Contoh Perhitungan Pondasi

Design Pondasi
A. Data teknis
	Gaya aksial Perlu/vertikal (Pu)				=	5706.4	kg
	Gaya aksial Perlu/horizontal (Ph)				=	0	kg
	Momen perlu (M1)				=	70	kgm
	f'c				=	21	Mpa
	fy				=	240	Mpa
	Lebar kolom pondasi(b)				=	200	mm
	Tinggi kolom pondasi (h)				=	300	mm
B. Hasil Analisa
		Lebar pondasi (B)			=	500	mm
		Tebal pondasi (d)			=	150	mm
		Tulangan			=	dia8 - 200
		Jumlah tul.per lebar pondasi			=	4	bh
C. Analisa
	Data penyelidikan tanah
	g	(Wet density)			=	1.289	t/m3
	f	(angle of internal friction)			=	4.5	o
	c	(cohesion)			=	0.05	kg/cm2
	t	(timbunan tanah merah)			=	0	cm	B=500
		Analisa Terzaghi
	qult = 1,3.c.Nc.+q.Nq+0,4.g.B.Ng
	Tabel Tersaghi untuk nilai :f,Nc,Nq,Ng
		f			=	0	5
		Nc			=	5.7	7.3
		Nq			=	1	1.6
		Ng			=	0	0.5
	Untuk sudut geser (f)				=	4.5	o
		Nc			=	7.14
		Nq			=	1.54
		Ng			=	0.45
	qult = 1,3.c.Nc.+q.Nq+0,4.g.B.Ng				=	0.476	kg/cm2
	SF ( safety factor)				=	1.5
	qa tanah  = qult/SF				=	0.317	kg/cm2
	Daya dukung gelamf10-600				=	1.233	kg/cm2
	Rencana jumlah gelam				=	2	bh
	Daya dukung gelam totalf10-600				=	2.466	kg/cm2
	qa tanah + gelam				=	2.783	kg/cm2	=	0.2783	N/mm2
		Pu/B^2 <s'lentur
		B^2 ³ Pu/s'lentur,		B^2	=	205035	mm2
				B	=	453	mm
	1. One way action
			diambil lebar pondasi (B)		=	500	mm
	qu= Pu/B^2				=	0.22826	N/mm2
	Vu < Vc
	qu*B(0.5*B-0.5*b-d)£0.6*1/6(sqrt(f'c))*B*d				=
			Tebal pondasi (d)		³	148.862	mm
			diambil tebal pondasi (d)		=	150	mm
	qu= Pu/B^2<s'lentur				=	0.228	N/mm2	<	0.278	N/mm2	Ok!
	2. Two way action
	bc =B/b				=	2.5
	bo =2*((B+d)+(B+d))				=	2600	mm
	Vc =(1+2/bc)*2*sqrtf'c*bo*d				=	6433936.3	N
	Vcmax=4*sqrt*f'c*bo*d				=	7148818	N
	Vc < Vcmax……..ambil Vc				=	6433936.3	N
	fVc=0.6*Vc				=	3860362	N
	Vu =qu*((B^2-(B+d)*(b+d))				=	5135.76	N	<	3860362	N	Ok!
	3. Tinjauan terhadap momen lentur
	M1 (momen akibat reaksi kolom)				=	70000	kgmm
	M2 (momen akibat gaya horisontal)				=	0	kgmm
	AS = Mu/f0,9*d*fy				=	27.0062	mm2
	dicoba tulangan f				=	8	mm
	Jarak				=	200	mm
	As perlu				=	251	mm2
	r =As perlu/B*d				=	0.00335	>	r min =	0.0006
	r min =1.4/FY				=	0.00058
	rb =b1*((0.85*f'c)/fy)+(600/(600+fy))				=	0.26322	>	r  =	0.0034
	r max =0.75*rb				=	0.19741
	Lengan momen (a)=(Aperlu*fy)/(0.85*f'c*b)				=	6.75838	mm
	fMn =fAs*fy(d-0.5a)				=	7075167	Nmm
					=	707.517	kgm	> Mu	70	kgm	Ok!