## Recommendation points

- Varieties of hip roof
- Scheme and main elements
- The main elements of the rafter system
- Reinforcement elements of the hip rafter system
- Calculation of the rafter system
- The choice of the angle of inclination of the longitudinal and end slopes
- Calculation of the main rafter elements
- Calculation of the roof area

Hip roofs have many benefits. They are beautiful, reliable in all weather conditions, the four-sided design allows you to effectively insulate the house from the roof side. The device of the rafter system presents some complexity. We will deal with its schemes and calculations in this article..

Hip roofs, sometimes called Dutch and Danish, are distinguished by their quality, reliability and spectacular European design. The rafter base of such roofs consists of many basic and reinforcing elements that require sketches or three-dimensional drawings, accurate calculations and execution.

## Varieties of hip roof

Hip roofs, in addition to the basic classical design, consisting of two trapezoidal slopes and two triangular end hips, also include their varieties:

- Half-hip gable.
- Half-hip four-slope.
- Tent.
- Hip-pediment.
Semi-hinged gable roof

Semi-hinged hipped roof

Hipped roof

Hip-gable roof

Each type has its own rafter system scheme. Next, we consider and calculate the classic hip roof.

## Scheme and main elements

To calculate the rafter system, you need to familiarize yourself with its basic scheme, main and auxiliary elements.

## The main elements of the rafter system

The main elements are (see the figure below):

- Mauerlat. It is a timber fixed along the perimeter of the outer walls with an indent from the outer edge. Wall mounted. Mauerlat distributes the load from the pressure of the rafters, connects the rafter system with the walls of the house, is the basis of the roof.
- Skate. Top crossbar for fastening the rafters of the roof slopes. The height of the ridge is based on the angle of inclination of the slopes. Gives the system rigidity and strength.
- The central rafters of the slopes. The ends of the ridge are supported on the sides of the Mauerlat. There are 4 such elements in the system. – 2 pcs. on every slope.
- Central rafters of the hips. The ends of the ridge are supported on the end sides of the Mauerlat. There are 2 such elements in the system. – 1 pc. on each hip.
- Slant legs (diagonal, angled rafters). Connect the corners of the Mauerlat with the ends of the ridge. Are part of the supporting structure. There are 4 of them in the rafter system.
- Intermediate rafters of the slopes. They are installed parallel to the central rafters of the slope between them with the same pitch, leaning on the side of the Mauerlat and the ridge bar. If the length of the skate is insignificant, it may not be used.
- Shortened rafters of the slopes. They are installed parallel to the central rafters of the slopes and have variable length – the closer to the corner, the shorter. Lean on the side of the Mauerlat and on the legs. The number of elements depends on the installation step.
- Shortened hip rafters or rafters. They are installed parallel to the center rafters of the hips and have variable length – the closer to the corner, the shorter. Lean on the end part of the Mauerlat and on the slant legs. The number of elements depends on the installation step.
Diagram and main elements of the rafter system

You can read more about attaching rafters to the Mauerlat in our article.

## Reinforcement elements of the hip rafter system

The above elements are basic, basic. Other elements are designed to strengthen the main ones and are used in critical buildings, for example, for residential buildings:

- Vertical racks for supporting the ridge bar. Lean on the crossbars (see below), laid parallel to the end of the house or a bed, located along the longitudinal axis of the building (if there is a capital wall under it).
- Crossbars or tightening. The rafter legs of the slopes are tied in pairs. Serve as a support for racks and diagonal struts (see below). They can serve as floor beams if they are built into the Mauerlat or installed directly into the longitudinal walls of the house. If the puffs are placed closer to the ridge, they will form the basis of the attic ceiling..
- Diagonal braces (braces). They are used to increase the rigidity of the system if the length of the rafters is more than 4.5 m.The use of struts allows you to reduce the cross-section of the rafters, which they reinforce.
- Sprengel. Beam installed in the corners of the Mauerlat. Serves for mounting a stand that supports and reinforces the forefoot leg.
- Wind beam. Serves to resist deformation of rafter legs in gusty, strong winds. Fastened to the rafters of the slopes from the inside, obliquely, on one or both sides – depending on the wind load in the construction area.
- Filly. An element of a smaller section than the rafters themselves. Extends the rafter leg to organize the overhang of the roof in the event that a single element does not work due to limited length of lumber or for reasons of economy.
Reinforcement elements

## Calculation of the rafter system

The calculation of the system includes the choice of the angle of inclination of the slopes and hips and the calculation of the lengths of its main and auxiliary elements.

## The choice of the angle of inclination of the longitudinal and end slopes

The choice of the angle of the slopes and hips ranges from 25–45 ° and depends on the desire to have an attic space, the adopted roofing material, the assessment of static (roof weight) and dynamic (wind, snow) loads.

In hipped roofs, the angle of inclination of the hips and slopes is the same. In hip roofs, they also often take the same angles in terms of aesthetics, but they can be different if this is the architect’s idea..

Recommendations for the use of roofing materials

For a better understanding of the calculation algorithm, consider, as an example, a hip roof of a house with sides of 8 and 12 m, and a ridge height of 2.5 m.The slope angle of the slopes is 35 °, and the hip angle is 45 °.

## Calculation of the main rafter elements

The classic hip roof consists of two trapezoidal slopes connected in a ridge, and two hips – end slopes in the form of triangles.

First you need to remember some formulas from the school algebra curriculum. This is the ratio of the lengths of the sides of a right-angled triangle, expressed in terms of the trigonometric function of the angle and the Pythagorean theorem.

Trigonometric functions of an acute angle of a right triangle

Pythagorean theorem

Let’s depict the frame of the truss system in axonometric form:

We will calculate the main elements of the rafter system.

1. Calculate the length of the central hip rafter CD, which is the height of the isosceles triangle (hip) and the hypotenuse of the right triangle, the height of which is equal to the height of the ridge (CE = 2.5 m). Hip angle? = 45 °. Sin 45 ° = 0.71 (according to the Bradis table).

Roof angletg ?sin ?5 ° 0.09 0.09 10 ° 0.18 0.17 15 ° 0.27 0.26 20 ° 0.36 0.34 25 ° 0.47 0.42 30 ° 0.58 0.50 35 ° 0.70 0.57 40 ° 0.84 0.64 45 ° 1.00 0.71 50 ° 1.19 0.77 55 ° 1.43 0.82 60 ° 1.73 0.87 According to the trigonometric ratio:

- СD = CE / sin? = 2.5 / 0.71 = 3.52 m
2. Determine the length of the ridge K. For this, from the previous triangle we find the length of the base ED, using the Pythagorean theorem:

House length: BL = 12 m.

Skate length:

- CF = 12 – 2.478 x 2 = 7.044 m
3. The length of the corner rafters CA can also be obtained from the Pythagorean theorem for the triangle ACD. Half the width of the house AD = 8/2 = 4 m, CD = 3.52 m:

4. The length of the central rafters of the slope GF is the hypotenuse of the triangle, the legs of which are the height of the ridge H (CE) and half the width of the house AD:

The intermediate rafters of the ramps are of the same length. Their number depends on the pitch and section of the beams and is determined by calculating the total load, including the weather.

These tables correspond to the atmospheric loads of the Moscow region.

Rafters step, cmRafter length, m3.03.54.04.55.05.56.0215 100×150 100×175 100×200 100×200 100×250 100×250 – 175 75×150 75×200 75×200 100×200 100×200 100×200 100×250 140 75×125 75×125 75×200 75×200 75×200 100×200 100×200 110 75×150 75×150 75×175 75×175 75×200 75×200 100×200 90 50×150 50×175 50×200 75×175 75×175 75×200 75×200 60 40×150 40×175 50×150 50×150 50×175 50×200 50×200 Let’s compare the maximum, average and minimum cross-section of a bar with a length of 4.717 m (look at the values for 5.0 m).

When cutting

100×250 mmthe step will be 215 cm. With a ridge length of 7.044 m, the number of intermediate rafters will be: 7.044 / 2.15 = 3.28 segments. Rounding up – up to 4. The number of intermediate rafters of one slope – 3 pieces.Volume of lumber for both slopes:

- 0.1 0.25 4.717 3 2 = 0.708 m
^{3}When cutting

75×200 mmthe step will be 140 cm. With a ridge length of 7.044 m, the number of intermediate rafters will be: 7.044 / 1.4 = 5.03 segments. The number of intermediate rafters of one slope – 4 pieces.Volume of lumber for both slopes:

- 0.075 0.2 4.717 4 2 = 0.566 m
^{3}When cutting

50×175 mmthe step will be 60 cm. With a ridge length of 7.044 m, the number of intermediate rafters will be: 7.044 / 0.6 = 11.74 segments. Rounding up – up to 12. The number of intermediate rafters of one slope – 11 pieces.Volume of lumber for both slopes:

- 0.05 * 0.175 * 4.717 * 11 * 2 = 0.908 m
^{3}Therefore, for our geometry, the optimal option from the point of view of the economy would be a section of 75×200 mm with a step of 1.4 m.

5. To calculate the lengths of the shortened rafters of the MN slope, you will again have to remember the school curriculum, namely the rule of similarity of triangles.

Similarity of triangles on three sides

The large triangle, which we need to strengthen with shortened rafters, has known dimensions: GF = 4.717 m, ED = 2.478 m.

If the shortened rafters are installed with the same pitch as the intermediate ones, their number will be 1 piece in each corner:

- 2.478 m / 1.4 m = 1.77 pcs.
That is, two segments are formed with one shortened rafter in the middle. A small triangle will have a leg half the ED:

- BN = 2.478 / 2 = 1.239 m
We compose the proportion of such triangles:

Based on this ratio:

At this height, the rafter section is taken according to the table – 75×125 mm. The total number of shortened rafters of both slopes – 4 pcs..

6. Determination of the length of the shortened rafters of the hips (rafters) is also carried out from the ratio of similar triangles. Since the length of the central rafters of the hips is CD = 3.52 m, the step between the shortened rafters may be larger. With AD = 4 m, shortened rafters with a step of 2 m will have one on each side of the central rafter of the hips:

- (2 3.52) / 4 = 1.76 m
At this height, the rafter section is taken as 75×125 mm. The total number of shortened rafters of both hips – 4 pieces.

Attention! In our calculations, we did not take into account the overhang.

## Calculation of the roof area

This calculation boils down to determining the areas of the trapezoid (slope) and triangle (hip).

Area of a trapezoid and a triangle

Let’s do the calculation for our example.

1. The area of one hip with CD = 3.52 m and AB = 8.0 m, taking into account the overhang of 0.5 m:

- S = ((3.52 + 0.5) (8 + 2 0.5)) / 2 = 18.09 m
^{2}2. The area of one slope at BL = 12 m, CF = 7.044 m, ED = 2.478 m, taking into account overhangs:

- S = (2.478 + 0.5) ((12.0 + 2 0.5) + 7.044) / 2 = 29.85 m
^{2}Total roof area:

- S
_{?}= (18.09 + 29.85) 2 = 95.88 m^{2}Advice! When buying material, consider cutting and inevitable losses. The material produced by elements of a large area is not the best option for hip roofs.