#Given import math b = 40.0 #mm - The width 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 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 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 y = 140 #mm The point of interest, the distance of element from com n = y/(c) # The ratio of the distances from nuetral axis to the elements 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 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 o = 60 #degrees - the plane angle stress_theta = stress_L_1/2 + stress_L_1*(math.cos(math.radians(o)))/2 - stress_xy*(math.sin(math.radians(o))) #Mpa by direct application of equations stress_shear = -stress_L_1*(math.sin(math.radians(o)))/2 - stress_xy*(math.cos(math.radians(o))) #Mpa Shear stress print "a)The principle stresses are ",round(stress_max_1,2),"MPa,",round(stress_max_2,2),"Mpa" print "b)The stresses on inclines plane ",round(stress_theta,2),"Mpa noraml, ",round(stress_shear,2),"Mpa shear "
a)The principle stresses are 16.67 MPa, -16.67 Mpa b)The stresses on inclines plane 11.11 Mpa noraml, -7.06 Mpa shear
#Given M = 10 #KN-m moment 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 S = M*(10**6)/stress_allow #mm3 #lets arbitarly assume h = 2b #S = b*(h**2)/6 h = pow(12*S,0.333) #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 if stress_shear<stress_allow_shear: print "The stress developed ",round(stress_shear,2)," is in allowable ranges for ",round(A,2),"mm2 area" else: print "The stress developed",stress_shear," is in non allowable ranges",A,"area" Area_allow = v*(10**3)/shear_allow_per #mm - the allowable area print "The minimum area is ",Area_allow ,"mm2"
The stress developed 0.4 is in allowable ranges for 30077.85 mm2 area The minimum area is 5714.28571429 mm2
#Given stress_allow = 24 #ksi - The maximum allowable stress stress_allow_shear = 14.5 #ksi- The maximum allowable shear stress M_max = 36 #k-ft The maximum moment l = 16 #in-The length of the rod w = 2 #k/ft - The force distribution on the rod 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 l =0.245 #in - W8x24 crossesction length #Caliculations stress_xy = v_max/A #ksi the approximate shear stress if stress_xy < stress_allow_shear: print "W8x24 gives the allowable ranges of shear stress" else: print "W8x24 doesnot gives the allowable ranges of shear stress" k = 7.0/8 #in the distance from the outer face of the flange to the webfillet #at+kt should not exceed 0.75 of yeild stress #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 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 print "lengths of the bearing at A ",round(a,3),"in" print "lengths of the bearing at B",round(a_1,3),"in"
W8x24 gives the allowable ranges of shear stress lengths of the bearing at A 0.092 in lengths of the bearing at B 2.12 in
#given 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 = pow(((M_ver**2)+(M_hor**2)),0.5) #K-in The resultant d = pow((16*(((M**2)+(T**2))**0.5)/(stress_allow_shear*3.14)),0.333) #in, The suggested diameter from derivation print "The suggested diameter is",round(d,2),"in"
The suggested diameter is 2.66 in