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Continuous Beams

Problem 877 | Continuous Beam by Moment Distribution Method

Problem 877
By means of moment-distribution method, solve the moment at R2 and R3 of the continuous beam shown in Fig. P-815.
 

815-continuous-beam-triangular-concentrated-loads.gif

 

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Problem 872 | Continuous Beam with Spring End-Support

Problem 872
Repeat Problem 871 assuming the loadings on the spans are interchanged.
 

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Problem 871 | Continuous Beam with Spring End-Support

Problem 871
The continuous beam in Figure P-871 is supported at its left end by a spring whose constant is 300 lb/in. For the beam, E = 1.5 × 106 psi and I = 115.2 in.4. Compute the load on the spring and its deflection.
 

871-continuous-beam-spring-support.gif

 

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Problem 869 | Deflection by Three-Moment Equation

Problem 869
Find the value of EIδ at the center of the first span of the continuous beam in Figure P-869 if it is known that M2 = -980 lb·ft and M3 = -1082 lb·ft.
 

869-continuous-beam.gif

 

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Problem 448 - Beam of Three Segments Joined by Internal Hinges

Problem 448
A beam carrying the loads shown in Figure P-448 is composed of three segments. It is supported by four vertical reactions and joined by two frictionless hinges. Determine the values of the reactions.
 

448-beam-with-internal-hinges.gif

 

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The Moment Distribution Method

Moment distribution is based on the method of successive approximation developed by Hardy Cross (1885–1959) in his stay at the University of Illinois at Urbana-Champaign (UIUC). This method is applicable to all types of rigid frame analysis.
 

008-carry-over-moment.gif

 

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Problem 855 | Continuous Beams with Fixed Ends

Problem 855
If the distributed load in Prob. 854 is replaced by a concentrated load P at midspan, determine the moments over the supports.
 

855-i-span.gif

 

Answers:
$M_1 = \dfrac{PL}{8} \cdot \dfrac{1}{\alpha + 2} = M_4$

$M_2 = -\dfrac{PL}{8} \cdot \dfrac{2}{\alpha + 2} = M_3$
 

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Problem 854 | Continuous Beams with Fixed Ends

Problem 854
Solve for the moment over the supports in the beam loaded as shown in Fig. P-854.
 

854-i-span.gif

 

Answers:
$M_1 = \dfrac{w_o L^2}{12} \cdot \dfrac{1}{\alpha + 2} = M_4$

$M_2 = -\dfrac{w_o L^2}{12} \cdot \dfrac{2}{\alpha + 2} = M_3$
 

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Problem 853 | Continuous Beams with Fixed Ends

Problem 853
For the continuous beam shown in Fig. P-853, determine the moment over the supports. Also draw the shear diagram and compute the maximum positive bending moment. (Hint: Take advantage of symmetry.)
 

853-shear-diagram.gif

 

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Problem 852 | Continuous Beams with Fixed Ends

Problem 852
Find the moments over the supports for the continuous beam in Figure P-852. Use the results of Problems 850 and 851.
 

852-fixed-ended-continuous-beam.gif

 

Answers
$M_1 = -146.43 ~ \text{N}\cdot\text{m}$

$M_2 = -307.14 ~ \text{N}\cdot\text{m}$

$M_3 = -521.43 ~ \text{N}\cdot\text{m}$
 

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