Building Extension Task A building firm has been commissioned to build an extension to the rear of a detached property. The extension is to be 5m wide, extend 4m from the rear of the property and rise to a height of 2.2m at the eaves. The extension is to have a pitched roof to match the house and be constructed of Fibre Cement Slate (technical data for slates is provided in Fig 1). The two upper windows at the rear of the property have stone sills which are 4m above ground level at the lowest point. The owners have informed the builders that the rear of the property is often exposed to severe weather conditions. Given all the information provided, examine the proposed extension and comment on both the roof pitch and the number of slates required. Fig 1 - Technical Data Size of Slate 600 x 300 Minimum Pitch (100mm lap) Moderate exposure 23.5⁰ Severe exposure 29.5⁰ Minimum Pitch (110mm lap) Moderate exposure 20.0⁰ Severe exposure 22.5⁰ Coverage capacity (net slates/m2 ) 13.4 at 100 lap 13.6 at 110 lap Eaves

Building Extension Task (Answer sheet) Roof Pitch Trigonometry can be used to calculate the height of the extension where it abuts the house. Calculations should employ the angles for the minimum pitch required for severe weather conditions. If calculations are carried out for the minimum pitch required with 100mm lap (29.5⁰) it will be apparent that the height of the extension will overlap the upstairs windows. Calculations for the minimum pitch required for 110mm lap (22.5⁰) should indicate that the height of the extension will fall below the window sill, therefore it will be necessary to lap the slates 110mm and pitch the roof at least 22.5⁰. Number of slates required The area of the roof is required which requires the length of the roof pitch to be calculated; this can be achieved by employing trigonometry or Pythagoras. The number of slates can be calculated by multiplying the area by the coverage capacity for 110mm lap. tan 29.5 ⁰ = o/4 o = tan29.5 ⁰ x 4 = 2.26m o 29.5⁰ 4m tan 22.5 ⁰ = o/4 o = tan22.5 ⁰ x 4 = 1.66m h o 22.5⁰ cos22.5 ⁰ = 4/h h = 4/cos22.5 ⁰ = 4.33m 4m Slates required = 4.33 x 5 x 13.6 = 295 slates