bend. Category Archives: Diagram problems Longitudinal force in a beam

count beam for bending there are several options:
1. Calculation of the maximum load that it will withstand
2. Selection of the section of this beam
3. Calculation of the maximum allowable stresses (for verification)
let's consider general principle of beam section selection on two supports loaded with a uniformly distributed load or a concentrated force.
To begin with, you will need to find a point (section) at which there will be a maximum moment. It depends on the support of the beam or its termination. Below are diagrams of bending moments for schemes that are most common.

After finding the bending moment, we must find the modulus Wx of this section according to the formula given in the table:

Further, when dividing the maximum bending moment by the moment of resistance in a given section, we get maximum stress in the beam and this stress we must compare with the stress that our beam of a given material can generally withstand.

For plastic materials(steel, aluminum, etc.) the maximum voltage will be equal to material yield strength, a for fragile(cast iron) - tensile strength. We can find the yield strength and tensile strength from the tables below.

Let's look at a couple of examples:
1. [i] You want to check if an I-beam No. 10 (St3sp5 steel) 2 meters long rigidly embedded in the wall can withstand you if you hang on it. Let your mass be 90 kg.
First, we need to choose a calculation scheme.

This diagram shows that the maximum moment will be in the termination, and since our I-beam has the same section along the entire length, then the maximum voltage will be in the termination. Let's find it:

P = m * g = 90 * 10 = 900 N = 0.9 kN

M = P * l = 0.9 kN * 2 m = 1.8 kN * m

According to the I-beam assortment table, we find the moment of resistance of I-beam No. 10.

It will be equal to 39.7 cm3. Convert to cubic meters and get 0.0000397 m3.
Further, according to the formula, we find the maximum stresses that we have in the beam.

b = M / W = 1.8 kN/m / 0.0000397 m3 = 45340 kN/m2 = 45.34 MPa

After we have found the maximum stress that occurs in the beam, we can compare it with the maximum allowable stress equal to the yield strength of steel St3sp5 - 245 MPa.

45.34 MPa - right, so this I-beam can withstand a mass of 90 kg.

2. [i] Since we got quite a large supply, we will solve the second problem, in which we will find the maximum possible mass that the same I-beam No. 10, 2 meters long, can withstand.
If we want to find the maximum mass, then the values ​​of the yield strength and the stress that will occur in the beam, we must equate (b \u003d 245 MPa \u003d 245,000 kN * m2).

A longitudinal-transverse bend is a combination of a transverse bend with compression or tension of a beam.

When calculating for longitudinal-transverse bending, the bending moments in the cross sections of the beam are calculated taking into account the deflections of its axis.

Consider a beam with hinged ends, loaded with some transverse load and a compressive force 5 acting along the axis of the beam (Fig. 8.13, a). Let us denote the deflection of the beam axis in the cross section with the abscissa (we take the positive direction of the y axis downwards, and, therefore, we consider the deflections of the beam to be positive when they are directed downwards). The bending moment M, acting in this section,

(23.13)

here is the bending moment from the action of the transverse load; - additional bending moment from the force

The total deflection y can be considered to consist of the deflection arising from the action of only the transverse load, and an additional deflection equal to that caused by the force .

The total deflection y is greater than the sum of the deflections arising from the separate action of the transverse load and the force S, since in the case of the action of only the force S on the beam, its deflections are equal to zero. Thus, in the case of longitudinal-transverse bending, the principle of independence of the action of forces is not applicable.

When a tensile force S acts on the beam (Fig. 8.13, b), the bending moment in the section with the abscissa

(24.13)

The tensile force S leads to a decrease in the deflections of the beam, i.e., the total deflections y in this case are less than the deflections caused by the action of only the transverse load.

In the practice of engineering calculations, longitudinal-transverse bending usually means the case of the action of a compressive force and a transverse load.

With a rigid beam, when the additional bending moments are small compared to the moment, the deflections y differ little from the deflections . In these cases, it is possible to neglect the influence of the force S on the magnitudes of the bending moments and the deflections of the beam and calculate it for central compression (or tension) with transverse bending, as described in § 2.9.

For a beam whose rigidity is low, the influence of the force S on the values ​​of the bending moments and deflections of the beam can be very significant and cannot be neglected in the calculation. In this case, the beam should be calculated for longitudinal-transverse bending, meaning by this the calculation for the combined action of bending and compression (or tension), performed taking into account the influence of the axial load (force S) on the bending deformation of the beam.

Consider the methodology for such a calculation using the example of a beam hinged at the ends, loaded with transverse forces directed in one direction and with a compressive force S (Fig. 9.13).

Substitute in the approximate differential equation of an elastic line (1.13) the expression of the bending moment M according to the formula (23.13):

[the minus sign in front of the right side of the equation is taken because, in contrast to formula (1.13), here the downward direction is considered positive for deflections], or

Hence,

To simplify the solution, let us assume that the additional deflection varies sinusoidally along the length of the beam, i.e. that

This assumption makes it possible to obtain sufficiently accurate results when a transverse load is applied to the beam, directed in one direction (for example, from top to bottom). Let us replace the deflection in formula (25.13) by the expression

The expression coincides with the Euler formula for the critical force of a compressed rod with hinged ends. Therefore, it is denoted and called the Euler force.

Hence,

The Euler force should be distinguished from the critical force calculated by the Euler formula. The value can be calculated using the Euler formula only if the rod flexibility is greater than the limit; the value is substituted into the formula (26.13) regardless of the flexibility of the beam. The formula for the critical force, as a rule, includes the minimum moment of inertia of the cross section of the rod, and the expression for the Euler force includes the moment of inertia about that of the main axes of inertia of the section, which is perpendicular to the plane of action of the transverse load.

From formula (26.13) it follows that the ratio between the total deflections of the beam y and the deflections caused by the Action of only the transverse load depends on the ratio (the magnitude of the compressive force 5 to the magnitude of the Euler force).

Thus, the ratio is a criterion for the rigidity of the beam in longitudinal-transverse bending; if this ratio is close to zero, then the stiffness of the beam is large, and if it is close to one, then the stiffness of the beam is small, i.e., the beam is flexible.

In the case when , deflection, i.e., in the absence of force S, deflections are caused only by the action of a transverse load.

When the value of the compressive force S approaches the value of the Euler force, the total deflections of the beam increase sharply and can be many times greater than the deflections caused by the action of only a transverse load. In the limiting case at, deflections y, calculated by formula (26.13), become equal to infinity.

It should be noted that formula (26.13) is not applicable for very large deflections of the beam, since it is based on an approximate expression for curvature. This expression is applicable only for small deflections, and for large deflections it must be replaced by the same curvature expression (65.7). In this case, the deflections y at at would not be equal to infinity, but would be, although very large, but finite.

When a tensile force acts on the beam, formula (26.13) takes the form.

From this formula, it follows that the total deflections are less than the deflections caused by the action of only the transverse load. With a tensile force S numerically equal to the value of the Euler force (i.e., at ), the deflections y are half the deflections

The largest and smallest normal stresses in the cross section of a beam with hinged ends at longitudinal-transverse bending and compressive force S are equal to

Consider a two-bearing I-section beam with a span. The beam is loaded in the middle with a vertical force P and is compressed by an axial force S = 600 (Fig. 10.13). Cross-sectional area of ​​the beam moment of inertia, moment of resistance and modulus of elasticity

The transverse braces connecting this beam with adjacent beams of the structure exclude the possibility of the beam becoming unstable in the horizontal plane (i.e., in the plane of least rigidity).

The bending moment and deflection in the middle of the beam, calculated without taking into account the influence of the force S, are equal to:

The Euler force is determined from the expression

Deflection in the middle of the beam, calculated taking into account the influence of the force S on the basis of formula (26.13),

Let us determine the greatest normal (compressive) stresses in the average cross section of the beam according to the formula (28.13):

from where after transformation

Substituting into expression (29.13) various values ​​of P (in), we obtain the corresponding stress values. Graphically, the relationship between determined by expression (29.13) is characterized by the curve shown in fig. 11.13.

Let us determine the allowable load P, if for the beam material and the required safety factor, therefore, the allowable stress for the material

From fig. 11.23 it follows that the stress occurs in the beam under load and the stress - under load

If we take the load as the permissible load, then the stress safety factor will be equal to the specified value. However, in this case, the beam will have an insignificant load safety factor, since stresses equal to from will arise in it already at Rot

Consequently, the load safety factor in this case will be equal to 1.06 (since e. is clearly insufficient.

In order for the beam to have a safety factor equal to 1.5 in terms of load, the value should be taken as the permissible value, while the stresses in the beam will be, as follows from Fig. 11.13, approximately equal

Above, the strength calculation was carried out according to the allowable stresses. This provided the necessary margin of safety not only in terms of stresses, but also in terms of loads, since in almost all cases considered in the previous chapters, the stresses are directly proportional to the magnitudes of the loads.

With longitudinal-transverse bending of the stress, as follows from Fig. 11.13 are not directly proportional to the load, but change faster than the load (in the case of a compressive force S). In this regard, even a slight accidental increase in load in excess of the calculated one can cause a very large increase in stresses and destruction of the structure. Therefore, the calculation of compressed-bent rods for longitudinal-transverse bending should be carried out not according to the allowable stresses, but according to the allowable load.

By analogy with formula (28.13), let us compose the strength condition when calculating the longitudinal-transverse bending according to the allowable load.

Compressed-curved rods, in addition to calculating the longitudinal-transverse bending, must also be calculated for stability.

UDC 539.52

LIMIT LOAD FOR A CLAMPED BEAM LOADED BY A LONGITUDINAL FORCE, ASYMMETRICLY DISTRIBUTED LOAD AND SUPPORT MOMENTS

I.A. Monakhov1, Yu.K. Bass2

department of building production Building faculty Moscow State Machine-Building University st. Pavel Korchagin, 22, Moscow, Russia, 129626

2Department of Building Structures and Constructions Faculty of Engineering Peoples' Friendship University of Russia st. Ordzhonikidze, 3, Moscow, Russia, 115419

The article develops a technique for solving problems of small deflections of beams made of an ideal rigid-plastic material under the action of asymmetrically distributed loads, taking into account the preliminary tension-compression. The developed technique is used to study the stress-strain state of single-span beams, as well as to calculate the ultimate load of beams.

Key words: beam, nonlinearity, analytical.

In modern construction, shipbuilding, mechanical engineering, the chemical industry and other branches of technology, the most common types of structures are rods, in particular beams. Naturally, to determine the real behavior of bar systems (in particular, beams) and their strength resources, it is necessary to take into account plastic deformations.

The calculation of structural systems, taking into account plastic deformations using the model of an ideal rigid-plastic body, is the simplest, on the one hand, and quite acceptable from the point of view of design practice requirements, on the other. If we keep in mind the region of small displacements of structural systems, then this is due to the fact that the bearing capacity (“ultimate load”) of ideal rigid-plastic and elastic-plastic systems turns out to be the same.

Additional reserves and a more rigorous assessment of the bearing capacity of structures are revealed as a result of taking into account geometric nonlinearity when they are deformed. At present, taking into account geometric nonlinearity in the calculations of structural systems is a priority not only from the point of view of the development of the theory of calculation, but also from the point of view of the practice of designing structures. Acceptability of solutions to problems of structural analysis under conditions of smallness

displacements is quite uncertain, on the other hand, practical data and properties of deformable systems allow us to assume that large displacements are realistically achievable. It suffices to point to the structures of construction, chemical, shipbuilding and machine-building facilities. In addition, the model of a rigid-plastic body means that elastic deformations are neglected, i.e. plastic deformations are much greater than elastic ones. Since displacements correspond to deformations, it is appropriate to take into account large displacements of rigid-plastic systems.

However, geometrically nonlinear deformation of structures in most cases inevitably leads to the occurrence of plastic deformations. Therefore, the simultaneous consideration of plastic deformations and geometrical nonlinearity in the calculations of structural systems and, of course, rod ones, is of particular importance.

This article deals with small deflections. Similar problems were solved in the works.

We consider a beam with pinched supports, under the action of a stepped load, edge moments and a preliminarily applied longitudinal force (Fig. 1).

Rice. 1. Beam under distributed load

The beam equilibrium equation for large deflections in dimensionless form has the form

d2 t / , h d2 w dn

-- + (n ± w)-- + p \u003d ^ - \u003d 0, dx ax ax

x 2w p12 M N ,g,

where x==, w=-, p=--, t=--, n=-, n and m are internal normal

I to 5xЪk b!!bk 25!!k

force and bending moment, p - transverse uniformly distributed load, W - deflection, x - longitudinal coordinate (origin on the left support), 2k - cross-sectional height, b - cross-sectional width, 21 - beam span, 5^ - yield strength material. If N is given, then the force N is a consequence of the action p at

available deflections, 11 = = , the line above the letters means the dimension of the values.

Consider the first stage of deformation - "small" deflections. The plastic section arises at x = x2, in it m = 1 - n2.

The expressions for the deflection rates have the form - deflection at x = x2):

(2-x), (x > X2),

The solution of the problem is divided into two cases: x2< 11 и х2 > 11.

Consider the case x2< 11.

For zone 0< х2 < 11 из (1) получаем:

Px 111 1 P11 k1p/1 m = + k1 p + p/1 -k1 p/1 -±4- + -^41

x - (1 - p2) ± a,

(, 1 , p/2 k1 p12L

Px2 + k1 p + p11 - k1 p11 -+ 1 ^

X2 = k1 +11 - k111 - + ^

Taking into account the occurrence of a plastic hinge at x = x2, we obtain:

tx \u003d x \u003d 1 - n2 \u003d - p

(12 k12 L k +/ - k1 - ^ + k "A

k, + /, - k, /, -L +

(/ 2 k/ 2 A k1 + /1 - k1/1 - ^ + M

Considering the case x2 > /1, we get:

for zone 0< х < /1 выражение для изгибающих моментов имеет вид

k p-p2 + car/1+p/1 -k1 p/1 ^ x-(1-P12)±

and for zone 11< х < 2 -

^ p-rC + 1^ L

x - (1 - p-) ± a +

(. rg-k1 p1-L

Kx px2 + kx p+

0, and then

I2 12 1 h h x2 = 1 -- + -.

The equality follows from the plasticity condition

where we get the expression for the load:

k1 - 12 + M L2

K1/12 - k2 ¡1

Table 1

k1 = 0 11 = 0.66

table 2

k1 = 0 11 = 1.33

0 6,48 9,72 12,96 16,2 19,44

0,5 3,24 6,48 9,72 12,96 16,2

Table 3

k1 = 0.5 11 = 1.61

0 2,98 4,47 5,96 7,45 8,94

0,5 1,49 2,98 4,47 5,96 7,45

Table 5 k1 = 0.8 11 = 0.94

0 2,24 3,56 4,49 5,61 6,73

0,5 1,12 2,24 3,36 4,49 5,61

0 2,53 3,80 5,06 6,33 7,59

0,5 1,27 2,53 3,80 5,06 6,33

Table 3

k1 = 0.5 11 = 2.0

0 3,56 5,33 7,11 8,89 10,7

0,5 1,78 3,56 5,33 7,11 8,89

Table 6 k1 \u003d 1 11 \u003d 1.33

0 2,0 3,0 4,0 5,0 6,0

0,5 1,0 2,0 3,0 4,0 5,0

Table 7 Table 8

k, = 0.8 /, = 1.65 k, = 0.2 /, = 0.42

0 2,55 3,83 5,15 6,38 7,66

0,5 1,28 2,55 3,83 5,15 6,38

0 7,31 10,9 14,6 18,3 21,9

0,5 3,65 7,31 10,9 14,6 18,3

By setting the load factor k1 from 0 to 1, the bending moment a from -1 to 1, the value of the longitudinal force n1 from 0 to 1, the distance /1 from 0 to 2, we obtain the position of the plastic hinge according to formulas (3) and (5), and then we obtain the value of the ultimate load according to formulas (4) or (6). The numerical results of the calculations are summarized in tables 1-8.

LITERATURE

Basov Yu.K., Monakhov I.A. Analytical solution of the problem of large deflections of a rigid-plastic pinched beam under the action of a local distributed load, support moments and longitudinal force // Vestnik RUDN University. Series "Engineering Research". - 2012. - No. 3. - S. 120-125.

Savchenko L.V., Monakhov I.A. Large deflections of physically nonlinear round plates. Bulletin of INGECON. Series "Technical Sciences". - Issue. 8(35). - St. Petersburg, 2009. - S. 132-134.

Galileev S.M., Salikhova E.A. Investigation of natural vibration frequencies of structural elements made of fiberglass, carbon fiber and graphene // Bulletin of INGECON. Series "Technical Sciences". - Issue. 8. - St. Petersburg, 2011. - P.102.

Erkhov M.I., Monakhov A.I. Large deflections of a prestressed rigid-plastic beam with hinged supports under a uniformly distributed load and edge moments // Bulletin of the Department of Building Sciences of the Russian Academy of Architecture and Building Sciences. - 1999. - Issue. 2. - S. 151-154. .

THE LITTLE DEFLECTIONS OF THE PREVIOUSLY INTENSE IDEAL PLASTIC BEAMS WITH THE REGIONAL MOMENTS

I.A. Monakhov1, U.K. Basov2

"Department of Building production manufacture Building Faculty Moscow State Machine-building University Pavla Korchagina str., 22, Moskow, Russia,129626

Department of Bulding Structures and Facilities Enqineering Faculty Peoples" Friendship University of Russia Ordzonikidze str., 3, Moskow, Russia, 115419

In the work up the technique of the decision of problems about the little deflections of beams from ideal hard-plastic material, with various kinds of fastening, for want of action of the asymmetrically distributed loads with allowance for of preliminary stretching-compression is developed. The developed technique is applied for research of the strained-deformed condition of beams, and also for calculation of a deflection of beams with allowance for of geometrical nonlinearity.

Key words: beam, analytic, nonlinearity.

Bending moment, transverse force, longitudinal force- internal forces arising from the action of external loads (bending, transverse external load, tension-compression).

Plots- graphs of changes in internal forces along the longitudinal axis of the rod, built on a certain scale.

Plot ordinate shows the value of the internal force at a given point of the section axis.

17. Bending moment. Rules (order) for constructing a diagram of bending moments.

Bending moment- internal force arising from the action of an external load (bending, eccentric compression - extension).

The order of plotting bending moments:

1. Determination of the support reactions of this design.

2. Determination of sections of this design, within which the bending moment will change according to the same law.

3. Make a section of this structure in the vicinity of the point that separates the sections.

4. Discard one of the parts of the structure, divided in half.

5. Find the moment that will balance the action on one of the remaining parts of the structure of all external loads and coupling reactions.

6. Apply the value of this moment, taking into account the sign and the selected scale, on the diagram.

Question number 18. Transverse force. Construction of a diagram of transverse forces using a diagram of bending moments.

Shear forceQ- internal force arising in the rod under the influence of external load (bending, transverse load). The transverse force is directed perpendicular to the axis of the rod.

The diagram of the transverse forces Q is built on the basis of the following differential dependence: ,i.e. The first derivative of the bending moment along the longitudinal coordinate is equal to the transverse force.

The sign of the shear force is determined based on the following position:

If the neutral axis of the structure on the diagram of moments rotates clockwise to the axis of the diagram, then the diagram of shear forces has a plus sign, if against - minus.

Depending on the diagram M, the diagram Q can take one form or another:

1. If the diagram of the moments has the form of a rectangle, then the diagram of the transverse forces is equal to zero.

2. If the diagram of the moments is a triangle, then the diagram of the transverse forces has the form of a rectangle.

3. If the diagram of moments has the form of a square parabola, then the diagram of transverse forces has a triangle and is built according to the following principle

Question number 19. Longitudinal strength. A method for constructing a plot of longitudinal forces using a plot of transverse forces. Sign rule.

Shear force N- internal force arising from central and eccentric tension-compression. The longitudinal force is directed along the axis of the rod.

In order to build a diagram of longitudinal forces, you need:

1. Cut out the knot of this design. If we are dealing with a one-dimensional structure, then make a section in the section of this structure that interests us.

2. Remove from the Q diagram the values ​​of the forces acting in the immediate vicinity of the cut node.

3. Give direction to the transverse force vectors, based on what sign the given transverse force has on the Q diagram according to the following rules: if the transverse force has a plus sign on the Q diagram, then it must be directed so that it rotates this node clockwise, if the shear force has a minus sign, counterclockwise. If an external force is laid to the knot, then it must be left and the knot should be considered together with it.

4. Balance the knot with longitudinal forces N.

5. Rule of signs for N: if the longitudinal force is directed towards the section, then it has a minus sign (works in compression). If the longitudinal force is directed away from the section, it has a plus sign (works in tension).

Question number 20M, Q, N.

1. In the section where the concentrated force F is applied, on the diagram Q there will be a jump equal to the value of this force and directed in the same direction (when plotting the diagram from left to right), and the diagram M will have a fracture directed towards the force F .

2. In the section where the concentrated bending moment is applied on the diagram M, there will be a jump equal to the value of the moment M; there will be no change in the Q plot. In this case, the direction of the jump will be down (when plotting from left to right), if the concentrated moment acts clockwise, and up, if counterclockwise.

3. If in the area where there is a uniformly distributed load, the shear force in one of the sections is zero (Q=M"=0), then the bending moment in this section takes on the extreme value M extra - maximum or minimum (here the tangent to the diagram M horizontal).

4. To check the correctness of the construction of the diagram M, you can use the method of cutting nodes. In this case, the moment applied in the knot must be left when cutting the knot.

The correctness of plotting Q and M can be checked by duplicating the method of cutting nodes using the section method and vice versa.

Posted on 13/11/2007 12:34 pm

So beam

1. beam; run; crossbar

2. beam

3. timber; crossbar, traverse

4. rocker (weights)

5. boom or boom (crane) handle

beam and column - beam-rack construction; end [end] frame of a metal frame

beam carrying transverse loads - beam loaded with transverse forces [transverse load]

beam fixed at both ends - beam with pinched ends

beam loaded unsymmetrically - a beam loaded with an asymmetric load (acting outside the plane of symmetry of the section and causing oblique bending)

beam made of precast hollow blocks - a beam assembled from hollow [box-shaped] sections (with a tension of longitudinal reinforcement)

beam on elastic foundation - beam on an elastic foundation

beams placed monolithically with slabs - beams concreted together with floor slabs

beam precast on site

beam subjected to (both) transverse and axial loads - a beam loaded with transverse and longitudinal forces; beam subjected to transverse and axial loads

beam supported on a girder - a beam based on a run; beam supported by purlin

beam with overhangs - cantilever beam

beam with rectangular section - rectangular beam

beam with symmetrical (cross) section - beam of symmetrical (cross) section

beam with unsymmetrical (cross) section - beam of asymmetrical (cross) section

beam of constant depth — beamconstant height

beam of one span - single-span beam

beam of uniform strength

anchor beam - anchor beam

angle beam - metal corner; angle steel

annular beam - annular beam

arch(ed) beam

2. convex beam with belts of different curvature

baffle beam - visor beam

balance beam - balance beam; balance beam

bamboo-reinforced concrete beam - concrete beam reinforced with bamboo

basement beam - basement beam

bedplate beam - beam [edge] of the base plate

bending test beam - beam (-sample) (beam-sample¦ beam) for bending testing

Benkelman beam - Benkelman beam, deflection meter

bind beam - pile nozzle

bisymmetrical beam - a beam with a section symmetrical about two axes

block beam - prestressed reinforced concrete beam from separate blocks [sections] (connected by reinforcement tension)

bond beam - connecting [reinforcing] beam (reinforced concrete beam that reinforces a stone wall and prevents the formation of cracks in it)

boundary beam - rafter beam; edge beam

box beam - box-shaped beam; box beam

braced beam - trussed beam

bracing beam - bracing beam; spacer

brake beam - brake beam

breast beam - jumper [beam] over a wide opening in the wall

brick beam - ordinary brick jumper (reinforced with steel bars)

bridge beam - bridge beam, bridge run

bridging beam - cross beam (between floor beams)

broad-flange(d) beam

buffer beam - buffer beam, bumper

built-in beam - built-in (in masonry) beam; beam with pinched ends

built-up beam - composite beam

camber beam

1. beam with a convex upper chord

2. beam, slightly curved upwards (to create a building lift)

candle beam - a beam that supports candles or lamps

cantilever beam

1. cantilever beam, console

2. beam with one or two consoles

capping beam

1. cap; nozzle (bridge supports)

2. grillage strip pile foundation

cased beam

1. steel beam embedded in concrete

2. steel beam with an outer shell (usually decorative)

castellated beam - perforated beam

castella Z beam - perforated Z profile

ceiling beam - ceiling beam; beam protruding from the ceiling; false ceiling beam

channel beam - channel beam

chief beam - main beam, run

circular beam - circular beam

collar beam - increased tightening of hanging rafters

composite beam - composite beam

compound beam - composite beam

conjugate beam - conjugate beam

constant-section beam - beam of constant section

continuous beam - continuous beam

crane lifting beam

crane runway beam

cross beam

1. cross beam

2. hydr. hat beam

curved beam

1. beam with a curved axis (in the plane of loading)

2. curved (in plan) beam

deck beam - a beam that supports the deck; deck rib

deep beam - beam-wall

double-T beam

1. double "T" shaped precast concrete beam

2. precast concrete panel with two ribs

doubly symmetrical beam - a beam of symmetrical section with two axes of symmetry

dragging beam - a piece of timber that supports the slanting rafter leg at the bottom; trimmer

drop-in beam - hanging beam; beam supported (at both ends) by cantilevers

eaves beam - under rafter beam (outer row of columns)

edge beam

1. edge beam

2. side stone

elastically restrained beam - elastically restrained beam, beam with elastically restrained ends

encastre beam - a beam with pinched ends

externally reinforced concrete beam

false beam - false beam

fish(ed) beam

1. wooden composite beam with side metal butt plates

2. beam with convex curvilinear chords

fixed(-end) beam - beam with fixed ends

flitch(ed) beam - composite wood-metal beam (consisting of a medium steel strip and two side boards bolted together)

floor beam

1. floor beam; floor beam, lag

2. transverse beam of the bridge carriageway

3. landing beam

footing beam - tightening the raftertrusses (at the level of the ends of the rafter legs)

foundation beam - foundation beam, rand beam

framework beam - crossbar of the frame (frame structure)

free beam - free-supported beam on two supports

gantry beam - crane beam

Gerber beam - hinged beam, Gerber beam

glue(d) laminated (timber) beamglued beam

grade beam - foundation beam, rand beam

grillage beams - grillage beams

ground beam

1. foundation beam, grillage; rand beam

2. lower trim of the frame wall; sill

H beam - wide-shelf beam, wide-shelf I-beam

hammer beam

haunched beam - beam with haunches

high strength concrete beam - a beam made of high-strength reinforced concrete

hinged beam - hinged beam

hollow beam - hollow beam; box [tubular] beam

hollow prestressed concrete beam - hollow prestressed concrete beam

horizontally curved beam - curved beam

hung-span beam - multi-span cantilever beam, Gerber beam

hybrid beam - steelcomposite beam (made of steels of different grades)

I beam - I-beam, I-beam

inverted T beam - tee (reinforced concrete) beam with a wall facing up

jack beam - rafter beam

jesting beam - decorative [ornamental] beam

joggle beam - a composite beam of wooden beams connected in height by reciprocal protrusions and grooves

jointed beam

1. monolithic reinforced concrete beam, concreted with butt joints

2. precast concrete beam, assembled from separate sections

keyed beam - a beam of bars with connections on prismatic keys

L beam - L-shaped beam

laminated beam - laminated beam

laterally-unsupported beam - a beam without lateral bracing

lattice beam - lattice [through] beam

leveling beam - a rail for checking the evenness of the road surface

lifting beam - lifting beam

link beam - jumper (above the opening in the wall)

longitudinal beam - longitudinal beam

main beam - main beam

modified I beam - precast concrete beam with collars protruding from the upper flange (for connection with the upper cast-in-situ reinforced concrete slab)

multispan beam - multispan beam

nailed beam - a composite wooden beam with nailed joints; nail beam

needle beam

1. beam for temporary wall support (when strengthening the foundation)

2. upper thrust run of the spoke shutter

outrigger beam - beam of an outrigger [additional] support (crane, excavator)

overhead runway beam - beam crane

parallel flanges beam - beam with parallel mi shelves

partition beam - a beam that carries a partition

precast beam - precast concrete beam

precast toe beam - precast support beam (e.g. supporting brick facing)

prestressed concrete beam - prestressed concrete beam

prestressed precast concrete beam

prismatic beam - prismatic beam

propped cantilever beam - a beam with one pinched and other hinged ends

rectangular beam - rectangular beam

reinforced concrete beam - reinforced concrete beam

reinforced floor beam - reinforced concrete ribbed floor beam

restrained beam - beam with pinched ends

ridge beam - ridge beam, ridge beam

ring beam - ring beam

rolled beam with cover plates

rolled I beam - rolled [hot-rolled] I-beam

rolled steel beam - rolled steel beam

roof beam - roof beam

runway beam - beam crane

sandwich beam - composite beam

secondary beam - secondary [auxiliary] beam

simple beam - simple [single-span freely supported] beam

simple-span beam - single-span beam

simply supported beam - freely supported beam

single web beam - (composite) beam with one web

slender beam

soldier beam - steel rack for fastening the walls of trenches or bolts

spandrel beam

1. foundation beam, rand beam

2. frame beam supporting the [bearing] outer wall

spreader beam - distribution beam

statically determinate beam - statically determinate beam

statically indeterminate beam - statically indeterminate beam

steel beam - steel beam

steel binding beam - steel spacer, steel connecting beam

stiff beam - rigid beam

stiffening beam - stiffening beam

straight beam - straight [rectilinear] beam

reinforced beam - reinforced beam

strut-framed beam - trussed beam

supporting beam - supporting [supporting] beam

suspended-span beam - suspended [hanging] beam of a cantilever span (bridge)

T beam - tee beam

tail beam - a shortened wooden floor beam (at the opening)

tee beam - tee beam

tertiary beam - a beam supported by auxiliary beams

test beam

through beam - continuous multi-span beam

tie beam

1. tightening (rafters, arches) at the level of supports

2. distribution foundation beam (distributes off-center load)

top beam - increased tightening of the rafters

top-running crane beam - supporting crane beam (moving along the upper belt of crane beams)

transverse beam - transverse beam

trolley I beam - reeling (I-beam) beam

trussed beam

1. truss with parallel chords, girder truss

2. truss beam

uniformly loaded beam - a beam loaded with a uniformly distributed load; uniformly loaded beam

unjointed beam

1. monolithic reinforced concrete beam without working seam

2. steel beam without a joint in the web

upstand beam - ribbed floor beam protruding above the slab

valley beam - rafter beam of the middle row of columns; valley supporting beam

vibrating beam

vibrating leveling beam

vibratory beam

wall beam - steel anchor for attaching wooden beams or ceilings to the wall

welded I beam - welded I-beam

wide-flanked beam - wide-shelf beam, wide-shelf I-beam

wind beam - increased tightening of hanging rafters

wood I beam - wooden I-beam

AZM

Used photo from the materials of the press service of ASTRON Buildings