Question

1) BEAM ANALYSIS (20pts) Part III Analysis (45pts) The cantilever beam (W140*60) BC is attached to the steel cable AB as shown. Knowing the cable is initially taut, determine the tension in the cable caused by the distributed load shown and the reactions at the right end. A -A = 255 mm$^2$ E = 200 \times 10$^9$ Pa I = 216 \times 10$^6$ mm$^4$ = 216 \times 10$^{-6}$ m$^4$ 3 m EI = (200 \times 10$^9$)(216 \times 10$^{-6}$) = 43.2 \times 10$^6$ N \cdot m$^2$ = 43200 kN \cdot m$^2$ 20 kN/m B C 3m 6 m Steps: 1) Use singular function to express the distributed load.

          1) BEAM ANALYSIS (20pts)
Part III Analysis (45pts)
The cantilever beam (W140*60) BC is attached to the steel cable AB as shown. Knowing the
cable is initially taut, determine the tension in the cable caused by the distributed load shown and
the reactions at the right end.
A
-A = 255 mm$^2$
E = 200 \times 10$^9$ Pa  I = 216 \times 10$^6$ mm$^4$ = 216 \times 10$^{-6}$ m$^4$
3 m
EI = (200 \times 10$^9$)(216 \times 10$^{-6}$) = 43.2 \times 10$^6$ N \cdot m$^2$ = 43200 kN \cdot m$^2$
20 kN/m
B
C
3m
6 m
Steps: 1) Use singular function to express the distributed load.

1) BEAM ANALYSIS (20pts)
Part III Analysis (45pts)
The cantilever beam (W140*60) BC is attached to the steel cable AB as shown. Knowing the
cable is initially taut, determine the tension in the cable caused by the distributed load shown and
the reactions at the right end.
A
-A = 255 mm^2
E = 200 ×10^9 Pa I = 216 ×10^6 mm^4 = 216 ×10^-6 m^4
3 m
EI = (200 ×10^9)(216 ×10^-6) = 43.2 ×10^6 N ·m^2 = 43200 kN ·m^2
20 kN/m
B
C
3m
6 m
Steps: 1) Use singular function to express the distributed load.

Added by Lourdes M.

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