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

 

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

 

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

 

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

 

The Moment Distribution Method

Propped Beam Reactions by 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

 

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

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

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

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

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

 

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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