SCOTTISH EXECUTIVE

Building Regulations: Technical Standards

FLOORS

Ground floors

Note: in using the tables for floors it is first necessary to calculate the ratio P/A, where:

P is the floor perimeter length in metres; and

A is the floor area in square metres.

Table E9 to Part J Appendix E: Insulation thickness for solid floors in contact with the ground

	*Insulation thickness (mm) for: U-value* of 0.25 W/m²K**
	Thermal conductivity of insulant (W/mK)
	P/A	0.02	0.025	0.03	0.035	0.04	0.045	0.05
	A	B	C	D	E	F	G	H
1	1.00	62	77	93	108	124	139	155
2	0.90	61	76	91	107	122	137	152
3	0.80	60	75	90	105	119	134	149
4	0.70	58	73	87	102	116	131	145
5	0.60	56	70	84	98	111	125	139
6	0.50	52	66	79	92	105	118	131
7	0.40	47	59	71	83	95	107	119
8	0.30	39	49	59	69	79	88	98
9	0.20	24	30	36	42	48	54	60
	U-value of 0.35 W/m²K
10	1.00	39	49	58	68	78	88	97
11	0.90	38	48	57	67	76	86	95
12	0.80	37	46	55	65	74	83	92
13	0.70	35	44	53	62	70	79	88
14	0.60	33	41	49	58	66	74	82
15	0.50	30	37	44	52	59	67	74
16	0.40	25	31	37	43	49	55	61
17	0.30	16	21	25	29	33	37	41
18	0.20	1	1	1	2	2	2	2
	U-value of 0.45 W/m²K
19	1.00	26	33	39	46	53	59	66
20	0.90	25	32	38	44	51	57	63
21	0.80	24	30	36	42	48	54	60
22	0.70	22	28	34	39	45	51	56
23	0.60	20	25	30	35	40	45	50
24	0.50	17	21	25	30	34	38	42
25	0.40	12	15	18	21	24	27	30
26	0.30	4	5	6	6	7	8	9
27	<0.27	0	0	0	0	0	0	0

Table E10 to Part J Appendix E: Insulation thickness for suspended timber ground floors

	Insulation thickness (mm) for: U-value of 0.25 W/m²K
	Thermal conductivity of insulant (W/mK)
	P/A	0.02	0.025	0.03	0.035	0.04	0.045	0.05
	A	B	C	D	E	F	G	H
1	1.00	95	110	126	140	155	170	184
2	0.90	93	109	124	138	153	167	181
3	0.80	91	106	121	136	150	164	178
4	0.70	88	103	118	132	145	159	173
5	0.60	85	99	113	126	139	153	166
6	0.50	79	92	106	118	131	143	156
7	0.40	71	83	95	107	118	129	140
8	0.30	57	68	78	88	97	106	116
9	0.20	33	39	46	52	58	64	69
	U-value of 0.35 W/m²K
10	1.00	57	67	77	87	96	106	115
11	0.90	55	66	75	85	94	103	112
12	0.80	53	63	73	82	91	100	109
13	0.70	51	60	69	78	87	95	104
14	0.60	47	56	64	73	81	89	97
15	0.50	42	50	57	65	72	79	87
16	0.40	34	41	47	53	60	66	72
17	0.30	22	26	31	35	39	43	47
18	0.20	1	1	2	2	2	2	3
	U-value of 0.45 W/m²K
19	1.00	37	44	51	57	64	70	77
20	0.90	35	42	49	55	62	68	74
21	0.80	33	40	46	53	59	65	70
22	0.70	31	37	43	49	54	60	65
23	0.60	27	33	38	43	49	54	58
24	0.50	22	27	32	36	40	44	49
25	0.40	15	18	22	25	28	31	34
26	0.30	4	5	6	7	8	9	10
27	<0.27	0	0	0	0	0	0	0

Notes:

1. The table is derived for suspended timber floors for which the proportion of timber is 12%, which corresponds to 48 mm wide timbers at 400 mm centres.
2. For other proportions of timber the U-value can be calculated using the procedure in Appendix A.

Table E11 to Part J Appendix E: Insulation thickness for suspended concrete beam and block ground floors

	Insulation thickness (mm) for: U-value of 0.25 W/m²K
	Thermal conductivity of insulant (W/mK)
	P/A	0.02	0.025	0.03	0.035	0.04	0.045	0.05
	A	B	C	D	E	F	G	H
1	1.00	60	75	90	104	119	134	149
2	0.90	59	73	88	103	118	132	147
3	0.80	58	72	86	101	115	130	144
4	0.70	56	70	84	98	112	126	140
5	0.60	54	67	80	94	107	121	134
6	0.50	50	63	75	88	101	113	126
7	0.40	45	57	68	79	91	102	113
8	0.30	37	46	56	65	74	84	93
9	0.20	22	27	33	38	43	49	54
	U-value of 0.35 W/m²K
10	1.00	37	46	55	64	74	83	92
11	0.90	36	45	54	63	72	81	90
12	0.80	35	43	52	61	69	78	87
13	0.70	33	41	50	58	66	74	83
14	0.60	31	38	46	54	61	69	77
15	0.50	27	34	41	48	55	62	69
16	0.40	22	28	34	39	45	50	56
17	0.30	14	18	21	25	29	32	36
18	0.20	0	0	0	0	0	0	0
	U-value of 0.45 W/m²K
19	1.00	24	30	36	42	48	54	60
20	0.90	23	29	35	41	46	52	58
21	0.80	22	28	33	39	44	50	55
22	0.70	20	25	31	36	41	46	51
23	0.60	18	23	27	32	36	41	45
24	0.50	15	18	22	26	29	33	37
25	0.40	10	12	15	17	19	22	24
26	0.30	2	2	2	3	3	4	4
27	<0.28	0	0	0	0	0	0	0

Example 10 - Solid floor in contact with the ground

Determine the thickness of the insulation layer required to achieve a U-value of 0.45 W/m²K for the ground floor slab shown below.

diagram

It is proposed to use insulation with a thermal conductivity of 0.02 W/mK.

The overall perimeter length of the slab is: (10 + 4 + 4 + 2 + 6 + 6) = 32 m

The floor area of the slab is: (6 x 6) + (4 x 4) = 52 m²

The ratio: formula

Using Table E9, column B, row 23 indicates 20 mm of insulation is required.

The insulation may be located above or below the concrete slab.

Example 11 - Suspended timber floor

If the floor shown above was of suspended timber construction, the perimeter length and floor area would be the same, yielding the same ratio of:

formula

To achieve a U-value of 0.45W/m²K, using insulation with a thermal conductivity of 0.04 W/mK, Table E10 column F, row 23 indicates that the insulation thickness between the joists should be not less than 49 mm.

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