Skip to main content

Moment distribution: frame, 150 kN, 60 kN/m

Analyse the frame shown in figure below by using the moment distribution method. [Figure: frame; column AB (2I) of 10 m (8 m + 2 m marks), A fixed; 150 kN horizontal at B; beam B-C-D with 50 kN loads at 2 m from B and at 4 m, spans 2 m + 4 m + 2 m; BC (2I) and C-D (I) of 5 m with 60 kN/m UDL, D roller; column CE (I) of 6 m, E fixed.]

Assumptions (figure partly illegible): column AB (2I) is 10 m high, fixed at A; the girder is B-C-D with BC = 4 m (2I) carrying a 50 kN load at mid-span (2 m from B), CD = 5 m (I) carrying 60 kN/m, and D on a roller; column CE (I, 6 m) runs down from C to its fixed base E. A second 50 kN load acts vertically at C (it passes through the axis of column CE and produces no bending moment). The 150 kN load at B acts to the right. The roller at D allows horizontal movement, so the frame sways; the moment distribution is done for the non-sway case and the sway case.

Distribution factors (k=EI/Lk=EI/L; the hinged-end member CD gets the factor 3/4)

  • Joint B: relative stiffness Σk\Sigma k = 0.7000; BA: k=EI/Lk=EI/L = 0.2000, DF = 0.2857; BC: k=EI/Lk=EI/L = 0.5000, DF = 0.7143
  • Joint C: relative stiffness Σk\Sigma k = 0.8167; CB: k=EI/Lk=EI/L = 0.5000, DF = 0.6122; CD: k=EI/Lk=EI/L = 0.1500 (far end hinged: 3/4 factor), DF = 0.1837; CE: k=EI/Lk=EI/L = 0.1667, DF = 0.2041

Case I: sway prevented by an imaginary support at B

Fixed-end moments: BC ∓50×48=∓25\mp\dfrac{50\times4}{8}=\mp25 kNm; CD ∓60×5212=∓125\mp\dfrac{60\times5^2}{12}=\mp125 kNm (modified at C for the hinge at D: -187.50 kNm).

ABBABCCBCDCEEC
DF-0.2860.7140.6120.1840.204-
FEM0.000.00-25.0025.00-187.500.000.00
Balance0.007.1417.8699.4929.8533.160.00
Carry-over3.570.0049.748.930.000.0016.58
Balance0.00-14.21-35.53-5.47-1.64-1.820.00
Carry-over-7.110.00-2.73-17.770.000.00-0.91
Balance0.000.781.9510.883.263.630.00
Carry-over0.390.005.440.980.000.001.81
Balance0.00-1.55-3.88-0.60-0.18-0.200.00
Carry-over-0.780.00-0.30-1.940.000.00-0.10
Further cycles (converged)-0.05-0.090.391.140.380.420.21
Final M-3.97-7.947.94120.64-155.8335.1917.59

The horizontal reaction of the imaginary support (positive to the right) is R1=−157.61R_1=-157.61 kN; this is the force that the imaginary support must supply to prevent sway and that the sway correction has to remove.

Case II: sway Δ\Delta with no loads

Give the girder a sway to the right such that the fixed-end moment in AB is −100-100 kNm at each end. Since FEM∝EI/L2FEM\propto EI/L^2 and both columns move equally, column CE (EI, 6 m) gets −100×(1/36)(2/100)=−138.89-100\times\dfrac{(1/36)}{(2/100)}=-138.89 kNm.

ABBABCCBCDCEEC
DF-0.2860.7140.6120.1840.204-
FEM-100.00-100.000.000.000.00-138.89-138.89
Balance0.0028.5771.4385.0325.5128.340.00
Carry-over14.290.0042.5235.710.000.0014.17
Balance0.00-12.15-30.37-21.87-6.56-7.290.00
Carry-over-6.070.00-10.93-15.180.000.00-3.64
Balance0.003.127.819.302.793.100.00
Carry-over1.560.004.653.900.000.001.55
Balance0.00-1.33-3.32-2.39-0.72-0.800.00
Carry-over-0.660.00-1.20-1.660.000.00-0.40
Further cycles (converged)0.110.220.971.120.250.280.14
Final M-90.78-81.5681.5693.9721.28-115.25-127.07

The horizontal reaction of the imaginary support is R2=57.62R_2=57.62 kN.

Combination

R1+kR2=0⇒k=−−157.6157.62=2.7353R_1+kR_2=0\Rightarrow k=-\dfrac{-157.61}{57.62}=2.7353

EndCase Ik × Case IIFinal M (kNm)
AB-3.97-248.31-252.28
BA-7.94-223.09-231.03
BC7.94223.09231.03
CB120.64257.04377.68
CD-155.8358.20-97.63
CE35.19-315.23-280.05
EC17.59-347.57-329.97

Checks: Joint B: 0.000.00; Joint C: 0.000.00; hinge D: MDC=0M_{DC}=0.

MemberBM at first endBM at second endMax within member
AB-252.28231.03-
BC231.03-377.68-
CD-97.630.00141.86 at 2.83 m
CE-280.05329.97-

Answer: MAB=−252.28M_{AB}=-252.28, MBA=−231.03M_{BA}=-231.03, MBC=231.03M_{BC}=231.03, MCB=377.68M_{CB}=377.68, MCD=−97.63M_{CD}=-97.63, MCE=−280.05M_{CE}=-280.05, MEC=−329.97M_{EC}=-329.97 kNm (clockwise end moments +).

✓ Answer checked by Claude Opus 5.5 (high effort)

All Theory of Structures II chapter 3 questions →