Parallel axis theorm +I_e = b_e*(l_e**3)/12 + A_e*((y_e-y_com)**2) //Parallel axis theorem +I_h = b_h*(l_h**3)/12 - A_h*((y_h-y_com)**2) //Parallel axis theorem I_total = I_e - I_h printf("\n The moment of inertia of total system is %e mm^4",I_total) diff --git a/3776/CH6/EX6.4/Ex6_4.sce b/3776/CH6/EX6.4/Ex6_4.sce index 99f0edaa9..98157a320 100644 --- a/3776/CH6/EX6.4/Ex6_4.sce +++ b/3776/CH6/EX6.4/Ex6_4.sce @@ -5,7 +5,7 @@ b = 300 //mm - breath F = 20 //KN _ the force applied on the beam F_d = 0.75 //KN-m - The force distribution d = 2 //mt - the point of interest from the free end -//caliculations +//calculations //From moment diagram M = F*d - F_d*d*1 I = b*(l**3)/12 //mm4 - Bending moment diagram diff --git a/3776/CH6/EX6.5/Ex6_5.sce b/3776/CH6/EX6.5/Ex6_5.sce index 551491f53..6e8121ffb 100644 --- a/3776/CH6/EX6.5/Ex6_5.sce +++ b/3776/CH6/EX6.5/Ex6_5.sce @@ -7,17 +7,17 @@ l_1 = 1 //in l_2 = 3 //in b_1 = 4 //in b_2 = 1 //in -A_1 = l_1* b_1 //in2 - area of part_1 +A_1 = l_1* b_1 //sq.in - area of part_1 y_1 = 0.5 //in com distance from ab -A_2 =l_2*b_2 //in2 - area of part_1 +A_2 =l_2*b_2 //sq.in - area of part_1 y_2 = 2.5 //in com distance from ab -A_3 = l_2*b_2 //in2 - area of part_1 +A_3 = l_2*b_2 //sq.in - area of part_1 y_3 = 2.5 //in com distance from ab y_net = (A_1*y_1 +A_2*y_2 + A_3*y_3)/(A_1+A_2+A_3) //in - The com of the whole system c_max = (4-y_net) //in - The maximum distace from com to end c_min = y_net //in - the minimum distance from com to end -I_1 = b_1*(l_1**3)/12 + A_1*((y_1-y_net)**2) //Parallel axis theorm +I_1 = b_1*(l_1**3)/12 + A_1*((y_1-y_net)**2) //Parallel axis theorem I_2 = b_2*(l_2**3)/12 + A_2*((y_2-y_net)**2) I_3 = b_2*(l_2**3)/12 + A_2*((y_2-y_net)**2) I_net = I_1 + I_2 + I_3 //in^4 - the total moment of inertia diff --git a/3776/CH6/EX6.8/Ex6_8.sce b/3776/CH6/EX6.8/Ex6_8.sce index 244deb69e..51d04bb08 100644 --- a/3776/CH6/EX6.8/Ex6_8.sce +++ b/3776/CH6/EX6.8/Ex6_8.sce @@ -2,20 +2,20 @@ clear //Given //Given //We will divide this into two parts -E_w = 10.0 //Gpa - Youngs modulus of wood -E_s = 200.0 //Gpa - Youngs modulus of steel +E_w = 10.0 //GPa - Youngs modulus of wood +E_s = 200.0 //GPa - Youngs modulus of steel M = 30.0 //K.N-m _ applied bending moment n = E_s/E_w l_1 = 250 //mm l_2 = 10 //mm b_1 = 150.0 //mm b_2 = 150.0*n //mm -A_1 = l_1* b_1 //mm2 - area of part_1 +A_1 = l_1* b_1 //sq.mm - area of part_1 y_1 = 125.0 //mm com distance from top -A_2 =l_2*b_2 //mm2 - area of part_1 +A_2 =l_2*b_2 //sq.mm - area of part_1 y_2 = 255.0 //mm com distance from top y_net = (A_1*y_1 +A_2*y_2)/(A_1+A_2) //mm - The com of the whole system from top -I_1 = b_1*(l_1**3)/12.0 + A_1*((y_1-y_net)**2) //Parallel axis theorm +I_1 = b_1*(l_1**3)/12.0 + A_1*((y_1-y_net)**2) //Parallel axis theorem I_2 = b_2*(l_2**3)/12.0 + A_2*((y_2-y_net)**2) I_net = I_1 + I_2 //mm4 - the total moment of inertia c_s= y_net // The maximum distance in steel diff --git a/3776/CH6/EX6.9/Ex6_9.sce b/3776/CH6/EX6.9/Ex6_9.sce index 74d9c8acf..1c99a03a6 100644 --- a/3776/CH6/EX6.9/Ex6_9.sce +++ b/3776/CH6/EX6.9/Ex6_9.sce @@ -1,12 +1,12 @@ clear -//Given +//Given M = 50000 //ft-lb , positive bending moment applied -N = 9 // number of steel bars -n = 15 // The ratio of steel to concrete -A_s = 30 //in2 area of steel in concrete +N = 9 // number of steel bars +n = 15 // The ratio of steel to concrete +A_s = 30 //sq.in area of steel in concrete //(10*y)*(y/2) = 30*(20-y) //y**2 + 6*y -120 -//solving quadractic equation +//solving quadractic equation // a = 1 @@ -21,16 +21,16 @@ sol2 = (-b+sqrt(d))/(2*a) y = sol2 // Nuetral axis is found l_1 = y //in- the concrete below nuetral axis is not considered b_1 = 10 //in - width -A_1 = l_1* b_1 //in2 - area of concrete -y_1 = y/2 //in com of the concrete -y_2 = 20-y //in com of the transformed steel -I_1 = b_1*(l_1**3)/12.0 + A_1*((y_1-y)**2) //in^4 parallel axis theorm +A_1 = l_1* b_1 //sq.in - area of concrete +y_1 = y/2 //in com of the concrete +y_2 = 20-y //in com of the transformed steel +I_1 = b_1*(l_1**3)/12.0 + A_1*((y_1-y)**2) //in^4 parallel axis theorem I_2 = A_s*((y_2)**2) //in^4 first part is neglected I_net = I_1 + I_2 //in^4 - the total moment of inertia -c_c= y //in The maximum distance in concrete -stress_concrete = M*12*c_c/I_net //psi - The maximum stress in concrete -c_s= 20- y -stress_steel =n*M*12*c_s/I_net //psi - The maximum stress in concrete +c_c= y //in The maximum distance in concrete +stress_concrete = M*12*c_c/I_net //psi - The maximum stress in concrete +c_s= 20- y +stress_steel =n*M*12*c_s/I_net //psi - The maximum stress in concrete printf("\n The maximum stress in concrete %0.2f psi",stress_concrete) // printf("\n The stress in steel %0.2f psi",stress_steel) -printf("\n answer varies due to rounding off errors")
\ No newline at end of file +// answer varies due to rounding off errors |