Question 14.3: The following information is given for proportioning a canti...

The following information is given for proportioning a cantilever footing with reference to Fig. 14.2.

Column Loads: Q1=1455kN,Q2=1500kNQ_{1}=1455 kN , Q_{2}=1500 kN.

Size of column: 0.5 x 0.5 m.

Lc=6.2m,qs=384kN/m2L_{c}=6.2 m , q_{s}=384 kN / m ^{2}

It is required to determine the size of the footings for columns 1 and 2.

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Assume the width of the footing for column 1=B1=2m1=B_{1}=2 m.

First trial

 Try e=0.5m. Now, LR=6.20.5=5.7m\text { Try } e=0.5 m \text {. Now, } L_{R}=6.2-0.5=5.7 m.

Reactions

 

R1=Q11+eLR=14551+0.55.7=1583R_{1}=Q_{1} 1+\frac{e}{L_{R}}=14551+\frac{0.5}{5.7}=1583

 

R2=Q2Q1eLR=15001455×0.55.7=1372kNR_{2}=Q_{2}-\frac{Q_{1} e}{L_{R}}=1500-\frac{1455 \times 0.5}{5.7}=1372 kN

 

Size of footings – First trial

 

Col. 1. Area of footing A1=1583384=4.122sq.mA_{1}=\frac{1583}{384}=4.122 sq . m

 

Col. 2. Area of footing A2=1372384=3.57sqmA_{2}=\frac{1372}{384}=3.57 sq \cdot m

 

Try 1.9 x 1.9m

Second trial

 

New value of e=B12b12=220.52=0.75me=\frac{B_{1}}{2}-\frac{b_{1}}{2}=\frac{2}{2}-\frac{0.5}{2}=0.75 m

 

New LR=6.200.75=5.45mL_{R}=6.20-0.75=5.45 m

 

R1=14551+0.755.45=1655kNR_{1}=14551+\frac{0.75}{5.45}=1655 kN

 

R2=15001455×0.755.45=1300kNR_{2}=1500-\frac{1455 \times 0.75}{5.45}=1300 kN

 

A1=1655384=4.31sqm or 2.08×2.08mA_{1}=\frac{1655}{384}=4.31 sq \cdot m \text { or } 2.08 \times 2.08 m

 

A2=1300384=3.38sqm or 1.84×1.84mA_{2}=\frac{1300}{384}=3.38 sq \cdot m \text { or } 1.84 \times 1.84 m

 

Check e=B12b12=1.040.25=0.790.75me=\frac{B_{1}}{2}-\frac{b_{1}}{2}=1.04-0.25=0.79 \approx 0.75 m

 

Use 2.08 x 2.08 m for Col. 1 and 1.90 x 1.90 m for Col. 2.

Note: Rectangular footings may be used for both the columns.

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