   # Determine the reactions and draw the shear and bending moment diagrams for the beams shown in Figs. PD.1 through PD.8 using the three-moment equation.

#### Solutions

Chapter
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Chapter D, Problem 7P
Textbook Problem
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## Determine the reactions and draw the shear and bending moment diagrams for the beams shown in Figs. PD.1 through PD.8 using the three-moment equation. To determine

Calculate the reactions for the given beam.

Sketch the shear and bending moment diagrams for the given beam.

### Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions using the three equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero (Fx=0), consider the forces acting towards right side as positive (+) and the forces acting towards left side as negative ().
• For summation of forces along y-direction is equal to zero (Fy=0), consider the upward force as positive (+) and the downward force as negative ().
• For summation of moment about a point is equal to zero (Matapoint=0), consider the clockwise moment as negative and the counter clockwise moment as positive.

Calculation:

Find the degree of indeterminacy of the structure:

Degree of determinacy of the beam is equal to the number of unknown reactions minus the number of equilibrium equations.

The beam is supported by 5 support reactions and the number of equilibrium equations is 3.

Therefore, the degree of indeterminacy of the beam is i=2.

Select the bending moment MB and MC at the interior supports B and C as redundant.

Consider the supports A, B, and C as l, c, and r respectively.

Express the general three-moment equation as shown below:

MlLlIl+2Mc(LlIl+LrIr)+MrLrIr=[PlLl2klIl(1kl2)PrLr2krIr(1kr2)wlLl34IlwrLr34Ir6E(ΔlΔcLl+ΔrΔcLr)]        (1)

Here, Mc is the bending moment at support c, Ml,Mr are the bending moments at the adjacent supports to the left and to the right of c, E is the modulus of elasticity, Ll,Lr are the lengths of the spans to the left and to the right of c, Il,Ir are the moments of inertia of the spans to the left and to the right of c, Pl,Pr are the concentrated loads acting on the left and the right spans, kl or kr is the ratio of distance of Pl or Pr from the left or right support to the span length, wl,wr are the uniformly distributed loads to the left and right spans, Δc is the settlement of support c, Δl,Δr is the settlement of adjacent supports to the left and to the right of c,

Apply three-moment equation at joint B,

The moment at A is MA=0.

Substitute 24 ft for Ll, 24 ft for Lr I for Il, 2I for Ir, 0 for Pl, 0 for Pr, 0 for kl, 0 for kr, 2.5k/ft for wl, 2.5k/ft for wr, and 0 for Δc,Δl,Δr in Equation (1).

0+2MB(24I+242I)+MC(242I)=[2.5(243)4I2.5(243)8I]72MBI+12MCI=12,960I72MB+12MC=12,960        (2)

Apply three-moment equation at joint C,

The moment at E is ME=0.

Substitute 24 ft for Ll, 24 ft for Lr 2I for Il, I for Ir, 0 for Pl, 40 k for Pr, 0 for kl, 0.5 for kr, 2.5k/ft for wl, 0 for wr, and 0 for Δc,Δl,Δr in Equation (1).

MB(242I)+2MC(242I+24I)=[2.5(24)38I40(24)2(0.5)I(10.52)]12MBI+72MCI=12,960I12MB+72MC=12,960        (3)

Solve Equation (2) and Equation (3).

MB=154.3k-ftMC=154.3k-ft

Sketch the span end moments and shears for span AB, BC, and CDE as shown in Figure 1.

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