Analyze the structure calculate the horizontal displacement of joint B.
Explanation of Solution
Given information:
The Young’s modulus E is constant and equals to
Calculation:
Sketch the free body diagram of frame as shown in Figure 1.
Find the stiffness of each member connected to joint B as follows;
Find the total stiffness at joint B as follows;
Find the stiffness of each member connected to joint C as follows;
Find the total stiffness at joint C as follows;
Find the distribution factors at joint B using the relation;
Find the distribution factors at joint C using the relation;
Refer the appendix Table A.4 to find the fixed end moments.
Find the fixed end moment at each end of the member BC as follows;
Joint | A | B | C | D | ||
Member | AB | BA | BC | CB | CD | DC |
DF | 0.43 | 0.57 | 0.55 | 0.45 | ||
FEM | -32 | 32 | ||||
Balancing | 13.76 | 18.24 | -17.6 | -14.4 | ||
CO | -8.8 | 9.12 | -7.2 | |||
Balancing | 3.78 | 5.02 | -5.02 | -4.1 | ||
CO | -2.51 | 2.51 | -2.1 | |||
Balancing | 1.08 | 1.43 | -1.38 | -1.13 | ||
CO | -0.69 | 0.72 | -0.57 | |||
Balancing | 0.3 | 0.39 | -0.4 | -0.32 | ||
CO | -0.2 | 0.2 | -0.16 | |||
Balancing | 0.09 | 0.11 | -0.11 | -0.09 | ||
19 | -19 | 20.06 | -20.06 | -10.03 |
Sketch the free body diagram as shown in Figure 2.
Show the members AB, BC, and CD as in Figure 3.
Taking moment about B ,
Summation of vertical force is equal to zero.
Taking moment about A ,
Taking moment about D,
Summation of horizontal force is equal to zero.
Find the fixed end moment at each end of the member AB as follows;
Consider the
Find the fixed end moment at each end of the member CD as follows;
Joint | A | B | C | D | ||
Member | AB | BA | BC | CB | CD | DC |
DF | 0.43 | 0.57 | 0.55 | 0.45 | ||
FEM | -86.8 | -86.8 | -48.22 | -48.22 | ||
Balancing | 86.8 | 37.32 | 49.48 | 26.52 | 21.7 | |
CO | 43.4 | 13.26 | 24.74 | 10.85 | ||
Balancing | -24.36 | -32.3 | -13.61 | -11.13 | ||
CO | -6.81 | -16.15 | -5.57 | |||
Balancing | 2.93 | 3.88 | 8.88 | 7.27 | ||
CO | 4.44 | 1.94 | 3.64 | |||
Balancing | -1.91 | -2.53 | -1.07 | -0.87 | ||
CO | -0.54 | -1.27 | -0.44 | |||
Balancing | 0.23 | 0.31 | 0.7 | 0.57 | ||
CO | 0.35 | 0.16 | 0.29 | |||
Balancing | -0.15 | -0.2 | -0.09 | -0.06 | ||
CO | -0.05 | -0.1 | -0.03 | |||
Balancing | 0.02 | 0.03 | 0.06 | 0.04 | ||
0 | -29.32 | 29.32 | 30.75 | -30.75 | -39.98 |
Show the member AB,BC, CD, as shown in figure 4.
Consider span AB:
Take moment about B is Equal to zero.
Summation of forces along x-direction is Equal to zero.
Consider span BC:
Take moment about B is Equal to zero.
Summation of forces along y-direction is Equal to zero.
Consider span CD.
Take moment about C is Equal to zero.
Summation of forces along x-direction is Equal to zero.
Consider the entire structure.
Summation of forces along x-direction is Equal to zero.
Take moment about A is Equal to zero.
Summation of forces along y-direction is Equal to zero.
Sketch the free body diagram of shear and moment as shown in Figure 5.
Sketch the deflected shape as shown in Figure 6.
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