Problem 827 | Continuous Beam by Three-Moment Equation | Strength of Materials Review 11/30/16, 8)35 AM
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Problem 827 | Continuous Beam by Three-Moment Equation
Problem 827
See Figure P-827.
Solution 827
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Apply three-moment equation to first and middle spans
6A1 ā1 6A2 b̄2
M1 L1 + 2M2 (L1 + L2 ) + M3 L2 + + =0
L1 L2
Where,
M1 = 0
L1 = L2 = 12 ft
6A1 ¯ 1 3 3 2
= o = (150)( ) = 34 560 lb ⋅
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Problem 827 | Continuous Beam by Three-Moment Equation | Strength of Materials Review 11/30/16, 8)35 AM
6A1 ā1 8 8
= w L3
60 o
= (150)(12 3 ) = 34 560 lb ⋅ ft2
L1 60
6A2 b̄2 Pb 400(9) 300(4)
= Σ (L 2 − b2 ) = (12 2 − 92 ) + (12 2 − 42 )
L2 L 12 12
6A2 b̄2
= 31 700 lb ⋅ ft2
L2
Thus,
0 + 2M2 (12 + 12) + M3 (12) + 34 560 + 31 700 = 0
48M2 + 12M3 = −66 260 ← equation (1)
Apply three-moment equation to middle and last spans
6A2 ā2 6A3 b̄3
M2 L2 + 2M3 (L2 + L3 ) + M4 L3 + + =0
L2 L3
Where,
L2 = L3 = 12 ft
L3 = 3 ft
M4 = −4(50)(2) = −400 lb ⋅ ft
6A2 ā2 Pb 400(3) 300(8)
= Σ (L 2 − a2 ) = (12 2 − 32 ) + (12 2 − 82 )
L2 L 12 12
6A2 ā2
= 29 500 lb ⋅ ft2
L2
6A3 b̄3 1 1
= w L3
4 o
= (50)(12 3 ) = 21 600 lb ⋅ ft2
L3 4
Thus,
M2 (12) + 2M3 (12 + 12) − 400(12) + 29 500 + 21 600 = 0
12M2 + 48M3 = −46 300 ← equation (2)
Solving equations (1) and (2) simultaneously
M2 = −1215.22 lb ⋅ ft answer
M3 = −660.78 lb ⋅ ft answer
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Problem 827 | Continuous Beam by Three-Moment Equation | Strength of Materials Review 11/30/16, 8)35 AM
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continuous beam triangular load uniformly varying load rectangular load uniformly distributed
load concentrated load moment over the support
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Strength of Materials
Chapter 01 - Simple Stresses
Chapter 02 - Strain
Chapter 03 - Torsion
Chapter 04 - Shear and Moment in Beams
Chapter 05 - Stresses in Beams
Chapter 06 - Beam Deflections
Chapter 07 - Restrained Beams
Chapter 08 - Continuous Beams
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Problem 827 | Continuous Beam by Three-Moment Equation | Strength of Materials Review 11/30/16, 8)35 AM
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