A 1.40m x 1.80m footing is used to support a total factored load of 750 kN. The footing thickness is 325 mm and the column size is 300mm x 300mm. Use effective steel covering of 75mm, fc’ = 21MPa and fy = 276 MPa. (See Figure 6)

Structural Analysis
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Chapter2: Loads On Structures
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A 1.40m x 1.80m footing is used to support a total factored load of 750 kN. The footing thickness is 325 mm and the column size is 300mm x 300mm. Use effective steel covering of 75mm, fc’ = 21MPa and fy = 276 MPa. (See Figure 6)

A 1.40m x 1.80m footing is used to support a total factored load of 750 kN. The footing thickness is 325 mm and the column size is
300mm x 300mm. Use effective steel covering of 75mm, fc' = 21MPa and fy = 276 MPa. (See Figure 6)
a) Calculate the design beam shear (maximum factored shear force at distance "d" from face of column).
b) Calculate the design punching shear (factored shear force at distance "d/2" from face of column).
c) Is the thickness of the footing sufficient for shear requirements? Support your answer.
d) Calculate the required area of steel in the long direction.
e) Calculate the required area of steel in the short direction.
|Pu = 750 kN
0.325 m
75 mm
- Short direction steel
0.30 m
0.30 m
-Long direction steel
1.80 m
Figure 6 Problem 6
1.40 m
Transcribed Image Text:A 1.40m x 1.80m footing is used to support a total factored load of 750 kN. The footing thickness is 325 mm and the column size is 300mm x 300mm. Use effective steel covering of 75mm, fc' = 21MPa and fy = 276 MPa. (See Figure 6) a) Calculate the design beam shear (maximum factored shear force at distance "d" from face of column). b) Calculate the design punching shear (factored shear force at distance "d/2" from face of column). c) Is the thickness of the footing sufficient for shear requirements? Support your answer. d) Calculate the required area of steel in the long direction. e) Calculate the required area of steel in the short direction. |Pu = 750 kN 0.325 m 75 mm - Short direction steel 0.30 m 0.30 m -Long direction steel 1.80 m Figure 6 Problem 6 1.40 m
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