OIQ Exam Question 2

Question from OIQ Technical Exam, obviously meant to be solved using moment distribution, but here we see how easy it is using slope deflection instead. This version users a newer version of 'sdutil' that computes end-shears as well as moments.



In [1]:

    
from IPython import display
display.SVG('oiq-frame-1.svg')









    Out[1]:

Solution 1

Actually, solution 1 does not use any special library modules at all (except for 'sympy').



In [2]:

    
from sympy import *



In [3]:

    
var('EI ta tb tc td')
Mab = (EI/4)*(4*ta + 2*tb) - 24*4**2/12
Mba = (EI/4)*(2*ta + 4*tb) + 24*4**2/12
Mbc = (EI/2/4)*(4*tb + 2*tc)
Mcb = (EI/2/4)*(2*tb + 4*tc)
Mcd = (EI/4)*(4*tc + 2*td) - 72*4/8
Mdc = (EI/4)*(2*tc + 4*td) + 72*4/8



In [4]:

    
eqns = [ta, Mba+Mbc, Mcb+Mcd, Mdc]
soln = solve( eqns, [ta, tb, tc, td] )
soln









    Out[4]:





{ta: 0.0,
 tb: -29.5172413793103/EI,
 tc: 49.1034482758621/EI,
 td: -60.551724137931/EI}



In [5]:

    
[m.subs(soln).n(4) for m in [Mab,Mba,Mbc,Mcb,Mcd,Mdc]]









    Out[5]:





[-46.76, 2.483, -2.483, 17.17, -17.17, 0]



In [6]:

    
Rd = (Mcd + 72*2)/4
Ra = (-(Mab + Mba) + 24*4*2)/4
Rc = 24*4 + 72 - Ra - Rd
Hd = (Mbc + 72*2 - Rd*4)/4
[r.subs(soln).n(4) for r in [Ra,Rc,Rd,Hd]]









    Out[6]:





[59.07, 77.22, 31.71, 3.672]

Solution 2

Here it is again, using the SD utilities:



In [7]:

    
from sympy import *
init_printing(use_latex='mathjax')
from sdutil2 import SD,FEF



In [8]:

    
var('EI theta_a theta_b theta_c theta_d')
M_ab, M_ba, V_ab, V_ba = SD(4,EI,theta_a,theta_b) + FEF.udl(4,24)
M_bc, M_cb, V_bc, V_cb = SD(4,EI/2,theta_b,theta_c)
M_cd, M_dc, V_cd, V_dc = SD(4,EI,theta_c,theta_d) + FEF.p(4,72,2)



In [9]:

    
eqns = [theta_a, M_ba+M_bc, M_cb+M_cd, M_dc]
soln = solve( eqns, [theta_a, theta_b, theta_c, theta_d] )
soln









    Out[9]:





$$\left \{ \theta_{a} : 0.0, \quad \theta_{b} : - \frac{29.5172413793103}{EI}, \quad \theta_{c} : \frac{49.1034482758621}{EI}, \quad \theta_{d} : - \frac{60.551724137931}{EI}\right \}$$

End moments and shears:



In [10]:

    
[m.subs(soln).n(4) for m in [M_ab,M_ba,M_bc,M_cb,M_cd,M_dc]]









    Out[10]:





$$\left [ -46.76, \quad 2.483, \quad -2.483, \quad 17.17, \quad -17.17, \quad 0\right ]$$



In [11]:

    
[m.subs(soln).n(4) for m in [V_ab,V_ba,V_bc,V_cb,V_cd,V_dc]]









    Out[11]:





$$\left [ 59.07, \quad -36.93, \quad -3.672, \quad -3.672, \quad 40.29, \quad -31.71\right ]$$

Reactions:



In [12]:

    
R_d = -V_dc
R_c = V_cd - V_ba
R_a = V_ab
H_d = -V_cb
H_a = V_cb
M_a = M_ab
[r.subs(soln).n(4) for r in [R_a,R_c,R_d,H_d,H_a,M_a]]









    Out[12]:





$$\left [ 59.07, \quad 77.22, \quad 31.71, \quad 3.672, \quad -3.672, \quad -46.76\right ]$$

Equilibrium check



In [13]:

    
## sum vertical forces
(R_a + R_c + R_d - 24*4 - 72).subs(soln)









    Out[13]:





$$0$$



In [14]:

    
## sum horizontal forces
(H_a+H_d).subs(soln)









    Out[14]:





$$0$$



In [15]:

    
## sum moments about a
(M_a + 24*4*2 - R_c*4 + 72*6 - R_d*8 - H_d*4).subs(soln)









    Out[15]:





$$0$$



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