Explanation Given: The flexural rigidity, E I of the beam is constant. Concept Used: Write the expression for the fixed end moments in a beam with uniformly distributed load. F E M = w l 2 12 .......(I). Here, the load on the beam is w and the length of the beam is l . Write the expression for the fixed end moments in a beam with a point load at mid-span. F E M = w l 8 .......(II) Write the expression for load matrix. Q = k D .......(III) Here, the load matrix is Q , the stiffness is k and the displacement is D . Calculations: The beam with complete reaction at the nodes is shown below. Figure (1) Consider the loading at node 1 and 2 as shown below. Figure (2) Calculate the actual moment on node 1 . Substitute 2 k / ft for w and 12 ft for l in Equation (I). M 1 = 2 k / ft × ( 1 2 ft ) 2 12 = 24 k-ft Calculate the moment at node 2 . Substitute 2 k / ft for w and 12 ft for l in Equation (I). − M 2 = 2 k / ft × ( 1 2 ft ) 2 12 M 2 = − 24 k-ft Write the force matrix from the known force. Q = [ 24 − 24 0 0 − 8 ] 1 2 3 4 5 Write the displacement matrix for the known displacement The nodes 6 , 7 and 8 are constrained, thus the displacement is zero. D = [ 0 0 0 ] 6 7 8 Calculate the member stiffness matrix for member 1. k 1 = E I [ 12 L 3 6 L 2 − 12 L 3 6 L 2 6 L 2 4 L − 6 L 2 2 L − 12 L 3 − 6 L 2 12 L 3 − 6 L 2 6 L 2 2 L − 6 L 2 4 L ] 8 1 7 2 .......(IV) Substitute 12 ft for L in Equation (IV). k 1 = E I [ 12 12 3 6 12 2 − 12 12 3 6 12 2 6 12 2 4 12 − 6 12 2 2 12 − 12 12 3 − 6 12 2 12 12 3 − 6 12 2 6 12 2 2 12 − 6 12 2 4 12 ] 8 1 7 2 = E I [ 0.00695 0.04167 − 0.00695 0.04167 0.04167 0.333 − 0.04167 0.167 − 0.00695 − 0.04167 0.00695 − 0.04167 0.04167 0.167 − 0.04167 0.333 ] 8 1 7 2 Calculate the member stiffness matrix for member 2. k 2 = E I [ 12 L 3 6 L 2 − 12 L 3 6 L 2 6 L 2 4 L − 6 L 2 2 L − 12 L 3 − 6 L 2 12 L 3 − 6 L 2 6 L 2 2 L − 6 L 2 4 L ] 7 2 6 3 .......(V) Substitute 12 ft for L in matrix (V). k 2 = E I [ 12 12 3 6 12 2 − 12 12 3 6 12 2 6 12 2 4 12 − 6 12 2 2 12 − 12 12 3 − 6 12 2 12 12 3 − 6 12 2 6 12 2 2 12 − 6 12 2 4 12 ] 7 2 6 3 = E I [ 0.00695 0.04167 − 0.00695 0.04167 0.04167 0.333 − 0.04167 0.167 − 0.00695 − 0.04167 0.00695 − 0.04167 0.04167 0.167 − 0.04167 0.333 ] 7 2 6 3 Calculate the member stiffness matrix for member 3. k 3 = E I [ 12 L 3 6 L 2 − 12 L 3 6 L 2 6 L 2 4 L − 6 L 2 2 L − 12 L 3 − 6 L 2 12 L 3 − 6 L 2 6 L 2 2 L − 6 L 2 4 L ] 6 3 5 4 .......(VI) Substitute 8 ft for L in Equation (V). k 3 = E I [ 12 8 3 6 8 2 − 12 8 3 6 8 2 6 8 2 4 8 − 6 8 2 2 8 − 12 8 3 − 6 8 2 12 8 3 − 6 8 2 6 8 2 2 8 − 6 8 2 8 ] 6 3 5 4 = E I [ 0.023 0.094 − 0.023 0.094 0.094 0.5 − 0.094 0.25 − 0.023 − 0.094 0.023 − 0.094 0.094 0.25 − 0.094 0.5 ] 6 3 5 4 Calculate the stiffness matrix for the beam. K = k 1 + k 2 + k 3 .......(VII) Substitute E I [ 0.00695 0.04167 − 0.00695 0.04167 0.04167 0.333 − 0.04167 0.167 − 0.00695 − 0.04167 0.00695 − 0.04167 0.04167 0

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ISBN: 9780134713649

Author: HIBBELER

Publisher: PEARSON

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Equilibrium and Support Reactions

Civil engineers deal with structures that are composed of several members connected or with individual members such as beams and columns. To analyze such structures, the condition of statics must be applied. Under the static condition, the members are co…

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Chapter 15, Problem 15.4P

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The reactions at the supports. 1, 2 and 3.

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