diff options
Diffstat (limited to '3776')
67 files changed, 490 insertions, 422 deletions
diff --git a/3776/CH1/EX1.1/Ex1_1.sce b/3776/CH1/EX1.1/Ex1_1.sce index f0d2002bc..ea165c7b0 100644 --- a/3776/CH1/EX1.1/Ex1_1.sce +++ b/3776/CH1/EX1.1/Ex1_1.sce @@ -11,7 +11,7 @@ BC = 1.0 //m CF = 2.5 //m contact_area = 200*200 // sq.mm , The contact area at c -//caliculations +//calculations //Balancing forces in the x direction: // Balncing the moments about C and B: Fx = 0 diff --git a/3776/CH1/EX1.2/Ex1_2.sce b/3776/CH1/EX1.2/Ex1_2.sce index 90d76b556..04492aa43 100644 --- a/3776/CH1/EX1.2/Ex1_2.sce +++ b/3776/CH1/EX1.2/Ex1_2.sce @@ -1,24 +1,24 @@ clear -//Given +//Given load_distributed = 20 //kN/sq.m, This is the load distributed over the pier -H = 2 // m, Total height -h = 1 //m , point of investigation -base = 1.5 //m The length of crossection in side veiw +H = 2 // m, Total height +h = 1 //m , point of investigation +base = 1.5 //m The length of crossection in side view top = 0.5 //m ,The length where load is distributed on top -base_inv = 1 //m , the length at the point of investigation -area = 0.5*1 //m ,The length at a-a crossection +base_inv = 1 //m , the length at the point of investigation +area = 0.5*1 //m ,The length at a-a cross-section density_conc = 25 //kN/sq.m -//caliculation of total weight +//calculation of total weight -v_total = ((top+base)/2)*top*H //sq.m ,The total volume +v_total = ((top+base)/2)*top*H //sq.m ,The total volume w_total = v_total* density_conc //kN , The total weight -R_top = (top**2)*load_distributed //kN , THe reaction force due to load distribution +R_top = (top**2)*load_distributed //kN , THe reaction force due to load distribution reaction_net = w_total + R_top -//caliculation of State of stress at 1m +//calculation of State of stress at 1m v_inv = ((top+base_inv)/2)*top*h //sq.m ,The total volume from 1m to top w_inv = v_inv*density_conc //kN , The total weight from 1m to top reaction_net = w_inv + R_top //kN Stress = reaction_net/area //kN/sq.m printf("\n The total weight of pier is %0.3f kN",w_total) -printf("\n The stress at 1 m above is %0.1f kN/m**2",Stress) +printf("\n The stress at 1 m above is %0.1f kN/sq.m",Stress) diff --git a/3776/CH1/EX1.3/Ex1_3.sce b/3776/CH1/EX1.3/Ex1_3.sce index 05d597de5..60497acef 100644 --- a/3776/CH1/EX1.3/Ex1_3.sce +++ b/3776/CH1/EX1.3/Ex1_3.sce @@ -4,15 +4,15 @@ clear d_pins = 0.375 //inch load1 = 3 //kips AB_x = 6 //inch,X-component -AB_y = 3 //inch,Y-component +AB_y = 3 //inch,Y-component BC_y = 6 //inch,Y-component BC_x = 6 //inch,X-component -area_AB = 0.25*0.5 //inch*2 -area_net = 0.20*2*(0.875-0.375) //inch*2 -area_BC = 0.875*0.25 //inch*2 -area_pin = d_pins*2*0.20 //inch*2 +area_AB = 0.25*0.5 //inch*2 +area_net = 0.20*2*(0.875-0.375) //inch*2 +area_BC = 0.875*0.25 //inch*2 +area_pin = d_pins*2*0.20 //inch*2 area_pin_crossection = 2*3.14*((d_pins/2)**2) -//caliculations +//calculations slope = AB_y/ AB_x //For AB slope = BC_y/ BC_x //For BC @@ -27,18 +27,18 @@ F_C_x = -(load1*BC_x)/(BC_y + AB_y ) //kips, F_C_x X-component of F_c F_A= ((5**0.5)/2)*F_A_x //kips F_A_y = 0.5*F_A_x //kips -//X,Y components of F_C +//X,Y components of F_C F_C= (2**0.5)*F_C_x //kips F_C_y = F_C_x //kips T_stress_AB = F_A/area_AB //ksi , Tensile stress in main bar AB stress_clevis = F_A/area_net //ksi ,Tensile stress in clevis of main bar AB -c_strees_BC = F_C/area_BC //ksi , Comprensive stress in main bar BC +c_stress_BC = F_C/area_BC //ksi , Compressive stress in main bar BC B_stress_pin = F_C/area_pin //ksi , Bearing stress in pin at C To_stress_pin = F_C/area_pin_crossection //ksi , torsion stress in pin at C printf("\n Tensile stress in main bar AB: %0.1f ksi",T_stress_AB) printf("\n Tensile stress in clevis of main bar AB: %0.1f ksi",stress_clevis) -printf("\n Comprensive stress in main bar BC: %0.1f ksi",-c_strees_BC) +printf("\n Compressive stress in main bar BC: %0.1f ksi",-c_stress_BC) printf("\n Bearing stress in pin at C: %0.2f ksi",-B_stress_pin) printf("\n torsion stress in pin at C: %0.2f ksi",-To_stress_pin) diff --git a/3776/CH1/EX1.6/Ex1_6.sce b/3776/CH1/EX1.6/Ex1_6.sce index ba7d7e038..11a4a83f1 100644 --- a/3776/CH1/EX1.6/Ex1_6.sce +++ b/3776/CH1/EX1.6/Ex1_6.sce @@ -5,10 +5,10 @@ frequency = 10 //Hz stress_allow = 200 //MPa R = 0.5 //m -//caliculations +//calculations // w = 2*%pi*frequency //rad/sec a = (w**2)*R //sq.m/sec F = mass*a //N -A_req = F/stress_allow //sq.m , The required area for aloowing stress -printf("\n The required size of rod is: %0.2f sq.m",A_req) +A_req = F/stress_allow //sq.mm , The required area for allowing stress +printf("\n The required size of rod is: %0.2f sq.mm",A_req) diff --git a/3776/CH1/EX1.7/Ex1_7.sce b/3776/CH1/EX1.7/Ex1_7.sce index a9cde7839..b4256e15b 100644 --- a/3776/CH1/EX1.7/Ex1_7.sce +++ b/3776/CH1/EX1.7/Ex1_7.sce @@ -6,21 +6,69 @@ L_n_2 = 15 //kips ,live load 2 stress_allow = 22 //ksi phi = 0.9 //probalistic coefficients y_stress = 36 //ksi,Yeild strength -//According to AISR +//According to AISR //a p_1 = D_n + L_n_1 //kips since the total load is sum of dead load and live load p_2 = D_n + L_n_2 //kips, For second live load -Area_1 = p_1/stress_allow //in*2 ,the allowable area for the allowed stress -Area_2 = p_2/stress_allow //in*2 -printf("\n the allowable area for live load %0.3f is %0.3f in*2",L_n_1,Area_1) -printf("\n the allowable area for live load %0.3f is %0.3f in*2",L_n_2,Area_2) +Area_1 = p_1/stress_allow //sq.in ,the allowable area for the allowed stress +Area_2 = p_2/stress_allow //sq.in +printf("\n the allowable area for live load %0.3f is %0.3f sq.in",L_n_1,Area_1) +printf("\n the allowable area for live load %0.3f is %0.3f sq.in",L_n_2,Area_2) //b //area_crossection= (1.2*D_n +1.6L_n)/(phi*y_stress) -area_crossection_1= (1.2*D_n +1.6*L_n_1)/(phi*y_stress) //in*2,crossection area for first live load -area_crossection_2= (1.2*D_n +1.6*L_n_2)/(phi*y_stress) //in*2,crossection area for second live load -printf("\n the crossection area for live load %0.3f is %0.3f in*2",L_n_1,area_crossection_1) -printf("\n the crossection area for live load %0.3f is %0.3f in*2",L_n_2,area_crossection_2) +area_crossection_1= (1.2*D_n +1.6*L_n_1)/(phi*y_stress) //sq.in,crossection area for first live load +area_crossection_2= (1.2*D_n +1.6*L_n_2)/(phi*y_stress) //sq.in,crossection area for second live load +printf("\n the crossection area for live load %0.3f is %0.3f sq.in",L_n_1,area_crossection_1) +printf("\n the crossection area for live load %0.3f is %0.3f sq.in",L_n_2,area_crossection_2) + +//c + +//calculating safety indices for a) + +mu_r1=1.05*Area_1*y_stress//kips +del_R=0.11 +mu_q1 = 6//kips +del_q1 = 0.093 + +mu_r2=1.05*Area_2 *y_stress//kips +mu_q2 = 20//kips +del_q2 = 0.189 + +beta_1 = log(mu_r1/mu_q1)/(del_R**2+del_q1**2)**0.5 +beta_2 = log(mu_r2/mu_q2)/(del_R**2+del_q2**2)**0.5 + + + +printf("\n Safety index for a) beta1 is %0.3f ",beta_1) +printf("\n Safety index for a) beta2 is %0.3f ",beta_2) + + + + + + + + +//calculating safety indices for b) + +mu_r1=1.05*area_crossection_1*y_stress//kips +del_R=0.11 +mu_q1 = 6//kips +del_q1 = 0.093 + +mu_r2=1.05*area_crossection_2*y_stress//kips +mu_q2 = 20//kips +del_q2 = 0.189 + +beta_1 = log(mu_r1/mu_q1)/(del_R**2+del_q1**2)**0.5 +beta_2 = log(mu_r2/mu_q2)/(del_R**2+del_q2**2)**0.5 + +beta_1 = log(mu_r1/mu_q1)/(del_R**2+del_q1**2)**0.5 +beta_2 = log(mu_r2/mu_q2)/(del_R**2+del_q2**2)**0.5 + +printf("\n Safety index for b) beta1 is %0.3f ",beta_1) +printf("\n Safety index for b) beta2 is %0.3f ",beta_2) diff --git a/3776/CH1/EX1.8/Ex1_8.sce b/3776/CH1/EX1.8/Ex1_8.sce index e30f1631e..3673b9dd1 100644 --- a/3776/CH1/EX1.8/Ex1_8.sce +++ b/3776/CH1/EX1.8/Ex1_8.sce @@ -1,13 +1,13 @@ clear //Given -A_angle = 2 //in*2 +A_angle = 2 //sq.in stress_allow = 20 //ksi, The maximum alowable stress F = stress_allow*A_angle //K, The maximum force AD = 3 //in, from the figure DC = 1.06 //in, from the figure strength_AWS = 5.56 // kips/in,Allowable strength according to AWS -//caliculations +//calculations //momentum at point "d" is equal to 0 R_1 = (F*DC)/AD //k,Resultant force developed by the weld R_2 = (F*(AD-DC))/AD //k,Resultant force developed by the weld diff --git a/3776/CH10/EX10.1/Ex10_1.sce b/3776/CH10/EX10.1/Ex10_1.sce index dfc409a95..63307cca9 100644 --- a/3776/CH10/EX10.1/Ex10_1.sce +++ b/3776/CH10/EX10.1/Ex10_1.sce @@ -1,10 +1,11 @@ clear -//Given +//Given dia = 400 //mm - The diameter of a pulley -E = 2001 //Gpa - Youngs modulus +E = 200 //GPa - Youngs modulus t = 0.6 //mm - The thickness of band -c = t/2 //mm - The maximum stress is seen -//Caliculations +c = t/2 //mm - The maximum stress is seen +//calculations stress_max = E*c*(10**3)/(dia/2) //MPa - The maximum stress on the crossection occurs at the ends printf("\n The maximum bending stress developed in the saw %0.3f MPa",stress_max) +// answer varies from the book diff --git a/3776/CH10/EX10.10/Ex10_10.sce b/3776/CH10/EX10.10/Ex10_10.sce index b9e6eac7c..403def9d1 100644 --- a/3776/CH10/EX10.10/Ex10_10.sce +++ b/3776/CH10/EX10.10/Ex10_10.sce @@ -1,8 +1,8 @@ clear -k = 24.0*(10**12) //N.mm2 Flexure rigidity -E = 200.0 //Gpa - Youngs modulus of the string +k = 24.0*(10**12) //N.sq.mm Flexure rigidity +E = 200.0 //GPa - Youngs modulus of the string l = 5000.0 //mm - The length of the string -C_A = 300.0 //mm2 - crossection area +C_A = 300.0 //sq.mm - crossection area P = 50.0 //KN - The force applies at the end a = 2000.0 //mm - The distance C-F x = 1//X - let it be a variable X diff --git a/3776/CH10/EX10.11/Ex10_11.sce b/3776/CH10/EX10.11/Ex10_11.sce index a514a0eb8..7fd39419e 100644 --- a/3776/CH10/EX10.11/Ex10_11.sce +++ b/3776/CH10/EX10.11/Ex10_11.sce @@ -8,11 +8,11 @@ I_Z = 315 //in^4 - the moment of inertia wrt Z axis I_y = 8.13 //in^4 - the moment of inertia wrt Y axis o = 5 // degrees - the angle of acting force P = 2000 //k the acting force -P_h = P*sin((%pi/180)*(o)) //k - The horizantal component of P +P_h = P*sin((%pi/180)*(o)) //k - The horizontal component of P P_v = P*cos((%pi/180)*(o)) //k - The vertical component of P -e_h = P_h*(L**3)/(3*E*I_y) // the horizantal component of deflection +e_h = P_h*(L**3)/(3*E*I_y) // the horizontal component of deflection e_v = P_v*(L**3)/(3*E*I_Z ) // the vertical component of deflection e = ((e_h**2 + e_v**2)**0.5) -printf("\n the horizantal component of deflection %0.3f in",e_h) +printf("\n the horizontal component of deflection %0.3f in",e_h) printf("\n the vertical component of deflection %0.3f in",e_v) printf("\n the resultant deflection %0.3f in",e) diff --git a/3776/CH10/EX10.13/Ex10_13.sce b/3776/CH10/EX10.13/Ex10_13.sce index e2688eb46..75bc3aa0d 100644 --- a/3776/CH10/EX10.13/Ex10_13.sce +++ b/3776/CH10/EX10.13/Ex10_13.sce @@ -6,7 +6,7 @@ m = 15.3 // mass of the falling body h = 75.0 //mm - The height of the falling body p = m*9.81 //N the force acted due to the body L = 1000.0 //mm The length of the cantilever -E = 200 //Gpa The youngs modulus of the material used +E = 200 //GPa The youngs modulus of the material used I = (l**4)/12 //mm - the moment of inertia k = 300 //N/mm -the stiffness of the spring //Rigid supports diff --git a/3776/CH10/EX10.15/Ex10_15.sce b/3776/CH10/EX10.15/Ex10_15.sce index 72f3ed920..62eb98e8c 100644 --- a/3776/CH10/EX10.15/Ex10_15.sce +++ b/3776/CH10/EX10.15/Ex10_15.sce @@ -1,17 +1,14 @@ clear //Given -E = 30*(10**3) //ksi - The youngs modulus of the material +E = 30*(10**3) //ksi - The youngs modulus of the material stress_y = 40 //ksi - yield stress -stress_max = 24.2 //ksi - the maximum stress -l = 2 //in - The length of the crossection +stress_max = 24.4 //ksi - the maximum stress +l = 2 //in - The length of the crossection b = 3 //in - the width of the crossection -h = 3 //in - the depth of the crossection -//lets check ultimate capacity for a 2 in deep section -M_ul = stress_y*b*(l**2)/4 //K-in the ultimate capacity -curvature = 2*stress_y/(E*(h/2) ) //in*-1 the curvature of the beam -curvature_max = stress_max/(E*(h/2)) //in*-1 The maximum curvature -printf("\n the ultimate capacity %0.3f k-in",M_ul) -printf("\n the ultimate curvature %0.3f in *-1",curvature_max) -printf("\n E given in equation is wrong") -printf("\n Actual E in question is 30*10**3") - +h = 2 //in - the depth of the crossection +//lets check ultimate capacity for a 2 in deep section +M_ul = stress_max*b*(l**2)/4 //K-in the ultimate capacity +curvature = 2*stress_y/(E*(h/2) ) //per inch the curvature of the beam +curvature_max = stress_max/(E*(b/2)) //per inch The maximum curvature +printf("\n the curvature in 11-in is %e per inch",curvature) +printf("\n the ultimate curvature %e per inch",curvature_max) diff --git a/3776/CH10/EX10.16/Ex10_16.sce b/3776/CH10/EX10.16/Ex10_16.sce index 7cce6d1d2..d1e701290 100644 --- a/3776/CH10/EX10.16/Ex10_16.sce +++ b/3776/CH10/EX10.16/Ex10_16.sce @@ -1,12 +1,12 @@ clear -//Given -l_ad = 1600 //mm - The total length of the beam +//Given +l_ad = 1600 //mm - The total length of the beam l_ab = 600 //mm - The length of AB l_bc = 600 //mm - The length of BC -e_1 = 0.24 //mm - deflection +e_1 = 0.24 //mm - deflection e_2 = 0.48 //mm - deflection -E = 35 //Gpa -//Caliculation +E = 35 //GPa +//calculation A_afe = -(l_ab+l_bc)*e_1*(10**-3)/(2*E) A_afe = -(l_ab)*e_2*(10**-3)/(4*E) @@ -15,4 +15,4 @@ x_1 = 1200 //com from B x_2 = 800 //com from B y_b = A_afe*x_1 + A_afe*x_2 //mm The maximum deflection at tip B printf("\n The maximum deflection at tip B %0.2f mm",y_b) -printf("\n The slope at the tip B %0.2f radians",y_1_b) +printf("\n The slope at the tip B %e radians",y_1_b) diff --git a/3776/CH11/EX11.11/Ex11_11.sce b/3776/CH11/EX11.11/Ex11_11.sce index 510361b42..ba17c5e84 100644 --- a/3776/CH11/EX11.11/Ex11_11.sce +++ b/3776/CH11/EX11.11/Ex11_11.sce @@ -1,59 +1,48 @@ clear // -P = 200.0 //K The force on the beam +P = 200.0 //K The force on the beam L = 15 //ft - The length of the rod -F_y = 50.0 //ksi +F_y = 50.0 //ksi F_a = F_y/(5.0/3) //ksi -AISC MANUAL ,allowable axial stress if axial force is alone F_b = F_a //Allowable compressive bending stress M_1 = 600.0 //k-in - The moment acting on the ends of the rod M_2 = 800.0 //k-in - the moment acting on the other end of teh rod -B_x = 0.264 //in - Extracted from AISC manual -E = 29*(10**3) -A = P/F_a + M_2*B_x/F_b //in2- The minimum area +B_x = 0.264 //in - Extracted from AISC manual +E = 29*(10**3) +A = P/F_a + M_2*B_x/F_b //sq.in- The minimum area printf("\n \n The minimum area is %0.2f in^2",A) -//we will select W10x49 section -A_s = 14.4 //in2 - The area of the section -r_min = 2.54 //in The minimum radius -r_x = 4.35 //in -f_a = P/A_s //Ksi- The computed axial stress -f_b = M_2*B_x/A_s //Computed bending stess -C_c = ((2*(%pi**2)*E/F_y)**0.5) //Slenderness ratio L/R -C_s = L*12/r_min // Slenderness ratio L/R of the present situation -if C_s <C_c then - printf("\n The following approch is solvable") -else - printf("\n The caliculation is not possible") - end -F_a_1 = 19.3 //Ksi - AISC lets try this -c_m = 0.6 - 0.4*(-M_1/M_2) -F_e = (12*(%pi**2)*E)/(23*(L*12/r_x)**2) -k = f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b //Condition mentioned in AISC -if k>1 then - printf("\n The following W10x49 section is not satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b %0.3f >1",k) -else - printf("\n The following W10x49 section is satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b %0.3f <1",k) - end -//trail - 2 -//Lets take W10 x 60 -A_s = 17.6 //in2 - The area of the section -r_min = 2.57 //in The minimum radius -r_x = 4.39 //in -f_a = P/A_s //Ksi- The computed axial stress -f_b = M_2*B_x/A_s //Computed bending stess -C_c = ((2*(%pi**2)*E/F_y)**0.5) //Slenderness ratio L/R -C_s = L*12/r_min // Slenderness ratio L/R of the present situation -if C_s <C_c then - printf("\n The following approch is solvable") -else - printf("\n The caliculation is not possible") - end -F_a_1 = 19.3 //Ksi - AISC lets try this -c_m = 0.6 - 0.4*(-M_1/M_2) -F_e = (12*(%pi**2)*E)/(23*(L*12/r_x)**2) -k = f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b //Condition mentioned in AISC -if k>1 then - printf("\n The following W10x49 section is not satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b %0.3f >1",k) -else - printf("\n The following W10x49 section is satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b %0.3f <1",k) - end -printf("\n small variation due to rounding off errors")
\ No newline at end of file + + + + +for i=1:2 + + st =['W10x49', 'W10x60'] + printf("\n we will select %s section",st(i)) + + A_s = [14.4, 17.6 ] //sq.in - The area of the section + r_min = [2.54 , 2.57 ] //in The minimum radius + r_x = [4.35 ,4.39] //in + f_a = P/A_s(i) //Ksi- The computed axial stress + f_b = M_2*B_x/A_s(i) //Computed bending stess + C_c = ((2*(%pi**2)*E/F_y)**0.5) //Slenderness ratio L/R + C_s = L*12/r_min(i) // Slenderness ratio L/R of the present situation + if C_s <C_c then + printf("\n Since calculated Le/r ratio is less than Cc, we can apply the second ASD formula") + else + printf("\n The calculation is not possible") + end + F_a_1 = 19.3 //Ksi - AISC lets try this + c_m = 0.6 - 0.4*(-M_1/M_2) + F_e = (12*(%pi**2)*E)/(23*(L*12/r_x(i))**2) + k = f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b //Condition mentioned in AISC + if k>1 then + printf("\n The following %s section is not satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b %0.3f >1",st(i),k) + else + printf("\n The following %s section is satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b %0.3f <1",st(i),k) + end + +end + + +printf("\n small variation due to rounding off errors") diff --git a/3776/CH11/EX11.2/Ex11_2.sce b/3776/CH11/EX11.2/Ex11_2.sce index a11f41c53..eb888504a 100644 --- a/3776/CH11/EX11.2/Ex11_2.sce +++ b/3776/CH11/EX11.2/Ex11_2.sce @@ -3,14 +3,14 @@ clear // h = 60 //mm - the length of the crossection b = 100 //mm - the width of hte crossection -E = 200 //Gpa - The youngs modulus +E = 200 //GPa - The youngs modulus stress_cr = 250 //MPa - The proportionality limit -//Caliculations +//calculations I = b*(h**3)/12 //mm3 The momentof inertia of the crossection -A = h*b //mm2 - The area of teh crossection -//From Eulier formula +A = h*b //sq.mm - The area of teh crossection +//From Euler formula r_min = ((I/A)**0.5) //mm - The radius of the gyration -//(l/r)**2= (%pi**2)*E/stress_cr //From Eulier formula +//(l/r)**2= (%pi**2)*E/stress_cr //From Euler formula l = (((%pi**2)*E*(10**3)/stress_cr)**0.5)*r_min //mm - the length after which the beam starts buckling printf("\n The length after which the beam starts buckling is %0.0f mm",l) diff --git a/3776/CH11/EX11.6/Ex11_6.sce b/3776/CH11/EX11.6/Ex11_6.sce index a095bf461..a2b8da993 100644 --- a/3776/CH11/EX11.6/Ex11_6.sce +++ b/3776/CH11/EX11.6/Ex11_6.sce @@ -2,36 +2,54 @@ clear //Given // L = 15 //ft - The length of the each rod -A = 46.7 //in2 - The length of the crossection +A = 46.7 //sq.in - The length of the crossection r_min = 4 //in - The radius of gyration stress_yp = 36 //ksi - the yielding point stress E = 29*(10**3) //ksi - The youngs modulus C_c = ((2*(%pi**2)*E/stress_yp)**0.5) //Slenderness ratio L/R -C_s = L*12/r_min // Slenderness ratio L/R of the present situation -//According to AISC formulas -if (C_s <C_c) then - printf ("a)The following approch is solvable") +C_s1 = L*12/r_min // Slenderness ratio L/R of the present situation +//According to AISC formulas +printf ("a)calculated Le/r ratio is %f",C_s1) +if (C_s1 <C_c) then + printf ("\n a)Since calculated Le/r ratio is less than Cc(126), we can apply the second ASD formula") else - print ("The caliculation is not possible") + print ("The calculation is not possible") end -F_S = 5.0/3 +3*C_s/(8*C_c) -(3*C_s**3)/(8*C_c**3) //Safety factor -Stress_all = (1 - (C_s**2)/(2*C_c**2))*stress_yp/F_S //The allowable strees +F_S = 5.0/3 +3*C_s1/(8*C_c) -(C_s1**3)/(8*C_c**3) //Safety factor +Stress_all = (1 - (C_s1**2)/(2*C_c**2))*stress_yp/F_S //The allowable stress printf("\n a) The allowable stress in this case is %0.2f kips",Stress_all) +printf("\n a) The allowable pressure in this case is %0.2f kips",Stress_all*A) + //Part - B //Given L = 40 //ft - The length of the each rod -A = 46.7 //in2 - The length of the crossection +A = 46.7 //sq.in - The length of the crossection r_min = 4 //in - The radius of gyration stress_yp = 36 //ksi - the yielding point stress E = 29*(10**3) //ksi - The youngs modulus C_c = ((2*(%pi**2)*E/stress_yp)**0.5) //Slenderness ratio L/R -C_s = L*12/r_min // Slenderness ratio L/R of the present situation -//According to AISC formulas +C_s = L*12/r_min // Slenderness ratio L/R of the present situation +//According to AISC formulas +printf ("\n b)calculated Le/r ratio is %f",C_s) if C_s <C_c then - printf("b) The following approch is solvable") + printf("\n b)Since calculated Le/r ratio is less than Cc, we can apply the second ASD formula") else - printf("The caliculation is not possible") + printf("The calculation is not possible") end -F_S = 5.0/3 +3*C_s/(8*C_c) -(3*C_s**3)/(8*C_c**3) //Safety factor -Stress_all = (1 - (C_s**2)/(2*C_c**2))*stress_yp/F_S //The allowable strees +F_S = 5.0/3 +3*C_s/(8*C_c) -(C_s**3)/(8*C_c**3) //Safety factor +Stress_all = (1 - (C_s**2)/(2*C_c**2))*stress_yp/F_S //The allowable stress printf("\n b) The allowable stress in this case is %0.2f kips",Stress_all) +printf("\n b) The allowable pressure in this case is %0.2f kips",Stress_all*A) +printf("\n Similarly, for a column fixed at one end") +C_s = L*12/r_min*.8 // Slenderness ratio L/R of the present situation +//According to AISC formulas +printf ("\n b)calculated Le/r ratio is %f",C_s) +if C_s <C_c then + printf("\n b)Since calculated Le/r ratio is less than Cc, we can apply the second ASD formula") +else + printf("The calculation is not possible") +end +F_S = 5.0/3 +3*C_s/(8*C_c) -(C_s**3)/(8*C_c**3) //Safety factor +Stress_all = (1 - (C_s**2)/(2*C_c**2))*stress_yp/F_S //The allowable stress +printf("\n b) The allowable stress in this case is %0.2f kips",Stress_all) +printf("\n b) The allowable pressure in this case is %0.0f kips",Stress_all*A) diff --git a/3776/CH11/EX11.7/Ex11_7.sce b/3776/CH11/EX11.7/Ex11_7.sce index a265e2104..f933f32cd 100644 --- a/3776/CH11/EX11.7/Ex11_7.sce +++ b/3776/CH11/EX11.7/Ex11_7.sce @@ -1,20 +1,23 @@ clear //Given // -L = 15 //ft - The length of the each rod -p = 200 //kips The concentric load applied +L = 15 //ft - The length of the each rod +p = 200 //kips The concentric load applied r_min = 2.10 //in - The radius of gyration stress_yp = 50 //ksi - the yielding point stress E = 29*(10**3) //ksi - The youngs modulus C_c = ((2*(%pi**2)*E/stress_yp)**0.5) //Slenderness ratio L/R C_s = L*12/r_min //Slenderness ratio L/R present situation +printf("\n C_s = %0.2f ",C_s) +printf("\n C_c = %0.2f ",C_c) + if C_s <C_c then - printf("a)The following approch is solvable") + printf ("\nSince calculated C_s is less than Cc, we can apply the second ASD formula.") else - printf("The caliculation is not possible") + printf("The calculation is not possible") end -F_S = 5.0/3 +3*C_s/(8*C_c) -(3*C_s**3)/(8*C_c**3) //Safety factor -Stress_all = (1 - (C_s**2)/(2*C_c**2))*stress_yp/F_S //The allowable strees -a = p/Stress_all //in2 the alloawble area of the beam -printf("\n The allowable stress in this case is %0.2f kips",Stress_all) -printf("\n This stress requires %0.2f in2",a) +F_S = 5.0/3 +3*C_s/(8*C_c) -(C_s**3)/(8*C_c**3) //Safety factor +Stress_all = (1 - (C_s**2)/(2*C_c**2))*stress_yp/F_S //The allowable stress +a = p/Stress_all //sq.in the allowable area of the beam +printf("\n The allowable stress in this case is %0.0f kips",Stress_all) +printf("\n This stress requires %0.2f sq.in",a) diff --git a/3776/CH11/EX11.8/Ex11_8.sce b/3776/CH11/EX11.8/Ex11_8.sce index 84736d445..64d8b2256 100644 --- a/3776/CH11/EX11.8/Ex11_8.sce +++ b/3776/CH11/EX11.8/Ex11_8.sce @@ -2,18 +2,19 @@ clear //Given // L = 15.0 //ft - The length of the each rod -A = 46.7 //in2 - The length of the crossection +A = 46.7 //sq.in - The length of the crossection r_min = 4 //in - The radius of gyration stress_yp = 36.0 //ksi - the yielding point stress E = 29*(10**3) //ksi - The youngs modulus lamda = L*12*((stress_yp/E)**0.5)/(4*(%pi)) //column slenderness ratio if lamda<1.5 then - printf("The following approach is right") + printf("Since lamda<1.5 we can apply the AISC LFRD formula") else printf("The following approach is wrong") end -stress_cr = (0.658**(lamda**2))*stress_yp //ksi - The critical stress -P_n = stress_cr*A //kips //Nominal compressive strength +stress_cr = (0.658**(lamda**2))*stress_yp //ksi - The critical stress +P_n = stress_cr*A //kips //Nominal compressive strength o = 0.85 //Resistance factor -p_u = o*P_n //kips ,column design compressive strength -printf("\n column design compressive strength %0.3f kips",p_u) +p_u = o*P_n //kips ,column design compressive strength +printf("\ncolumn design compressive strength %0.3f kips",p_u) +// small variation due to rounding off errors diff --git a/3776/CH11/EX11.9/Ex11_9.sce b/3776/CH11/EX11.9/Ex11_9.sce index 98dcadcc9..ecc7bb597 100644 --- a/3776/CH11/EX11.9/Ex11_9.sce +++ b/3776/CH11/EX11.9/Ex11_9.sce @@ -1,16 +1,16 @@ clear -//Given -//FOR FLANGS -l = 5 //in - The length of the flang -b = 5 //in - Teh width of the flang -t = 0.312 //in - the thickness of the flang +//Given +//FOR FLANGE +l = 5 //in - The length of the flange +b = 5 //in - Teh width of the flange +t = 0.312 //in - the thickness of the flange L = 20 //in - Length of the beam, Extracted from AISC manuals -A = 4.563 //in2 - The area of crossection of the beam -r = 1.188 //in - radius of the gyration, Extracted from AISC manuals -//b/t- value of the flang -k = (5 -t)/(2*t) //b/t ratio +A = 4.563 //sq.in - The area of crossection of the beam +r = 1.188 //in - radius of the gyration, Extracted from AISC manuals +//b/t- value of the flange +k = (5 -t)/(2*t) //b/t ratio //AISC, lets check maximum allowable stress for slang -Stressf_all = 23.1 - 0.79*k //ksi The maximum allowable stress in case of flang,AISC +Stressf_all = 23.1 - 0.79*k //ksi The maximum allowable stress in case of flange,AISC //web width thickness ratio k_2 = (5 -2*t)/(t) @@ -20,17 +20,16 @@ if k_2<16 then //a) Overall buckling investment k_31 = L/r //slenderness ratio Stressb_all = 20.2 - 0.126*k_31//ksi The maximum allowable stress in case of Buckling,AISC -p_allow = A*Stressf_all //kips The allowable concentric load +p_allow = A*Stressf_all //kips The allowable concentric load //b) Overall buckling investmen -L_2 = 60 //in +L_2 = 60 //in k_3 = L_2/r //slenderness ratio Stressb_all_2 = 20.2 - 0.126*k_3//ksi The maximum allowable stress in case of Buckling,AISC -p_allow_2 = A*Stressb_all_2 //kips The allowable concentric load +p_allow_2 = A*Stressb_all_2 //kips The allowable concentric load printf("\n The maximum allowable stress in case of web width %0.2f ksi",Stressw_all) -printf("\n The maximum allowable stress in case of flang %0.2f ksi",Stressf_all) +printf("\n The maximum allowable stress in case of flange %0.2f ksi",Stressf_all) printf("\n a) The maximum allowable load in case of Buckling %0.2f kips",p_allow) printf("\n b) The maximum allowable load in case of Buckling %0.2f kips",p_allow_2) -printf("\n small variation due to rounding off errors") - +// small variation due to rounding off errors diff --git a/3776/CH12/EX12.10/Ex12_10.sce b/3776/CH12/EX12.10/Ex12_10.sce index 68fe4e541..31260aee0 100644 --- a/3776/CH12/EX12.10/Ex12_10.sce +++ b/3776/CH12/EX12.10/Ex12_10.sce @@ -1,7 +1,7 @@ clear //Given -A_1 = 0.125 //in2 , The area of the crossection of AB -A_2 = 0.219 //in2 , The area of the crossection of BC +A_1 = 0.125 //sq.in , The area of the crossection of AB +A_2 = 0.219 //sq.in , The area of the crossection of BC l_1 = 3*(5**0.5) //in , The length of AB l_2 = 6*(2**0.5) //in , The length of BC p = 3 //k , Force acting on the system diff --git a/3776/CH12/EX12.3/Ex12_3.sce b/3776/CH12/EX12.3/Ex12_3.sce index f99a26393..889a51fee 100644 --- a/3776/CH12/EX12.3/Ex12_3.sce +++ b/3776/CH12/EX12.3/Ex12_3.sce @@ -11,7 +11,7 @@ l_ab = 2.5 //mt - The length of the rod l_bc = 2 //mt - The length of the rod A_ab = 5*(10**-4) //mt2 the areaof ab A_bc = 5*(10**-3) //mt2 the areaof bc -E = 70 //Gpa The youngs modulus of the material +E = 70 //GPa The youngs modulus of the material e_a =(p_ab*l_ab*F_ab/(A_ab*E) + p_bc*l_bc*F_bc/(A_bc*E))*(10**-6) //KN-m //Part -2 due to flexure I = 60*10**6 //mm4 - the moment of inertia diff --git a/3776/CH12/EX12.5/Ex12_5.sce b/3776/CH12/EX12.5/Ex12_5.sce index c7bb91a72..cd3e6dbc7 100644 --- a/3776/CH12/EX12.5/Ex12_5.sce +++ b/3776/CH12/EX12.5/Ex12_5.sce @@ -7,8 +7,8 @@ F_ab = 2500 //lb F_bc = -2500 //lb l_ab = 60 //in - The length of the rod l_bc = 60 //in - The length of the rod -A_ab = 0.15 //in2 the areaof ab -A_bc = 0.25 //in2 the areaof bc +A_ab = 0.15 //sq.in the areaof ab +A_bc = 0.25 //sq.in the areaof bc E = 30*(10**6) //psi The youngs modulus of the material //Part_a e_a =p_ab*l_ab*F_ab/(A_ab*E) + p_bc*l_bc*F_bc/(A_bc*E) //lb-in the deflection @@ -16,7 +16,7 @@ e_a =p_ab*l_ab*F_ab/(A_ab*E) + p_bc*l_bc*F_bc/(A_bc*E) //lb-in the deflection p_bd = 1 //lb The recorded virtual loading F_bd = 1 //lb l_bd = 40 //in - The length of the rod -A_bd = 0.1 //in2 the areaof ab +A_bd = 0.1 //sq.in the areaof ab e_a_1 =p_ab*p_ab*l_ab/(A_ab*E) + p_bc*p_bc*l_bc/(A_bc*E) +p_bd*p_bd*l_bd/(A_bd*E) //lb-in the deflection //Since the produced defelection should compensate the other one x_d = e_a/e_a_1 diff --git a/3776/CH2/EX2.11/Ex2_11.sce b/3776/CH2/EX2.11/Ex2_11.sce index c4e15f631..4f9a017ec 100644 --- a/3776/CH2/EX2.11/Ex2_11.sce +++ b/3776/CH2/EX2.11/Ex2_11.sce @@ -1,21 +1,26 @@ clear -mass = 4 //kg +mass = 4 //kg dist = 1 //mt freely falling distance l = 1500 //mm length of rod d = 15 //mm diameter l_ab = 6.71 //inch l_bc = 8.29 //inch -E = 200 //GPA youngs modulus +E = 200 //GPA youngs modulus k = 4.5 // N/mm stiffness costant F = mass*9.81// The force applying -Area = 3.14*(d**2)/4 -// Two cases -//youngs modulus +Area = 3.14*(d**2)/4 +// Two cases +//youngs modulus e_y = F*l/(Area*E*(10**3)) // stiffness -e_f = F/k +e_f = F/k //total e = e_y +e_f k = 1+(2/(e*(10**-3))) stress_max_1 = F*(1+(k**0.5))/Area printf("\n The maximum stress is: %0.3f MPa",stress_max_1) + +//for the rod without washer +k2 = 1+(2/(e_y*(10**-3))) +stress_max_2 = F*(1+(k2**0.5))/Area +printf("\n The maximum stress for the rod without washer is: %0.3f MPa",stress_max_2) diff --git a/3776/CH2/EX2.12/Ex2_12.sce b/3776/CH2/EX2.12/Ex2_12.sce index b4a566a08..fb8e4ffb6 100644 --- a/3776/CH2/EX2.12/Ex2_12.sce +++ b/3776/CH2/EX2.12/Ex2_12.sce @@ -1,9 +1,10 @@ clear -flex_a = 1//f -flex_b = 2//f +flex_a = 1 //f +flex_b = 2 //f //removing lower support and solving FBD e = -2 -(2+1)//fp //e_1 = (2+1+1)*R //e_1 = -e Making the elongations zero since the both ends are fixed -R = e/(2+1+1.0) //P +R = -e/(2+1+1.0) //P +//since sum of forces are 0 printf("\n The reactions at bottom is %0.3f P",R) diff --git a/3776/CH2/EX2.19/Ex2_19.sce b/3776/CH2/EX2.19/Ex2_19.sce index 2f56f7092..cf1308495 100644 --- a/3776/CH2/EX2.19/Ex2_19.sce +++ b/3776/CH2/EX2.19/Ex2_19.sce @@ -3,8 +3,8 @@ clear l = 30 //in - The length of the rod p_1 = 80 //kips - The Force on the end p_2 = 125 //kips - The force on the other end -A_s = 0.5 //in2 - The crossection of the steel rod -A_a = 0.5 //in2 - The crossection of the aluminium +A_s = 0.5 //sq.in - The crossection of the steel rod +A_a = 0.5 //sq.in - The crossection of the aluminium E_a = 10*(10**6) //psi - The youngs modulus of the aluminium E_s = 30*(10**6) //psi - The youngs modulus of the steel //Internally stastically indeterminant diff --git a/3776/CH2/EX2.2/Ex2_2.sce b/3776/CH2/EX2.2/Ex2_2.sce index 9e6317279..88d9c4ad6 100644 --- a/3776/CH2/EX2.2/Ex2_2.sce +++ b/3776/CH2/EX2.2/Ex2_2.sce @@ -5,9 +5,9 @@ l_cd = 1500 //mm - length of rod cd p_ob = 100 //kN - Force in rods p_bc = -150 //KN p_cd = 50 //KN -A_ob = 1000 //mm2 - Area of rod ob -A_bc = 2000 //mm2 - Area of rod bc -A_cd = 1000 //mm2 - Area of rod cd +A_ob = 1000 //sq.mm - Area of rod ob +A_bc = 2000 //sq.mm - Area of rod bc +A_cd = 1000 //sq.mm - Area of rod cd E = 200.0 //GPA // the total deflection is algebraic sums of `deflection in each module e_1 = p_ob*l_ob/(A_ob*E) diff --git a/3776/CH2/EX2.4/Ex2_4.sce b/3776/CH2/EX2.4/Ex2_4.sce index 76ea23bb6..2a8f3fa43 100644 --- a/3776/CH2/EX2.4/Ex2_4.sce +++ b/3776/CH2/EX2.4/Ex2_4.sce @@ -1,21 +1,23 @@ clear -p_app = 3 //kips - applied force -P_A = 2.23 //kips +p_app = 3 //kips - applied force +P_A = 2.23 //kips p_B = -2.83 //kips - compressive force l_ab = 6.71 //inch l_bc = 8.29 //inch s_ab = 17.8 //ksi - tensile stress s_bc = -12.9 //ksi - compressive stress -E = 10.6 * (10**3) //ksi -youngs modulus +E = 10.6 * (10**3) //ksi -youngs modulus e_ab = s_ab*l_ab/E //elongation e_bc = s_bc*l_bc/E //contraction -x = -e_bc/e_ab //the Ratio of cosines of the deflected angles -// t_1 and t_2 be deflected angles +x = -e_bc/e_ab //the Ratio of cosines of the deflected angles +// t_1 and t_2 be deflected angles //t_2 = 180-45-26.6-t_1 the sum of angles is 360 //applying cos on both sides t_1=atand(1.29) e = e_ab/cosd((t_1)) //inch -k = p_app/e // kips/in vertical stiffness of the combination +e_t = e*cosd(11.2) +k = p_app/e_t // kips/in vertical stiffness of the combination printf("\n The vertical stiffness of the combination is %0.3f kips/inch",k) //answer in textbook is 167 +// answer varies due to rounding off errors diff --git a/3776/CH2/EX2.6/Ex2_6.sce b/3776/CH2/EX2.6/Ex2_6.sce index 844997aec..a5dcb9d0f 100644 --- a/3776/CH2/EX2.6/Ex2_6.sce +++ b/3776/CH2/EX2.6/Ex2_6.sce @@ -1,15 +1,15 @@ clear -dia = 50 //mm - diameter of aluminium -p = 100 // KN - instant force applid -dia_c = 0.1215 //mm- change in diameter +dia = 50 //mm - diameter of aluminium +p = 100 // KN - instant force applied +dia_c = 0.01215 //mm- change in diameter l_c = 0.219 //mm - change in length -l = 300 //mm - length -strain_dia = dia_c/dia // lateral strain -strain_l = l_c/l //longitudinal strain -po = strain_dia/strain_l // poission ratio -area = 3.14*dia*dia/4 //mm2 area -E = p*l/(area*l_c) //N/mm2 youngs modulus -printf("\n The lateral strain is: %0.3f no units",strain_dia) -printf("\n The longitudinal strain is: %0.3f no units",strain_l) -printf("\n The poissions ratio is: %0.3f no units",po) -printf("\n Youngs modulus: %0.2f N/sq.mm",E) +l = 300 //mm - length +strain_dia = -dia_c/dia // lateral strain +strain_l = -l_c/l //longitudinal strain +po = strain_dia/strain_l // Poisson ratio +area = 3.14*dia*dia/4 //sq.mm area +E = p*l/(area*l_c) //N/sq.mm youngs modulus +printf("\n The lateral strain is: %e mm/mm",strain_dia) +printf("\n The longitudinal strain is: %e no units",strain_l) +printf("\n The Poissons ratio is: %0.3f no units",po) +printf("\n Youngs modulus: %0.2f GPa",E) diff --git a/3776/CH2/EX2.7/Ex2_7.sce b/3776/CH2/EX2.7/Ex2_7.sce index c452c6448..0bfdaefa3 100644 --- a/3776/CH2/EX2.7/Ex2_7.sce +++ b/3776/CH2/EX2.7/Ex2_7.sce @@ -1,15 +1,16 @@ clear T = 12.9*(10**-6) ///F -t = 100.00 // F +t = 100.00 // F l_ab = 6.71 //inch l_bc = 8.49 //inch -e_ab = T*t*l_ab //in-elongation +e_ab = T*t*l_ab //in-elongation e_bc = T*t*l_bc //in-elongation -k = e_ab/e_bc // ratio of cosines of deflected angles -// t_1 and t_2 be deflected angles -//t_2 = 180-45-26.6-t_1 the sum of angles is 360 +k = e_ab/e_bc // ratio of cosines of deflected angles +// t_1 and t_2 be deflected angles +//t_2 = 45+26.6-t_1 the sum of angles is 360 //applying cos on both sides t_1 = atand(0.5) // e = e_bc/cosd(t_1) printf("\n The displacement in point B is : %e in",e ) +printf("\n It forms an angle of %f degrees with vertical",45-t_1 ) diff --git a/3776/CH3/EX3.2/Ex3_2.sce b/3776/CH3/EX3.2/Ex3_2.sce index 8a3a5388c..759864b7f 100644 --- a/3776/CH3/EX3.2/Ex3_2.sce +++ b/3776/CH3/EX3.2/Ex3_2.sce @@ -1,14 +1,14 @@ clear //Given a = 50 //mm - length of a cube -E = 200 // Gpa - the youngs modulus -v = 0.25 // no units- poissions ratio +E = 200 // GPa - the youngs modulus +v = 0.25 // no units- Poissons ratio pressure = 200 // MPa - pressure acting on all sides //pressure is a compressive stress -S_x = -200 // Gpa - The stress in X direction -S_y = -200 // Gpa - The stress in Y direction -S_z = -200 // Gpa - The stress in Z direction -//Caliculations +S_x = -200 // GPa - The stress in X direction +S_y = -200 // GPa - The stress in Y direction +S_z = -200 // GPa - The stress in Z direction +//calculations e = S_x*(10**-3)/E - v*S_y*(10**-3)/E-v*S_z*(10**-3)/E//mm - considering all three directions x = e*a //mmThe change in the dimension between parallel faces diff --git a/3776/CH3/EX3.3/Ex3_3.sce b/3776/CH3/EX3.3/Ex3_3.sce index 655ef3251..a63e8aa67 100644 --- a/3776/CH3/EX3.3/Ex3_3.sce +++ b/3776/CH3/EX3.3/Ex3_3.sce @@ -1,16 +1,16 @@ clear //Given -R = 1000 // mm - radius of the cylinder +R = 1000 // mm - radius of the cylinder t = 10 //mm - thickness of the cylinder -p_in = 0.80 //Mpa- Internal pressure -E = 200 //Mpa- youngs modulus -v = 0.25 // poission ratio -//caliculations +p_in = 0.80 //Mpa- Internal pressure +E = 200 //Mpa- youngs modulus +v = 0.25 // Poisson ratio +//calculations -Stress_1 = p_in*R/t //MPa -Hoop stress //From derived expressions -Stress_2 = p_in*R/(2*t) //Mpa- Longitudinal stress +Stress_1 = p_in*R/t //MPa -Hoop stress //From derived expressions +Stress_2 = p_in*R/(2*t) //Mpa- Longitudinal stress e = Stress_1*(10**-3)/E-v*Stress_2*(10**-3)/E -dia_change = e*R //mm- The change in daimeter of the cylinder -printf("\n The Hoop stress is %0.3f mm",Stress_1) -printf("\n The longitudinal stress is %0.3f",Stress_2) +dia_change = e*R //mm- The change in daimeter of the cylinder +printf("\n The Hoop stress is %0.3f MPa",Stress_1) +printf("\n The longitudinal stress is %0.3f MPa",Stress_2) printf("\n The change in daimeter of the cylinder is %0.3f mm",dia_change) diff --git a/3776/CH3/EX3.4/Ex3_4.sce b/3776/CH3/EX3.4/Ex3_4.sce index 550b6a893..f0d1c0d96 100644 --- a/3776/CH3/EX3.4/Ex3_4.sce +++ b/3776/CH3/EX3.4/Ex3_4.sce @@ -3,10 +3,10 @@ clear R = 1000 //mm - radius of the cylinder th = 10 //mm - thickness of the cylinder E = 200 //Mpa- youngs modulus -v = 0.25 // poission ratio +v = 0.25 // Poisson ratio p_in = 0.80 //Mpa- Internal pressure t = 10 //mm - thickness of the cylinder -//caliculations +//calculations Stress_1 = p_in*R/(2*t) //MPa -Hoop stress //From derived expressions Stress_2 = p_in*R/(2*t) //Mpa- Longitudinal stress diff --git a/3776/CH3/EX3.5/Ex3_5.sce b/3776/CH3/EX3.5/Ex3_5.sce index 4fe17f46f..78714b79b 100644 --- a/3776/CH3/EX3.5/Ex3_5.sce +++ b/3776/CH3/EX3.5/Ex3_5.sce @@ -1,15 +1,15 @@ clear //Given -p_in = 0.7 //MPa - internal pressure -n_bolts = 20 // number of bolts -dia = 650 //mm - bolt circle diameter -stress_allow = 125 //mm Maximum alowable stress +p_in = 0.7 //MPa - internal pressure +n_bolts = 20 // number of bolts +dia = 650 //mm - bolt circle diameter +stress_allow = 125 //MPa Maximum alowable stress Stress_conc = 2 //stress concentration -d = 25 //mm -//caliculations +d = 25 //mm +//calculations F = p_in*3.14*(((dia-2*d)/2)**2)*(10**6) //N F_each = F/n_bolts //N- force per each Bolt -A = Stress_conc*F_each/(stress_allow*(10**6)) //mm2 The bolt area +A = Stress_conc*F_each/(stress_allow*(10**6)) //sq.mm The bolt area Bolt_dia = 2*((A/3.14)**0.5) //mm the bolt daimeter printf("\n The diameter of each bolt is %0.1f mm",Bolt_dia) diff --git a/3776/CH4/EX4.16/Ex4_16.sce b/3776/CH4/EX4.16/Ex4_16.sce index d62fbcd29..0d9997618 100644 --- a/3776/CH4/EX4.16/Ex4_16.sce +++ b/3776/CH4/EX4.16/Ex4_16.sce @@ -1,12 +1,13 @@ clear -//Given +//Given dia_out = 10 //mm- outer diameter of shaft -dia_in = 8 //mm- inner diameter of shaft -c_out = dia_out/2 //mm - outer Radius of shaft -c_in = dia_in/2 //mm - inner radius of shaft -T = 40 //N/mm -Torque in the shaft -//caliculations - +dia_in = 8 //mm- inner diameter of shaft +c_out = dia_out/2 //mm - outer Radius of shaft +c_in = dia_in/2 //mm - inner radius of shaft +T = 40 //N/mm -Torque in the shaft +//calculations +dia_mean = (dia_out+dia_in)/2 +dia_diff = dia_out-dia_in J = 3.14*((dia_out**4)- (dia_in**4))/32 //mm4 shear_T_max = T*c_out*(10**3)/J // The maximum torsion shear in the shaft shear_T_min = T*c_in*(10**3)/J // The maximum torsion shear in the shaft diff --git a/3776/CH4/EX4.3/Ex4_3.sce b/3776/CH4/EX4.3/Ex4_3.sce index 7ad10bbe6..86dcaed1c 100644 --- a/3776/CH4/EX4.3/Ex4_3.sce +++ b/3776/CH4/EX4.3/Ex4_3.sce @@ -1,14 +1,15 @@ clear -//Given +//Given dia_out = 20 //mm- outer diameter of shaft -dia_in = 16 //mm- inner diameter of shaft -c_out = dia_out/2 //mm - outer Radius of shaft -c_in = dia_in/2 //mm - inner radius of shaft -T = 40 //N/mm -Torque in the shaft -//caliculations +dia_in = 16 //mm- inner diameter of shaft +c_out = dia_out/2 //mm - outer Radius of shaft +c_in = dia_in/2 //mm - inner radius of shaft +T = 40 //N/mm -Torque in the shaft +//calculations J = 3.14*((dia_out**4)- (dia_in**4))/32 //mm4 shear_T_max = T*c_out*(10**3)/J // The maximum torsion shear in the shaft shear_T_min = T*c_in*(10**3)/J // The maximum torsion shear in the shaft -printf("\n The maximum shear due to torsion is %0.2f MPa",shear_T_max) +printf("\n The maximum shear due to torsion is %e MPa",shear_T_max) + //answer in textbook is wrong printf("\n The minimum shear due to torsion is %0.0f MPa",shear_T_min) diff --git a/3776/CH4/EX4.4/Ex4_4.sce b/3776/CH4/EX4.4/Ex4_4.sce index 105784f04..d6f4a901b 100644 --- a/3776/CH4/EX4.4/Ex4_4.sce +++ b/3776/CH4/EX4.4/Ex4_4.sce @@ -1,16 +1,15 @@ clear //Given -hp = 10 // horse power of motor -f = 30 // given -shear_T = 55 //MPa - The maximum shearing in the shaft -//caliculations +hp = 10 // horse power of motor +f = 30 // given +shear_T = 55 //MPa - The maximum shear in the shaft +//calculations -T = 119*hp/f // N.m The torsion in the shaft +T = 119*hp/f // N.m The torsion in the shaft //j/c=T/shear_T=K k = T*(10**3)/shear_T //mm3 //c3=2K/3.14 -c = ((2*k/3)**0.33) //mm - The radius of the shaft -diamter = 2*c //mm - The diameter of the shaft -printf("\n The Diameter of the shaft used is %0.2f mm",diamter) +c = ((2*k/3)**0.33) //mm - The radius of the shaft +diameter = 2*c //mm - The diameter of the shaft +printf("\n The Diameter of the shaft used is %0.2f mm",diameter) printf("\n For practical purposes, a 16-mm shaft would probably be selected") - diff --git a/3776/CH4/EX4.5/Ex4_5.sce b/3776/CH4/EX4.5/Ex4_5.sce index 8d033dc9a..edd0d1fe6 100644 --- a/3776/CH4/EX4.5/Ex4_5.sce +++ b/3776/CH4/EX4.5/Ex4_5.sce @@ -1,23 +1,24 @@ clear -//Given +//Given hp = 200 //Horse power stress_sh = 10000 //psi- shear stress -rpm_1 = 20.0 // The rpm at which this shaft1 operates +rpm_1 = 20.0 // The rpm at which this shaft1 operates rpm_2 = 20000.0 // The rpm at which this shaft2 operates T_1= hp*63000.0/rpm_1 //in-lb Torsion due to rpm1 T_2= hp*63000/rpm_2 //in-lb Torsion due to rpm1 -//caliculations +//calculations //j/c=T/shear_T=K -k_1= T_1/stress_sh //mm3 +k_1= T_1/stress_sh //cu.in //c3=2K/3.14 -c_1= ((2*k_1/3)**0.33) //mm - The radius of the shaft -diamter_1 = 2*c_1 //mm - The diameter of the shaft -printf("\n The Diameter of the shaft1 is %0.2f mm",diamter_1) +//c_1= ((2*k_1/3)**0.33) //mm - The radius of the shaft +diamter_1 = (16*k_1/%pi)**(1/3) //mm - The diameter of the shaft +printf("\n The Diameter of the shaft1 is %0.2f in",diamter_1) //j/c=T/shear_T=K k_2= T_2/stress_sh //mm3 //c3=2K/3.14 -c_2= ((2*k_2/3)**0.33) //mm - The radius of the shaft -diamter_2 = 2*c_2 //mm - The diameter of the shaft -printf("\n The Diameter of the shaft2 is %0.3f mm",diamter_2) +//c_2= ((2*k_2/3)**0.33) //mm - The radius of the shaft +diamter_2 = (16*k_2/%pi)**(1/3) //mm - The diameter of the shaft + +printf("\n The Diameter of the shaft2 is %0.3f in",diamter_2) diff --git a/3776/CH4/EX4.7/Ex4_7.sce b/3776/CH4/EX4.7/Ex4_7.sce index 6b437ff50..2aa479f09 100644 --- a/3776/CH4/EX4.7/Ex4_7.sce +++ b/3776/CH4/EX4.7/Ex4_7.sce @@ -10,8 +10,8 @@ l_cd = 300 //mm - length of cd l_de = 500.0 //mm - length of de d_1 = 25 //mm - outer diameter d_2 = 50 //mm - inner diameter -G = 80 //Gpa -shear modulus -//Caliculations +G = 80 //GPa -shear modulus +//calculations J_ab = 3.14*(d_1**4)/32 //mm4 J_bc = 3.14*(d_1**4)/32 //mm4 diff --git a/3776/CH4/EX4.9/Ex4_9.sce b/3776/CH4/EX4.9/Ex4_9.sce index 5260a4497..31b57e941 100644 --- a/3776/CH4/EX4.9/Ex4_9.sce +++ b/3776/CH4/EX4.9/Ex4_9.sce @@ -1,30 +1,30 @@ clear -//given -//its a statistally indeterminant -//we will take of one of the support -//Given -T_ab = 0 //N.m - torsion in AB +//given +//its a statistally indeterminant +//we will take of one of the support +//Given +T_ab = 0 //N.m - torsion in AB T_bc = 150 //N.m - torsion in BC T_cd = 150 //N.m - torsion in CD T_de = 1150 //N.m - torsion in DE l_ab = 250 //mm - length of AB l_bc = 200 //mm - length of BC -l_cd = 300 //mm - length of cd +l_cd = 300 //mm - length of cd l_de = 500.0//mm - length of de -d_1 = 25 //mm - outer diameter +d_1 = 25 //mm - outer diameter d_2 = 50 //mm - inner diameter -//Caliculations +//calculations J_ab = 3.14*(d_1**4)/32 //mm4 J_bc = 3.14*(d_1**4)/32 //mm4 J_cd = 3.14*(d_2**4 - d_1**4)/32 //mm4 J_de = 3.14*(d_2**4 - d_1**4)/32 //mm4 -G = 80 //Gpa -shear modulus -rad = T_ab*l_ab/(J_ab*G)+ T_bc*l_bc/(J_bc*G)+ T_cd*l_cd/(J_cd*G)+ T_de*l_de/(J_de*G) +G = 80 //GPa -shear modulus +rad = T_ab*l_ab/(J_ab*G)+ T_bc*l_bc/(J_bc*G)+ T_cd*l_cd/(J_cd*G)+ T_de*l_de/(J_de*G) //now lets consider T_A then the torsion is only T_A // T_A*(l_ab/(J_ab*G)+ l_bc/(J_bc*G)+ l_cd/(J_cd*G)+ l_de/(J_de*G)) +rad = 0 -// since there will be no displacement -T_A =-rad/(l_ab/(J_ab*G)+ l_bc/(J_bc*G)+ l_cd/(J_cd*G)+ l_de/(J_de*G)) //Torsion at A +// since there will be no displacement +T_A =rad/(l_ab/(J_ab*G)+ l_bc/(J_bc*G)+ l_cd/(J_cd*G)+ l_de/(J_de*G)) //Torsion at A T_B = 1150 - T_A //n-m F_X = 0 torsion at B -printf("\n The Torsion at rigid end A is %0.2f N-m",T_A) -printf("\n The Torsion at rigid end B is %0.2f N-m",T_B) +printf("\n The Torsion at rigid end A is %0.0f N-m",T_A) +printf("\n The Torsion at rigid end B is %d N-m",T_B) diff --git a/3776/CH5/EX5.1/Ex5_1.sce b/3776/CH5/EX5.1/Ex5_1.sce index f223bc45a..7d38e393a 100644 --- a/3776/CH5/EX5.1/Ex5_1.sce +++ b/3776/CH5/EX5.1/Ex5_1.sce @@ -7,7 +7,7 @@ R_1 = 100 //N - The Force acting l_2 = 0.2 //mt -R_1 acting point the distance from 'a' R_2 = 160 //N The Force acting l_3 = 0.3 //mt -R_2 acting point the distance from 'a' -//caliculations +//calculations //F_X = 0 forces in x directions R_A_X = 0 // since there are no forces in X-direction diff --git a/3776/CH5/EX5.2/Ex5_2.sce b/3776/CH5/EX5.2/Ex5_2.sce index 53aac0bc3..d285b7978 100644 --- a/3776/CH5/EX5.2/Ex5_2.sce +++ b/3776/CH5/EX5.2/Ex5_2.sce @@ -2,8 +2,8 @@ clear //Given P_Max = 10 //N - the maximum distribution in a triangular distribution L = 3 //mt the total length of force distribution -L_X = 5 //mt - the horizantal length of the rod -//caliculations +L_X = 5 //mt - the horizontal length of the rod +//calculations F_y = P_Max*L*0.5 //N - The force due to triangular distribition L_com = 2*L /3 //mt - the resultant force acting as a result of distribution acting position diff --git a/3776/CH5/EX5.3/Ex5_3.sce b/3776/CH5/EX5.3/Ex5_3.sce index 44cb6144a..46e8c4366 100644 --- a/3776/CH5/EX5.3/Ex5_3.sce +++ b/3776/CH5/EX5.3/Ex5_3.sce @@ -2,26 +2,26 @@ clear //given F = 5 //K - force acting on the system tan1 = (4/3) // the Tan of the angle of force with x axis -l_ab = 12 //inch - the total length of ab +l_ab = 12 //inch - the total length of ab l = 3 // inch - Distance from 'a' -//caliculation +//calculation F_X = 4 //K F_Y = 3 //k //M_A = 0 momentum at point a is zero -// F_X*l- R_B_Y*l_ab = 0 +// F_X*l- R_B_Y*l_ab = 0 R_B_Y = F_X*l/l_ab //M_B= 0 momentum at point b is zero // R_A_Y*l_ab - F_X*(l_ab - l) R_A_Y = F_X*(l_ab - l)/l_ab - + //F_X = 0 forces in x directions -R_A_X = F_Y + R_B_Y +R_A_X = F_Y + R_B_Y R_B_X = R_B_Y // since the angle is 45 (180/%pi)* -//resultants -R_A = (R_A_X**2 + R_A_Y**2**0.5) -R_B = (R_B_X**2 + R_B_Y**2**0.5) +//resultants +R_A = (R_A_X**2 + R_A_Y**2)**0.5 +R_B = (R_B_X**2 + R_B_Y**2)**0.5 printf("The X,Y components and resultant of reaction force at A is %0.3f, %0.3f,%0.3f N",R_A_X,R_A_Y,R_A) printf("\nThe X,Y components and resultant of reaction force at B is %0.3f, %0.3f,%0.3f N",R_B_X,R_B_Y,R_B) diff --git a/3776/CH5/EX5.4/Ex5_4.sce b/3776/CH5/EX5.4/Ex5_4.sce index 2a3e57db7..2a3235419 100644 --- a/3776/CH5/EX5.4/Ex5_4.sce +++ b/3776/CH5/EX5.4/Ex5_4.sce @@ -2,8 +2,8 @@ clear //Given P_Max = 10 //N - the maximum distribution in a triangular distribution L = 3 //mt the total length of force distribution -L_X = 5 //mt - the horizantal length of the rod -//caliculations +L_X = 5 //mt - the horizontal length of the rod +//calculations F_y = P_Max*L*0.5 //N - The force due to triangular distribition L_com = 2*L /3 //mt - the resultant force acting as a result of distribution acting position diff --git a/3776/CH5/EX5.9/Ex5_9.sce b/3776/CH5/EX5.9/Ex5_9.sce index 245adc229..87a133ed4 100644 --- a/3776/CH5/EX5.9/Ex5_9.sce +++ b/3776/CH5/EX5.9/Ex5_9.sce @@ -2,8 +2,8 @@ clear //Given P_Max = 10 //N - the maximum distribution in a triangular distribution L = 3 //mt the total length of force distribution -L_X = 5 //mt - the horizantal length of the rod -//caliculations +L_X = 5 //mt - the horizontal length of the rod +//calculations F_y = P_Max*L*0.5 //N - The force due to triangular distribition L_com = 2*L /3 //mt - the resultant force acting as a result of distribution acting position 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 diff --git a/3776/CH7/EX7.1/Ex7_1.sce b/3776/CH7/EX7.1/Ex7_1.sce index 22b3a758c..4e1d1ef6f 100644 --- a/3776/CH7/EX7.1/Ex7_1.sce +++ b/3776/CH7/EX7.1/Ex7_1.sce @@ -7,14 +7,14 @@ l_1 = 50.0 //mm l_2 = 200.0 //mm b_1 = 200.0 //mm b_2 = 50.0 //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 = 25.0 //mm com distance -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 = 150.0 //in com distance y_net = (A_1*y_1 +A_2*y_2)/(A_1+A_2) //mm - The com of the whole system c_max = (4-y_net) //mm - The maximum distace from com to end c_min = y_net //mm - 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_net = I_1 + I_2 //mm4 - the total moment of inertia Q = A_1*(-y_1+y_net) //mm3 diff --git a/3776/CH7/EX7.2/Ex7_2.sce b/3776/CH7/EX7.2/Ex7_2.sce index 71c8e8755..9f6d25ae4 100644 --- a/3776/CH7/EX7.2/Ex7_2.sce +++ b/3776/CH7/EX7.2/Ex7_2.sce @@ -6,15 +6,15 @@ R_a = l*p/2 //KN -The reaction at a, Since the system is symmetry R_b = l*p/2 //KN -The reaction at b l_s = 10 //mm - The length of the screw shear_al = 2 //KN - The maximum load the screw can take -I = 2.36*(10**9) //mm2 The moment of inertia of the whole system +I = 2.36*(10**9) //sq.mm The moment of inertia of the whole system //We will divide this into two parts l_1 = 50.0 //mm l_2 = 50.0 //mm b_1 = 100.0 //mm b_2 = 200.0 //mm -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 = 200.0 //mm com distance -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 = 225.0 //in com distance Q = 2*A_1*y_1 + A_2*y_2 // mm**3 For the whole system q = R_a*Q*(10**3)/I //N/mm The shear flow diff --git a/3776/CH7/EX7.6/Ex7_6.sce b/3776/CH7/EX7.6/Ex7_6.sce index 89fca203f..7253fc27d 100644 --- a/3776/CH7/EX7.6/Ex7_6.sce +++ b/3776/CH7/EX7.6/Ex7_6.sce @@ -4,7 +4,7 @@ clear l = 10.0 // in - The height t = 0.1 // in - The width b = 5.0 //mm- The width of the above part -A = t* b //in2 - area of part +A = t* b //sq.in - area of part y_net = l/2 // The com of the system y_1 = l // The position of teh com of part_2 I_1 = t*(l**3)/12 //in^4 The moment of inertia of part 1 diff --git a/3776/CH7/EX7.8/Ex7_8.sce b/3776/CH7/EX7.8/Ex7_8.sce index 67646ba58..f39579fa7 100644 --- a/3776/CH7/EX7.8/Ex7_8.sce +++ b/3776/CH7/EX7.8/Ex7_8.sce @@ -2,7 +2,7 @@ clear //Given dia = 10.0 //mm - The diameter of the cylinder c = dia/2 //mm - the radius of the cylinder -A = 3.14*(c**2) //mm2 The area of the crossection +A = 3.14*(c**2) //sq.mm The area of the crossection y = 4*c/(3*3.14) //mm The com of cylinder I = 3.14*(c**4)/4 //mm4 - The moment of inertia of the cylinder j = 3.14*(dia**4)/32 //mm4 diff --git a/3776/CH7/EX7.9/Ex7_9.sce b/3776/CH7/EX7.9/Ex7_9.sce index 4822c44e0..2f8b342e6 100644 --- a/3776/CH7/EX7.9/Ex7_9.sce +++ b/3776/CH7/EX7.9/Ex7_9.sce @@ -2,8 +2,8 @@ clear //Given dia = 15 //mm - The diameter of the rod h = 0.5 //mt - The freely falling height -A = 3.14*(dia**2)/4 //mm2 The area of the crossection -E = 200 //Gpa -Youngs modulus +A = 3.14*(dia**2)/4 //sq.mm The area of the crossection +E = 200 //GPa -Youngs modulus L = 750 //mm - The total length of the rod G = 80 //gpa - Shear modulus N = 10 //number of live coils diff --git a/3776/CH8/EX8.1/Ex8_1.sce b/3776/CH8/EX8.1/Ex8_1.sce index 2510958ab..2fd755047 100644 --- a/3776/CH8/EX8.1/Ex8_1.sce +++ b/3776/CH8/EX8.1/Ex8_1.sce @@ -3,9 +3,9 @@ clear // // o = 22.5 //degrees , The angle of infetisimal wedge -A = 1 //mm2 The area of the element -A_ab = 1*(cos((%pi/180)*(o))) //mm2 - The area corresponds to AB -A_bc = 1*(sin((%pi/180)*(o))) //mm2 - The area corresponds to BC +A = 1 //sq.mm The area of the element +A_ab = 1*(cos((%pi/180)*(o))) //sq.mm - The area corresponds to AB +A_bc = 1*(sin((%pi/180)*(o))) //sq.mm - The area corresponds to BC S_1 = 3 //MN The stresses applying on the element S_2 = 2 //MN S_3 = 2 //MN diff --git a/3776/CH8/EX8.3/Ex8_3.sce b/3776/CH8/EX8.3/Ex8_3.sce index 6172f655f..9bda8dc65 100644 --- a/3776/CH8/EX8.3/Ex8_3.sce +++ b/3776/CH8/EX8.3/Ex8_3.sce @@ -1,16 +1,15 @@ clear -//Given +//Given // // S_x = -2 //MPa _ the noraml stress in x direction S_y = 4 //MPa _ the noraml stress in Y direction -c = (S_x + S_y)/2 //MPa - The centre of the mohr circle -point_x = -2 //The x coordinate of a point on mohr circle +c = (S_x + S_y)/2 //MPa - The centre of the mohr circle +point_x = 3 //The x coordinate of a point on mohr circle point_y = 4 //The y coordinate of a point on mohr circle -Radius = ((point_x-c)**2 + point_y**2**0.5) // The radius of the mohr circle +Radius = ((point_x)**2 + point_y**2)**0.5 // The radius of the mohr circle S_1 = Radius +1//MPa The principle stress S_2 = -Radius +1 //MPa The principle stress S_xy_max = Radius //MPa The maximum shear stress -printf("\n The principle stresses are %0.3f MPa %0.3f MPa",S_1,S_2) +printf("\n The principle stresses are %0.3f MPa, %0.3f MPa",S_1,S_2) printf("\n The maximum shear stress %0.3f MPa",S_xy_max) -printf("\n The maximum tensile stress which is the result of all stresses must act as shown in the figure") diff --git a/3776/CH8/EX8.4/Ex8_4.sce b/3776/CH8/EX8.4/Ex8_4.sce index 75a4498ff..fb3b4c6ff 100644 --- a/3776/CH8/EX8.4/Ex8_4.sce +++ b/3776/CH8/EX8.4/Ex8_4.sce @@ -3,18 +3,17 @@ clear // S_x = 3.0 //MPa _ the noraml stress in x direction S_y = 1.0 //MPa _ the noraml stress in Y direction -c = (S_x + S_y)/2 //MPa - The centre of the mohr circle +c = (S_x + S_y)/2 //MPa - The centre of the mohr circle point_x = 1 //The x coordinate of a point on mohr circle point_y = 3 //The y coordinate of a point on mohr circle -//Caliculations +//calculations -Radius = ((point_x-c)**2 + point_y**2**0.5) // The radius of the mohr circle -//22.5 degrees line is drawn -o = 22.5 //degrees -a = 71.5 - 2*o //Degrees, from diagram -stress_n = c + Radius*sin((180/%pi)*(o)) //MPa The normal stress on the plane -stress_t = Radius*cos((180/%pi)*(o)) //MPa The tangential stress on the plane +Radius = ((point_x)**2 + point_y**2)**0.5 // The radius of the mohr circle +//22.5 degrees line is drawn +o = 22.5 //degrees +a = 71.57 - 2*o //Degrees, from diagram +stress_n = c + Radius*sin((180/%pi)*(o)) //MPa The normal stress on the plane +ang = sind((-a)) +stress_t = Radius*ang //MPa The tangential stress on the plane printf("\n The normal stress on the 22 1/2 plane %0.2f MPa",stress_n) printf("\n The tangential stress on the 22 1/2 plane %0.2f MPa",stress_t) -printf("\n answer varies due to rounding off errors") - diff --git a/3776/CH8/EX8.7/Ex8_7.sce b/3776/CH8/EX8.7/Ex8_7.sce index a497ec56a..8013fda4e 100644 --- a/3776/CH8/EX8.7/Ex8_7.sce +++ b/3776/CH8/EX8.7/Ex8_7.sce @@ -3,7 +3,7 @@ clear e_x = -500 //10-6 m/m The contraction in X direction e_y = 300 //10-6 m/m The contraction in Y direction e_xy = -600 //10-6 m/m discorted angle -centre = (e_x + e_y)/2 //10-6 m/m +centre = (e_x + e_y)/2 //10-6 m/m point_x = -500 //The x coordinate of a point on mohr circle point_y = 300 //The y coordinate of a point on mohr circle Radius = 500 //10-6 m/m - from mohr circle @@ -11,5 +11,5 @@ e_1 = Radius +centre //MPa The principal strain e_2 = -Radius +centre //MPa The principal strain k = atan(300.0/900) // from geometry k_1 = (180/%pi)*(k) -printf("\n The principal strains are %0.3f um/m %0.3f um/m",e_1,e_2) +printf("\n The principal strains are %0.3f micro m/m %0.3f micro m/m",e_1,e_2) printf("\n The angle of principal plane %0.2f degrees",k_1) diff --git a/3776/CH8/EX8.8/Ex8_8.sce b/3776/CH8/EX8.8/Ex8_8.sce index 0c8fd0488..57bd71918 100644 --- a/3776/CH8/EX8.8/Ex8_8.sce +++ b/3776/CH8/EX8.8/Ex8_8.sce @@ -1,24 +1,25 @@ clear //Given -e_0 = -500 //10-6 m/m -e_45 = 200 //10-6 m/m +e_0 = -500 //10-6 m/m +e_45 = 200 //10-6 m/m e_90 = 300 //10-6 m/m -E = 200 //Gpa - youngs modulus of steel -v = 0.3 // poissions ratio -//Caliculations +E = 200 //GPa - youngs modulus of steel +v = 0.3 // Poissons ratio +//calculations e_xy = 2*e_45 - (e_0 +e_90 ) //10-6 m/m from equation 8-40 in text // from example 8.7 e_x = -500 //10-6 m/m The contraction in X direction e_y = 300 //10-6 m/m The contraction in Y direction e_xy = -600 //10-6 m/m discorted angle -centre = (e_x + e_y)/2 //10-6 m/m +centre = (e_x + e_y)/2 //10-6 m/m point_x = -500 //The x coordinate of a point on mohr circle point_y = 300 //The y coordinate of a point on mohr circle Radius = 500 //10-6 m/m - from mohr circle e_1 = Radius +centre //MPa The principle strain e_2 = -Radius +centre //MPa The principle strain -stress_1 = E*(10**-3)*(e_1+v*e_2)/(1-v**2) //MPa the stress in this direction -stress_2 = E*(10**-3)*(e_2+v*e_1)/(1-v**2) //MPa the stress in this direction +stress_1 = E*(10**-3)*(e_1+v*e_2)/(1-v**2) //MPa the stress in this direction +stress_2 = E*(10**-3)*(e_2+v*e_1)/(1-v**2) //MPa the stress in this direction printf("\n The principle stresses are %0.2f MPa %0.2f MPa",stress_1,stress_2) +// answer in textbook is wrong diff --git a/3776/CH9/EX9.4/Ex9_4.sce b/3776/CH9/EX9.4/Ex9_4.sce index f645233f8..1b4944c45 100644 --- a/3776/CH9/EX9.4/Ex9_4.sce +++ b/3776/CH9/EX9.4/Ex9_4.sce @@ -1,13 +1,13 @@ clear -//Given +//Given // b = 40.0 //mm - The width of the beam crossection -h = 300.0 //mm - The length of the beam crossection +h = 300.0 //mm - The length of the beam crossection V = 40.0 //KN - The shear stress in teh crossection -M = 10.0 //KN-m - The bending moment on K----K crossection +M = 10.0 //KN-m - The bending moment on K----K crossection c = h/2 //mm -The position at which maximum stress occurs on the crossection -I = b*(h**3)/12 //mmm4 - the moment of inertia -//Caliculations +I = b*(h**3)/12 //mmm4 - the moment of inertia +//calculations stress_max_1 = M*c*(10**6)/I //The maximum stress occurs at the end stress_max_2 = -M*c*(10**6)/I //The maximum stress occurs at the end @@ -16,13 +16,13 @@ n = y/(c) // The ratio of the distances from nuetral axis t stress_L_1 = n*stress_max_1 //The normal stress on elements L--L stress_L_2 = -n*stress_max_1 //The normal stress on elements L--L x = 10 //mm The length of the element -A = b*x //mm3 The area of the element +A = b*x //mm3 The area of the element y_1 = y+x/2 // the com of element from com of whole system -stress_xy = V*A*y_1*(10**3)/(I*b) //MPa - The shear stress on the element -//stresses acting in plane 30 degrees +stress_xy = V*A*y_1*(10**3)/(I*b) //MPa - The shear stress on the element +//stresses acting in plane 30 degrees o = 60 //degrees - the plane angle stress_theta = stress_L_1/2 + stress_L_1*(cos((%pi/180)*(o)))/2 - stress_xy*(sin((%pi/180)*(o))) //MPa by direct application of equations stress_shear = -stress_L_1*(sin((%pi/180)*(o)))/2 - stress_xy*(cos((%pi/180)*(o))) //MPa Shear stress - + printf("\n a)The principle stresses are %0.2f MPa %0.2f MPa",stress_max_1,stress_max_2) -printf("\n b)The stresses on inclines plane %0.2f MPa noraml, %0.2f MPa shear ",stress_theta,stress_shear) +printf("\n b)The stresses on inclined plane %0.2f MPa normal, %0.2f MPa shear ",stress_theta,stress_shear) diff --git a/3776/CH9/EX9.5/Ex9_5.sce b/3776/CH9/EX9.5/Ex9_5.sce index ee3f65e4e..432d3df4b 100644 --- a/3776/CH9/EX9.5/Ex9_5.sce +++ b/3776/CH9/EX9.5/Ex9_5.sce @@ -1,23 +1,24 @@ clear //Given M = 10 //KN-m moment -v = 8.0 //KN - shear Stress +v = 8.0 //KN - shear Stress stress_allow = 8 //MPa - The maximum allowable stress shear_allow_per = 1.4 //MPa - The allowable stress perpendicular to grain stress_allow_shear = 0.7 //MPa - The maximum allowable shear stress -//Caliculations +//calculations -S = M*(10**6)/stress_allow //mm3 +S = M*(10**6)/stress_allow //mm3 //lets arbitarly assume h = 2b //S = b*(h**2)/6 -h = (12*S**0.333) //The depth of the beam +h = (1.25*(10**6)*12)**(1/3) //The depth of the beam b = h/2 //mm The width of the beam -A = h*b //mm2 The area of the crossection , assumption -stress_shear = 3*v*(10**3)/(2*A) //MPa The strear stress +A = 140*240 //sq.mm The area of the crossection , assumption +stress_shear = 3*v*(10**3)/(2*A) //MPa The strear stress if stress_shear<stress_allow_shear then - printf("The stress developed %0.2f is in allowable ranges for %0.2f mm2 area",stress_shear,A) + printf("The stress developed %0.2f is in allowable ranges for %0.2f sq.mm area",stress_shear,A) else printf("The stress developed %0.3f is in non allowable ranges %0.3f area",stress_shear,A) end Area_allow = v*(10**3)/shear_allow_per //mm - the allowable area -printf("\n The minimum area is %0.3f mm2",Area_allow ) +printf("\n The minimum area is %0.3f sq.mm",Area_allow ) +//answer varies due to rounding off errors diff --git a/3776/CH9/EX9.6/Ex9_6.sce b/3776/CH9/EX9.6/Ex9_6.sce index a8a78e3ed..20e5c1797 100644 --- a/3776/CH9/EX9.6/Ex9_6.sce +++ b/3776/CH9/EX9.6/Ex9_6.sce @@ -9,11 +9,11 @@ A = l*w R_A = 6.4 //k - The reaction at A R_B = 25.6 //k - the reaction at B v_max = R_B-l*w //kips the maximum stress, from diagram -//W8x24 is used from the appendix table 3 and 4 +//W8x24 is used from the appendix table 3 and 4 l =0.245 //in - W8x24 crossesction length -//Caliculations +//calculations -stress_xy = v_max/A //ksi the approximate shear stress +stress_xy = v_max/A //ksi the approximate shear stress if stress_xy < stress_allow_shear then printf("W8x24 gives the allowable ranges of shear stress") else: @@ -24,7 +24,7 @@ k = 7.0/8 //in the distance from the outer face of the flange to the webfillet //a1t+2kt should not exceed 0.75 of yeild stress Stress_yp = 36 //ksi - The yeild stress t = 0.245 //in thickness of the web -//support a +//support a a = R_A/(0.75*Stress_yp*t)-k //in lengths of the bearings //support b a_1 = R_B/(0.75*Stress_yp*t)-2*k //in lengths of the bearings diff --git a/3776/CH9/EX9.8/Ex9_8.sce b/3776/CH9/EX9.8/Ex9_8.sce index b986095ec..f7b27f51e 100644 --- a/3776/CH9/EX9.8/Ex9_8.sce +++ b/3776/CH9/EX9.8/Ex9_8.sce @@ -4,8 +4,8 @@ hp = 63000 //horse power T = hp*20*(10**-3)/63 //k-in the torsion implies due to horse power stress_allow_shear = 6 //ksi- The maximum allowable shear stress M_ver = 6.72/2 //k-in the vertical component of the moment -M_hor = 9.10 //k-in the horizantal component of the moment -//Caliculations +M_hor = 9.10 //k-in the horizontal component of the moment +//calculations M = (((M_ver**2)+(M_hor**2))**0.5) //K-in The resultant d = ((16*(((M**2)+(T**2))**0.5)/(stress_allow_shear*3.14))**0.333) //in** The suggested diameter from derivation |