Structures made of wood and plastics. Course

Contents

Contents

Input data 3 pages

Frame layout of pages

Layout of crossbar pp

Determination of pp forces

Matching sections of elements of pages

Calculation of nodes of pages

Layout of column of pages

Determination of pp forces

Calculation of column of pages

Calculation of pinching unit pp

Output of pages

List of literature pages

Source Data

Ministry of Education of the Russian Federation

Far Eastern State Technical University

Department of Building Structures and Materials

Wood and Plastic Structures

Exchange Rate Project

The design of a one-story frame wooden building.

Source Data

Design a one-story production building from wooden structures with a width of l = 15 m and a length of 60 m. Wooden elements of trusses made of spruce boards, KB-3 glue. Stretched elements and knot parts made of class A-I steel of VStZps brand. The coating is insulated with a roll roof. Prefabricated structures, group A1.

Production building heated. Building class by degree of responsibility II. The building will be built in Komsomolsk-on-amur (IV snow district, III wind district). In an open area.

The span of the building is 12 m.

The length of the building is 60 m.

Height to the bottom of load-bearing structures is 7.2m.

The pitch of the columns is 6m.

Covering the building with a roll roof on glue plates and a wooden truss. Wall panels are plastic. The bearing structure of the coating is a triangular truss. Bearing structures are wooden (pine of grades I and II).

Frame layout.

In compliance with preset materials and method of manufacture, triangular metal-wooden large-panel trusses with upper belt made of plank blocks are accepted under roll roof. At the pitch of trusses B = 6 m, glue panels of box section measuring 6 X 1.5 m are laid on them.

The adopted coating panels, when rigidly attached to the upper belt, provide continuous bracing of the upper girders of the trusses against loss of stability from the plane. Therefore, rolling stiffness bonds are not required. Vertical connections are provided, located in the plane of support and intermediate posts and connecting trusses in pairs into a unit, convenient for installation.

Girders of transverse frames are single-span, on supports they are hingedly connected to extreme columns. Coating plates are glued with nominal width of 1490mm. Curtain wall panels are plastic curtain panels .

In the longitudinal direction, the rigidity of the building is provided by vertical connections installed in the upper belt in the extreme spans adjacent to the end of the building and in the middle part of the building for each row of columns. In the transverse direction, the rigidity of the building is provided by horizontal connections located directly at the end walls and in the middle of the building. The design of the end frame is a rigid unchanged system in its plane due to the braces installed in the spans between the end posts.

The covering trusses rest on plated posts (columns) anchored in the foundations, creating a frame frame of the building across. Wind load transverse to the building axis is perceived by frame frames. The longitudinal wind load to the axis of the building is transmitted through the end faceover below to the foundations, and at the top to the wind truss located in the plane of the upper belts of the trusses.

The design diagram of the transverse frame is double-ball.

Girder Layout

Wooden elements of trusses are plated from spruce boards, KB-3 glue. Stretched elements and knot parts made of class A-I steel of VStZps brand.

For roll roof with slope i = 1/10 we take trapezoidal metal-tread trusses with upper belt made of plank blocks in accordance with specified materials and method of manufacture. At the pitch of trusses B = 6 m, we lay glue panels of box section with a size of 6 × 1.5 m on them.

Vertical connections are provided, located in the plane of support and intermediate posts and connecting trusses in pairs into a unit, convenient for installation.

The girder is a trapezoidal braced wooden-metal truss, a collection consisting of two elements (half trusses) assembled on the construction site in a structure ready for installation (two trusses combined by vertical links). Design diagram of truss single-span free-running beam.

Determination of forces in crossbar elements

Defines the overall dimensions of the truss. At span l = 12m the design height of the truss in the skate hcp = 1/7, l = 1.7m. Upper belt length at i = 1:10 BG = = 6 m. To give a construction lift to the truss, we lower the support units by 15 cm. Then the height of the truss on the supports BA = 1.76 × 0.1 + 0.15 = 1, 25 m. Length of braces: DB = = 3.2m; DG = = 3.45 m. HP rack length = (1.7 + 1.1 )/2 = 1, 4m. Inclination angles to the horizon: upper belt - tg α1 = 0.1 and α1 = 5.71 °; BD brace - tg α3 = 1.1/3 = 0.36 and α2 = 19.79 °; the raskosa of DG is tg a3 = 1.7/3 = 0.56 and a3=29.29 °.

Standard snow load on the coating for the II district = 70 kg/m2. Overload factor as per para 5.7 [6] at is ncn = 1.5. Design snow load.

All loads are considered attached to the upper belt of the truss.

Fits cross sections of truss elements.

Upper belt. Calculated as compressed-bending rod for longitudinal force 01 = 02 = 9716 kgf and local transverse load

To reduce the calculated bending moment from the local transverse load Mg, the upper belt units are designed with an off-center transmission of longitudinal forces O with negative eccentricity, thereby achieving a discharge moment Me = Oe. Structurally, this is achieved by shifting the crumpling platforms in the nodes by a value of e = 130 mm relative to the geometric axis of the element. Calculated bending moment in the upper belt panel

When selecting the section of the belt, we accept the bending moment

We specify the calculated section width b = 150 mm (we accept 150 mm wide boards according to the grade) and find the required section height from the formula for calculating the rod for complex resistance

where: ξ=0.8 - the approximate coefficient considering increase in the moment at deformation of an element; tb = 1.15 - coefficient to resistance moment as per Table 18 [1]. From the above expression we find htr = 18 cm.

When resting a plank-glued rectangular upper belt with a part of a section on a steel shoe in support units and a frontal stop of elements in intermediate units, the local concentration on the supports of chipping stresses should be taken into account. We find the required section height based on the condition of maximum chipping stresses in support sections according to formula (6.6) and item 9.19 [91 (see annex 17):

where Q is the transverse force on the support equal to 990 = 1492 kgf; kCK - coefficient of concentration of chipping stresses, taken according to Annex 17 at hcc/h = 044 > 0.4, kCK - 2.1; 0.6 coefficient, taking into account non-gluing. From the above expression we find htr = 21.75 cm, which is more htr in terms of compression strength with bending. We take the height of the section of the belt h = 220 mm, composing it from 5 boards 44 mm thick (5 cm to the line).

Check the accepted section. Geometric characteristics: cross-sectional area

Fbr = bh = 15 × 22 = 330 cm2;

moment of resistance

W = (15 × 222 )/6 = 1210 cm3;

design flexibility

At hcm/h = 0.44 height of crushing platform hCM = 0.44 • 22 = 9.68 cm,

Then structurally eccentricity of longitudinal forces

see.

Find the minimum height of the crushing platforms of the end faces of the elements

.

The optimal eccentricity is obtained by equating the stresses in the belt along the middle of the panel and along the edges, from the formula

We finally accept e = 6 cm and the height of the crushing platforms is 10 cm, taking into account the clipping in the nodes to a depth of 10 cm:

hodr = 22 (2 × 6) = 10 cm.

We check the accepted cross section of the belt in the middle of the extreme panel at full load with snow load:

Mrach = 11259716 × 0.06 = 542 kgf × m;

Due to the high safety margin, one-way snow loading is not checked.

Lower belt.

Design force U2 = 10617kgf. Required cross-sectional area of metal belt

We accept the cross section of the belt from two equal-beam angles 50 × 5 mm with a total area of ​ ​ 4.8 × 2 = 9.6 > 5.06 cm2.

Stands.

Design compressive force V1 = 3008 kgf, design length lst = 1.4 m. We set the flexibilities? = 120 < [150], at which the height of the strut section

We accept posts from two boards 44 mm thick, 150 mm wide. Check accepted section 88 x 150 mm. Actual flexibility

,

Normal voltages

Braces.

Design force D1 = 1911 kgf, design length DG = 3.45 m. We set flexibility A = 120 < [150], then

We accept braces from two boards with a thickness of 44mm and a width of 150 mm. We check the section 8.8 × 15 = 132 cm.

,

Tension

Calculation of nodes.

Cornice knot.

End channel is selected along the bend from uniformly distributed load

Bending moment

Required resistance moment

W = M/R = 18225/2100 = 8.7 cm3.

Take channel No. 14 with Wy = 11 cm3 > 8.7 cm3. To maintain the height of the crush area hcm = 9.2 cm, weld a sheet with a height of hst = 9.2 cm, a width of 15 cm onto the channel wall. We find the thickness of the sheet δst due to the condition of its bending from the pressure of the upper belt end (excluding bending work of the channel wall)

The sheet is reinforced by a vertical stiffening rib bp × δp = 60 x 10 mm. We consider the 1 section with a size of 75 × 130 mm as a plate resting on a contour in which the bending moment in a strip with a width of 1 cm is equal to M = αg0a2 = 0,118 × 70.4 × 7.52 = 467 kgs⋅sm. Where: a - 0.118 coefficient with aspect ratios of the plate 13/7.5 = 1.7 [81].

Define wall thickness

we take δst = 13 mm.

Bending moment in stiffening rib

where: gp - edge load, gp = 1.16 • 70.4 • 7.5 = 612.5 kgf/cm.

Position of the center of gravity of the design section

Moment of inertia of section

Moment of resistance

WP = 230/7.5 = 30.7 cm3.

Required cross-sectional resistance moment

WTP = MP/R = 6475/2100 = 3.1 < Wp = 30.7 cm3.

The horizontal sheet is checked for bending from the support reactive pressure of the strut with the accepted section b x h = 150 × 88 mm.

Reactive pressure on sheet

Upper belt pressure on sheet

Design pressure on the right section of the sheet

gl = g1 - g2 = 46 4.3 = 41.7 kgf/cm3.

Bending moment in a plate resting on three ropes with aspect ratio 11/7 = 0.64 in a strip 1 cm wide [8]

Ml = αgla2 = 0.085 × 41.7 × 152 = 798 kgf ⋅ cm .

Required Sheet Thickness

Accept horizontal sheet with thickness δL = 16 mm.

The following length of seams is necessary for attachment of a channel to a fasonka manual welding by E42 electrodes with a height of seams of hsh = 6 mm from each party:

To attach the lower belt to the profile, the length of the joints with a height of hsh = 6 mm is determined by the formulas:

on the shell

on pen lsh = 10 cm.

Intermediate node B of the upper belt.

Design forces: 01 = 02 = 9716 kgf, V1 = 3008 kgf. Forces from one element of the upper belt on the other are transmitted by the frontal stop through the collision areas with hcm = 9.2 cm. The depth of the slot to create eccentricity e = 6 cm is 2e = 12 cm. The joint is overlapped on both sides by straps with a section of 132 × 170 mm on bolts d = 8 mm.

Forces from the upper belt are transmitted to the post through the crushing platform at the end of the post. Calculated resistance of wood to local crumbling across fibers is found by formula

Required crumbling area

Since the resistance of wood to crushing is satisfactory, additional knocks are not required.

Intermediate node D of lower belt.

Design forces: U1 = + 10298kgf, U2 = + 10617kgf, D1 = 1911 and + 416kgf, V1 = 3008kgf.

To attach the angles of the lower belt to the unit, the required length of welds with a height of hm = 6 mm for the OBD element: 180 mm over the shell, 100 mm over the shelf; for DD element 190 mm and 110 mm, respectively.

Compression force from brace D1 = 1911kgf is transmitted to metal diaphragms of the unit. Pressure on vertical diaphragm

Horizontal diaphragm is calculated for pressure from strut

We calculate section 1, resting on three sides. With an aspect ratio of 6.6/15 = 0.44, the coefficient α = 0.06 and M1 = 0.06 x 22.8 × 15 2 = 307.8 kgf • cm.

Required Sheet Thickness

take δtr = 11 mm.

The vertical rib supporting the horizontal diaphragm is calculated as a beam on two supports loaded with a concentrated force V1. We take the thickness of the rib δtr = 20 mm, then the required height of it

we take = 4 cm.

Skating knot G.

Separate semi-trusses coming to the construction site are connected to each other by paired wooden linings with a section of 80 x 100 mm on bolts d = 0.8 mm and metal flanges on bolts d = 12 mm. The required eccentricity is provided by a 120 mm slot.

The tensile force is perceived by two bolts d = 8 mm.

When the truss is one-sided loaded with snow in the unit, the transverse force Q = Pcn/2 = 2035/2 = 1017.5 kgf appears. This force causes a cut of four bolts d = 8 mm. Shear stress in bolts

To reduce the free length of the lower belt and its sagging, we provide a suspension of reinforcement steel d = 10 mm.

Reference node A.

The truss rests on the columns through banding bars, which act as horizontal spacers of vertical stiffness links between the columns. The height of the banding bar is selected according to the maximum flexibilities = 200 at a design length of 6m

; we accept hob.br = 120 mm.

The width of the banding bar is assigned equal to the width of the support rack - 8.8 cm.

The required length of the horizontal support sheet is subject to local crushing of the banding bar across the fibers at

Layout of the column.

The wall panels are glued three-layer with a total thickness (with skins) of 192 + 28 = 2O8 mm. Panel weight 31 kg/m2. Design load from panels 0.346 kN/m2 of wall area. Plated-glued covering trusses 150 mm wide, 1300 mm high on the support. Columns are designed from sawn softwood (pine, spruce). Wood of the third grade for columns.

Presets the section of columns. The maximum Flexibility for columns is 12O. At selection of the sizes of section of columns 1OO is expedient to be set by flexibility λ =. Then at λ = 1OO and struts located on top of columns. At building height H = 7.1 m we get

hc = H/13 = 7.1/13 = 0.546m; bk = H/29 = 7.1/29 = 0.245 m.

It is assumed that boards with a width of 275 mm and a thickness of 50 mm are used for the manufacture of columns. After milling (cutting), the thickness of the boards will be 44 mm. The width of the column after milling (cutting) of the preform blocks along the formation will be 2756 = 269 mm. Taking into account the accepted thickness of the boards after cutting, the height of the section of the columns will be Nk = 44 × 13 = 572 mm; bq = 269 mm.

Define Effort

Defines column loads. Load collection. The design diagram of the frame is given in Fig. 3.

Fig. 3. Design diagram of the frame.

We will determine the calculated vertical and horizontal loads acting on the column. Calculation of horizontal projection loads is given in Table 3. Column Loads:

With a small error, the height distribution scheme k can be simplified.

Determine horizontal loads acting on the frame taking into account the step S = 6 m

To simplify the determination of bending moments, we consider the force applied with eccentricity at a height of H/2.

Wind load transmitted from coating located outside the column:

Wind loads:

Defines the force in columns. Transverse frame of single-span building consisting of two columns rigidly pinched in foundations and hinged to girder in the form of beam is calculated for vertical and horizontal loads (Fig. 3). It is once a statically undefined system. With an infinitely large rigidity of the girder (conditional assumption), it is unnecessary unknown convenience to accept the longitudinal force in the girder, which is determined according to the known rules of construction mechanics.

Definition of bending moments (excluding combination factor):

from wind load: rigel force

bending moment at the foundation top level:

From Extra-Centered Wall Loads: Eccentricity of Wall Loads

bending moment acting on frame post:

Rigel force (tensile force):

bending moments at the foundation top level:

Determination of transverse forces (excluding cof. combinations):

from wind load

from off-center loading from walls

Determination of column forces, taking into account, where appropriate, combination factors:

First load combination

Moments at the foundation top level:

For calculation of columns for strength and stability of flat form of deformation we take values: M = Mlev = 41.33 kN· m; N = 73.21 kN.

The second combination of loads (with one time load factor, not taken into account).

Third combination of loads (factor is not taken into account since one time load):

bending moments in the foundation level:

With wood of the third grade and with accepted cross-sectional dimensions as per Table 3 of SNiP II-25-80

Rc = 11MPa;

Taking into account tn, msl = 1 and reliability factor αp = 0.95, we obtain

Rc = 11Would 1,2Would 1Would/0.95 = 13, 89MPa;

Hereinafter, when calculating strength and stability in the test formulas, it is convenient to write the values ​ ​ of N and Q to MN, and the value of M to MN· m.

At an epyura of the moments of a triangular outline (see item 4.17 Construction Norms and Regulations P2580) correction factor to ξ

kn = an + x (1-an) = 1.22 + 0.898 (11.22) = 1.02.

In this case, the moment era is close to triangular:

We leave the previously accepted section based on the need to limit flexibility.

The calculation for the stability of the flat deformation form is made according to formula (33) SNiP II2580. We accept that the spacers along the outer rows of columns (in a plane parallel to the outer walls) go only along the top of the columns. Then.

In the formula

index of degree p = 2 as for elements that do not have fixed stretched zone from deformation plane:

For the epure of triangular moments (see Table 2. Annex 4 SNiP II2580):

kf = 1,750, 75d = 1.75; d is O, since the moment at the top of the column is O:

Consequently, stability is ensured.

Calculation for stability from the plane as a centrally compressed rod, ¼ - 0.337 (see calculation for stability of a flat deformation form); N = 74.47 (for the second load combination):

stability is ensured.

Calculation of pinching node.

Calculation of the column pinching unit in the foundation. Calculation is made for the third combination of loadings (wind load + the minimum vertical loading calculated only taking into account constant load (constant load from covering weight from the body weight of a column is accepted with γ = 0.9. since at the same time the forces in anchors and weights will be more than with γ - 1.1... 1.3) and without taking into account the snow load). In this combination of loads, we obtain the maximum forces in the anchors and weights of the clamping unit structure shown in Fig. 4.

Select (preliminary) part sizes for subassemblies. We accept the column widening thickness (see Figure 4) equal to two plank thicknesses after milling. The sub-column is taken from concrete of class B25 (Rb = 14.5 MPa > Rcm)

Take thickness of steel anchor strip 10 mm (δa = 0.01 m). Taking into account the accepted extensions, we will receive

We accept height of broadenings to the equal width of a column low (hk. N) plus 150 mm, considering the structural design of the unit and the location of the weights at an angle of 45 °

Determination of forces in anchor strips and strands.

Forces in anchor strips and inclined strings, which provide column attachment to foundation, are calculated in case of decision of column pinching unit based on equilibrium of all forces acting on the unit. Design resistance of wood to crumpling is taken taking into account mn and αn.