Maximum & Minimum Pile Forces

P_{p max} = \frac{1, 111}{12} + \frac{(50 \cdot 12) 54}{19, 440} + \frac{(25 \cdot 12) 36}{10, 368} = 95.3 kips (ultimate)

P_{p max} = \frac{1, 111}{12} - \frac{(50 \cdot 12) 54}{19, 440} - \frac{(25 \cdot 12) 36}{10, 368} = 89.9 kips (ultimate)

P_{p max}= 95.3 kips < 1.6P = 160 kips - pile is adequate

Design of Pile Cap for Flexure

Critical section for flexure is at face of column,

For moment in minor direction (about x-x axis),

Self
				Weight = \frac{138 (43) (57)}{12^{3}} (\frac{150}{1, 000}) = 29.4 kips

M_{u} = 4 (160) \frac{36 + 3 - 0.5 (16)}{12} - 0.9 (29.4) \frac{43 / 2}{12} = 1, 606 kip-ft

A_{s} = \frac{1.7 {f^{'}}_{c} b d}{2 f_{y}} - \frac{1}{2} \sqrt{\frac{2.89 {({f^{'}}_{c} b d)}^{2}}{f_{y}^{2}} - \frac{6.8 {f^{'}}_{c} M_{u}}{f_{y}^{2}}}

A_{s} = 0.51 d - \sqrt{0.26 d^{2} - 0.0189 M_{u}}

A_{s} = [0.51 (46.3) - \sqrt{0.26 {(46.3)}^{2} - 0.0189 (\frac{1, 606 \cdot 12}{11.5})}] 11.5 = 8.1 {in}^{2}

Check minimum required steel for slab per ACI 318-14 (Sec. 24.4.3.2),

A_s,min1 = (0.0018×60,000/60,000)bh = 0.0018×(12 × 11.5)×57 = 14.16 in² Controls

A_s,min2 = 0.0014×bh = 0.0014×(12 × 11.5)×57 = 11.01 in²

Check minimum required steel for beam per ACI 318-14 (Sec.9.6.1.3),

4/3 A_s= 10.80 in²

200/f_ybd = 12×11.5×46.3×200 / 60,000 = 21.3 in²

Use 15 #9 bars

For moment in major direction (about y-y axis),

Self
				Weight = \frac{102 (59) (57)}{12^{3}} (\frac{150}{1, 000}) = 29.8 kips

M_{u} = 3 (160) [\frac{54 + 3 - 0.5 (20)}{12} - \frac{18 + 3 - 0.5 (20)}{12}] - 0.9 (29.8) \frac{59 / 2}{12} = 2, 258 kip-ft

A_{s} = [0.51 (47.4) - \sqrt{0.26 {(47.4)}^{2} - 0.0189 (\frac{2, 258 \cdot 12}{8.5})}] 8.5 = 10.92 {in}^{2}

Check minimum required steel for slab,

A_s,min = 0.0018bh = 0.0018 (12 · 8.5)57 = 10.47 in²

Check minimum required steel for beam,

4/3 A_s= 15.56 in² Controls

Controls

200/f_ybd = 12 · 8.5 · 47.4 · 200 / 60,000 = 16.12 in²

Use 15- #9 bars