## Example 81 Encastre Beam

An encastre beam is 8.0 m long and supports an unfactored load of 40 kN m as shown in Figure 8.3. Assuming that the yield stress py 460 N mm2 and a load factor X 1.7, determine the required plastic moment of resistance and plastic section modulus. The collapse load (40.0x1.7) 68.0 kN m The number of hinges required to induce collapse (lD+1) 3 (see Figure 8.1) The possible hinge positions are at the supports A and B and within the region of a distributed load since these are the positions where...

## Problem 411

4.3.2 Solutions McCaulay's and Equivalent UDL Methods for Topic Sialics Ily IIderm mule Htjnit- Ikllutlioii I'mlilciiiNiimkr 4,11 Tu cNd. 2 The hJiilitm of lite maximum dcflection si (he poiml of ens slope cum be determined from iMluinfrtrl 4) is frlli'uvs Aiiunid lli.i zero -lop Oscui wile J.i) S.iS Jt.O utd ik'jle I xvli it ill uLiil' Eto a + 24.3 -x> - 7.51* - if - - 2j* - 2S4.32 Solve the nsulth) cubic equadmi by Utal error ues i 3im (t . dtgblty Do the left oflhe mid-span) + 24. JS< 3-...

## Conditions for Full Collapse

There are three conditions which must be satisfied to ensure full collapse of a structure and the identification of the true collapse load, they are (i) the mechanism condition in which there must be sufficient plastic hinges to develop a mechanism, (i.e. number of plastic hinges > ID+1 ), (ii) the equilibrium condition in which the bending moments for any collapse mechanism must be in equilibrium with the applied collapse loads, (iii) the yield condition in which the magnitude of the bending...

## Jointed Frames

Topic Uni.1 Lund Met hud lor Singly-liei uihI iil) i'in jujnlid Krmttts Pnlilrni Nufflhtf 3,21 Fil'tNc. I Thccrpss-Sictitxiiil arc of members AEi, < ji EFmd F-'Ci is equal to 2W mm The cmos-sectfawl tcj or all oilier imbas is eqnii to 500 rum*. The Nippon at C sitdis by 11 nun, 205 kNAnra1 Sin -13-11 4.243 - 0.707 CtaQ < 3.0 4.20) - 0.707 AEm (200 x 205) 4lJ0v 10t FiN Conldtr (be vertical reaflloB support G (4 be redundant. The cquivaknl stem is lie suptrpiiiilian oi the sialics I ty...

## Instantaneous Centre of Rotation

In more complex frames it is convenient to use the 'instantaneous centre of rotation method' when developing a collapse mechanism. The technique is explained below in relation to a simple rectangular portal frame and subsequently in Example 8.7. Consider the asymmetric rectangular frame shown in Figure 8.25 in which there are two independent mechanisms, one beam and one sway. The frame requires three hinges to cause collapse. Both mechanisms can combine to produce a collapse mechanism with...

## Section Classification

In design codes the compression elements of structural members are classified into four categories depending upon their resistance to local buckling effects which may influence their load carrying capacity. The compression may be due to direct axial forces, bending moments, or a combination of both. There are two distinct types of element in a cross-section identified in the code 1. Outstand elements elements which are attached to an adjacent element at one edge only, the other edge being free,...

## E

In the case of the web of a hybrid section e should be based on the design strength pyf of the flanges. In addition to e, some limiting values also include parameters r1 and r2 which are stress ratios, these are not considered further here. The type of section e.g. universal beam, universal column, circular hollow sections, welded tubes, hot finished rectangular hollow sections, cold formed rectangular hollow sections etc. also influences the classification. The classifications given in codes...

## Kinematic Method

Consider each independent mechanism separately. tntcnial Work tone Mp (0+a) + Mt 0+ 0) + External Work Dome 1 20.0 x fit i ) + (20.0 x + 20.0 k j wM) (20.0 V 3.00 + (20.0 Si 6.0(3) + 20.0 x + ( 0.0 x S O0 Internal Wort Menial Work 6.70 2WG . Mt 42.9 kNm Combined Mechanism (iv) 2xmechanism (i) +mechanism (iii) which eliminates a hinge In'cnitil Work DOW - ty (0+ G) + M (9 + ft + 2.0MJJJ) A r (4 Oty + Mr (B + l .25 0) + 2 OA p 1.2 tfSMfi In'cnitil Work DOW - ty (0+ G) + M (9 + ft + 2.0MJJJ) A r...

## Problem 821

8.16 Solutions Plastic Analysis Rigid-Jointed Frames 3 topic I'LiMi Analysis - Ki iil JoluIciI frame* ) l'r hltm Number Kiiicmilic Mcth d Internal Work Done- l ier.i il Work Done Internal Work Dene M Ism-mill Worfi Done +A j ( * ,mo.i7i > (-is xiae i ism III Cninbined Henni UCl> nd Jinny In mechanism II the rxnm ion at joint I n meehamsnt 111 lite naiitw at ji> i at l Adding ctiuutirms for Mechiinisr ii l + ItJ - ill lowing for the hinge eliminated at ioinl I 12 Topic rlaslic Ani1> is -...

## Rigid Jointed Frames

Rigid-jointed frames are framed structures in which the members transmit applied loads by axial, shear, and bending effects. There are basically two types of frame to consider (i) statically determinate frames see Figure 5.1(a) and (ii) statically indeterminate frames see Figure 5.1(b). (to Stat teil ly-imtete mi I rate Fratries (to Stat teil ly-imtete mi I rate Fratries Rigid-joints (moment connections) are designed to transfer axial and shear forces in addition to bending moments between the...

## Propped Cantilevers

The fixed-end moment for propped cantilevers (i.e. one end fixed and the other end simply supported) can be derived from the standard values given for encastre beams as follows. Consider the propped cantilever shown in Figure 4.78, which supports a uniformly distributed load as indicated. The structure can be considered to be the superposition of an encastre beam with the addition of an equal and opposite moment to MB applied at B to ensure that the final moment at this support is equal to...

## Example 53 NoSway Rigid Jointed Frame

A rigid-jointed, two-bay rectangular frame is pinned at supports A, D and E and carries loading as indicated in Figure 5.15 Given that supports D and E settle by 3 mm and 2 mm respectively and that EI 102.5x103 kNm2 i) sketch the bending moment diagram and determine the support reactions, ii) sketch the deflected shape (assuming axially rigid members) and compare with the shape of the bending moment diagram (the reader should check the answer using a computer analysis solution). The final...

## Unit Load Method for Deflection of Beams

In Chapter 3, Section 3.5 the deflection of pin-jointed frames was calculated using the concept of strain energy and Castigliano's 1st Theorem. This approach can also be applied to structures such as beams and rigid-jointed frames in which the members are primarily subject to bending effects. In the case of pin-jointed frames the applied loads induce axial load effects and subsequent changes in the lengths of the members. In the case of beams and rigid-jointed frames, the corresponding applied...

## Example 51 Statically Determinate Rigid Jointed Frame

A asymmetric portal frame is supported on a roller at A and pinned at support D as shown in Figure 5.3. For the loading indicated i) determine the support reactions and ii) sketch the axial load, shear force and bending moment Apply the three equations of static equilibrium to the force system +Vir tLFV - 0 VA - 12.0 - ( 16,0 X 5,0) - 12.0 + V > - 0

## Elastic Assumptions

The laws of structural mechanics are well established in recognised elastic theory using the following assumptions the material is homogeneous which implies its constituent parts have the same physical properties throughout its entire volume. the material is isotropic which implies that the elastic properties are the same in all directions. the material obeys Hooke's Law, i.e. when subjected to an external force system the deformations induced will be directly proportional to the magnitude of...

## R

Distribution Fatten Juinl tiA- f')-ii In chif case, nincu lli-iire is mity oiii iiliinul join . only balancing OpetStiw i il1 Ml anwvtf wjll bg rcguirtd tohg Ihc clistrihulinn Ofthg momtnls._ Topic Moment Dislribuliun - fcSniy H.ij irt-Joiiitrd Krumrs I'rtliltiH uitibtf f it Fi I Topic Moment Dislribuliun - fcSniy H.ij irt-Joiiitrd Krumrs I'rtliltiH uitibtf f it Fi I * Sirac support C is nUcr, Uic iixcd-eiMj momenls flit (.ifa.- - 0JWa) ,n J aitd odd si C- tiv - - f- 2< U - (0,5 m 25 ,5) -...

## U M yrt

I. il n it iii Irnderntss < .2 .Ttif fs 17.15 h OVIOOO - 3,5(17.21 - 17.15V 1000 -0.0002 . A+fo + Ort _ 375 + (Q.0D0a+l)6a3l.l , Cflttprewtoq r sklan P, iptxAl) ( 6 .C x x loV'O1 - 4070 tN Topic liuck in > Inability PruMrm SufflW ( 2 Noli1 SiiHL' (lie nunc ciimu is used fur lmCei iho A-A nnd he B-R jL ts (i.e. a 3.5), (hp wmpiiisien resislanii will iornespond (l e whh the liigjietf sifiidimess value, ije. ihu on uhieh prodtiiei ihe kwesi u wlut Cuuldt if Iht- A-A a* is Buckling IthjMli Lm S...

## Application ofthe Method

All of the concepts outlined in Sections 4.7.1 to 4.7.8 are used when analysing indeterminate structures using the method of moment distribution. Consider the two separate beam spans indicated in Figure 4.87. Since the beams are not connected at the support B they behave independently as simply supported beams with separate reactions and bending moment diagrams, as shown in When the beams are continuous over support B as shown in Figure 4.89(a), a continuity moment develops for the continuous...

## Centre ofGravity and Centroid

The centre of gravity of an object is the point through which the force due to gravity on the total mass of the object is considered to act. The corresponding position on a plane surface (i.e. relating to the cross-sectional area) is known as the centroid both are indicated in Figure 2.19 Consider the cross-section A shown in Figures 2.20(a) and (b) which can be considered to be an infinite number of elemental areas each equal to 3A. The 1st moment of area (i.e. areaxperpendicular lever arm) of...

## Example 31 Pin Jointed Truss

A pin-jointed truss supported by a pinned support at A and a roller support at G carries three loads at joints C, D and E as shown in Figure 3.2. Determine the magnitude and sense of the forces induced in members X, Y and Z as indicated. Step 1 Evaluate the support reactions. It is not necessary to know any information regarding the frame members at this stage other than dimensions as shown in Figure 3.3, since only externally applied loads and reactions are involved. Apply the three equations...

## Critical Stress ocritical

In each case described in Sections 6.2.1 to 6.2.3 the critical load Pc (i.e. critical stressx cross-sectional area) must be estimated for design purposes. Since the critical stress depends on the slenderness it is convenient to quantify slenderness in mathematical terms Le is the effective buckling length, r is the radius of gyration and I and A are the second moment of area about the axis of bending and the cross-sectional area of the section as before. 6.3.1 Critical Stress for Short Columns...

## Example 83 Propped Cantilever

A propped cantilever is L m long and supports a collapse load of w kN m as shown in Figure 8.7. Determine the position of the plastic hinges and the required plastic moment The number of hinges required to induce collapse (ID+1) 2 (see Figure 8.1) The possible hinge positions are at the support A and within the region of a distributed load since these are the positions where the maximum bending moments occur. In this case the maximum moment under the distributed load does not occur at mid-span...

## Structural Degreesof Freedom

The degrees-of-freedom in a structure can be regarded as the possible components of displacements of the nodes including those at which some support conditions are provided. In pin-jointed, plane-frames each node, unless restrained, can displace a small amount 5 which can be resolved in to horizontal and vertical components 5H and 5V as shown in Figure 1.29. Each component of displacement can be regarded as a separate degree-of-freedom and in this frame there is a total of three...

## Jk

Topic Pla-JoJntod Frsrncs - JoJdI Rr otution in 'i.o +y.cJ W lrr -Jj,0i+3.( i -4,243 m Consider Eri.mi le CCOt Sil0 (3j(MM& 7) -0.il (, iW (9.OT.4& ) O.iWJ Smfi (MWWi ) Q.w Cot0- (3 jOfiMi ) (, in 'i.o +y.cJ W lrr -Jj,0i+3.( i -4,243 m Consider Eri.mi le CCOt Sil0 (3j(MM& 7) -0.il (, iW (9.OT.4& ) O.iWJ Smfi (MWWi ) Q.w Cot0- (3 jOfiMi ) (, Consider Erin II Ilz > ' Siu i ( m b) Lsa - itfSinjS- 6.0 * 0.9J9) m Collider irLutfk DCD' Sin0 (W t*) - - ti)ii f-m * Dl314> -Um CwO ( kd-...

## Moment Distribution for Rigid Jointed Frames with Sway

The frames in Section 5.2 are prevented from any lateral movement by the support conditions. In frames where restraint against lateral movement is not provided at each level, unless the frame, the supports and the loading are symmetrical it will sway and consequently induce additional forces in the frame members. Consider the frame indicated in Figure 5.18(a) in which the frame, supports and applied load are symmetrical. Consider the same frame in which the load has been moved such that is now...

## Strain Energy Bending Load Effects

A simply-supported beam subjected to a single point load is shown in Figure 4.39. An incremental length of beam dx, over which the bending moment can be considered to be constant, is indicated a distance 'x' from the left-hand support. where R is the radius of curvature and 1 R is the curvature of the beam, i.e. the rate of Assuming the moment is applied to the beam gradually, the relationship between the moment and the change in slope is as shown in Figure 4.40. The external werk-deni in he...

## Example 37 Singly Redundant Pin Jointed Frame

Using the data given, determine the member forces and support reactions for the pin-jointed frame shown in Figure 3.29. The cross-sectional area of all members is equal to 140 mm2. Assume E 205 kN mm2 All member lengths L 3.0 m AE (140x205) 28.7x103 kN Sin60 0.866 Cos60 0.5 Consider (he ipplied load sis two components 30.05inC0 25.9-3 kN The degree of indeterminacy ID (m+r)-2n (8+7)-(2x7) 1 Consider the vertical reaction at support F to be redundant. The equivalent system is the superposition...

## Distribution Factors

Consider a uniform two-span continuous beam, as shown in Figure 4.84. If an external moment M is applied to this structure at support B it will produce a rotation of the beam at the support part of this moment is absorbed by each of the two spans BA and BC, as indicated in Figure 4.85. Allied momtflt Rolalion ofbeam at suppon -W + The proportion of each moment induced in each span is directly proportional to the Allied momtflt Rolalion ofbeam at suppon -W + The proportion of each moment induced...

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## The invert of a matrix can be defined as

Where adj B is the adjoint of matrix B and is equal to the transpose of the co-factor The co-factor matrix is given by replacing each element in the matrix by its' co-factor, fl is the determinant of matrix j ,which can be calculated from PI - + fti.i ( - ( w x - i.iK hy * j) - ( j * KM

## Moment Distribution for NoSway Rigid Jointed Frames

The principles of moment distribution are explained in Chapter 4 in relation to the analysis of multi-span beams. In the case of rigid-jointed frames there are many instances where there more than two members meeting at a joint. This results in the out-of-balance moment induced by the fixed-end moments being distributed among several members. Consider the frame shown in Figure 5.11

## KNm

* Since support E is pinned, the fixed-end moments are (MCE-0.5MEC) at C and zero at E. Consider the settlement of supports D and E AB 3.0 mm and BC 1.0 mm Note the relative displacement between B and C i.e. 5bc (3.0-2.0) 1.0 mm Member bc -ftS + 25.6 + li.7 kWm - + J 7.8 + 25.6 + 434 kNm

## Example 54 Rigid Jointed Frame with Sway Frame

A rigid-jointed frame is fixed at support A, pinned at support H and supported on a roller at F as shown in Figure 5.23. For the relative EI values and loading given i) sketch the bending moment diagram, ii) determine the support reactions and iii) sketch the deflected shape (assuming axially rigid members) and compare with the shape of the bending moment diagram, (the reader should check the answer using a computer analysis solution). EI 10xl03 kNm2 rim i irn irim i 11 rim Bti m i 111 mi m i m...

## Statical Indeterminacy

Any plane-frame structure which is in a state of equilibrium under the action of an externally applied force system must satisfy the following three conditions the sum of the horizontal components of all applied forces must equal zero, the sum of the vertical components of all applied forces must equal zero, the sum of the moments (about any point in the plane of the frame) of all applied forces must equal zero. This is represented by the following 'three equations of static equilibrium' Sum of...

## Example 21 Plastic Crosssection Properties Section

Determine the position of the plastic neutral axis -Aplastic, the plastic section modulus Sxx and the shape factor u for the welded section indicated in Figure 2.46. (i) Position of plastic neutral axis (- plastic) A (90x10)+(90x15) 2250 mm2 A 2 (2250 2) 1125 mm2 (ii) Plastic section modulus (Sxx) (1st moment of area about the plastic neutral axis) Sxx (90x10)x20 + (15x15)x7.5) + (75x15)x37.5) 61.875x10 Shape factor u 1 u

## Secondary Stresses

As mentioned in Section 6.1, buckling is due to small imperfections within materials, application of load etc., which induce secondary bending stresses which may or may not be significant depending on the type of compression element. Consider a typical column as shown in Figure 6.5 in which there is an actual centre-line, reflecting the variations within the element, and an assumed centre-line along which acts an applied compressive At any given cross-section the point of application of the...

## Direct Stiffness Method

7.1 Direct Stiffness Method of Analysis The 'stiffness' method of analysis is a matrix technique on which most structural computer analysis programs are based. There are two approaches the indirect and the direct methods. The direct method as illustrated in this chapter requires the visual recognition of the relationship between structural forces displacements and the consequent element forces displacements induced by the applied load system. The indirect method is primarily for use in the...

## Problems Method of Tension Coefficients

The pin-jointed space-frames shown in Problems 3.11 to 3.16 have three pinned supports at A, B and C as indicated. In each case the supports A, B and C are in the same plane. Using the data given determine when the frames are subjected to the loading indicated. 3.4.4 Solutions Method of Tension Coefficients Topic I'in-Jiwiii'rt frame* - Method nf Ten mil Corlticirnts FroM Nunilwr A 11 Ns. I Not 4*1 temifiii cncfYicitni vsitiig.* indicate (tnsinn ntmbcn -vt ttiitot) cotllickut vihts hulif Jk...

## Problems Moment Distribution Continuous Beams

A series of continuous beams are indicated in Problems 4.28 to 4.32 in which the relative EI values and the applied loading are given. In each case i) determine the support reactions, ii) sketch the shear force diagram and iii) sketch the bending moment diagram. 4.7.12 Solutions Moment Distribution Continuous Beams Lirpir Moment Dislrilfuliuit - Cunlinuous lk- tins PI IIItki No whir iJI Putt Sc. J - 13,9 + M.6 - (J,0 x P hkj) 0 ffA - + 2.(58 kN Consider th( verticil equilibrium of the beam +vc...

## Perry Formula Strut

Buckling length Lb_b (1.0 x 4.0) 4.0 m > (1.0x3.0) 3.0 m > (0.85 x 5.0) 4.25 m The effective buckling length . . Lb-b 4.25 m Since 1.4Xc is the largest value this should be used to determine the value of pc using the Examples in structural analysis 510 limfllqg sLcndamcss h O.Sf J )65 0,2 x i x 205000 275)91 -17.15 r) fj(jJ - yiOQO - 5.5(63.02 - 7.1SytO0O 0.257 Critical value ejPjiir 1 W.4 M mmL Compression resistance K (to* AJ 19SA X 21* X ilfytf Ml kN A similar approach is taken when...

## Method of Joint Resolution

Considering the same frame using joint resolution highlights the advantage of the method of sections when only a few member forces are required. In this technique (which can be considered as a special case of the method of sections), sections are taken which isolate each individual joint in turn in the frame, e.g. In Figure 3.6 four sections are shown, each of which isolates a joint in the structure as indicated in Figure 3.7. Since in each case the forces are coincident, the moment equation is...

## Gable Mechanism

Another type of independent mechanism which is characteristic of pitched roof portal frames is the Gable Mechanism, as shown in Figure 8.23 with simple beam and sway mechanisms. In the beam and gable mechanisms the rafter of the frame is sloping and it is necessary to evaluate the displacement in the direction of the load. i.e. not necessarily perpendicular to the member as in previous examples. Consider the typical sloping member ABC shown in Figure 8.24(a) which is subject to a horizontal and...

## Simple Stress and Strain

The application of loads to structural members induce deformations and internal resisting forces within the materials. The intensity of these forces is known as the stress in the material and is measured as the force per unit area of the cross-sections which is normally given the symbol a when it acts perpendicular to the surface of a cross-section and t when it acts parallel to the surface. Different types of force cause different types and distributions of stress for example axial stress,...

## Example 421 Threespan Continuous Beam

A non-uniform, three span beam ABCDEF is fixed at support A and pinned at support F, as illustrated in Figure 4.90. Determine the support reactions and sketch the bending moment diagram for the applied loading indicated. The first step is to assume that all supports are fixed against rotation and evaluate the The values of the fixed-end moments for encastre beams are given in Appendix 2. The values of the fixed-end moments for encastre beams are given in Appendix 2. * Since support F is pinned,...

## Example 52 Statically Determinate Rigid Jointed Frame

A pitched-roof portal frame is pinned at supports A and H and members CD and DEF are pinned at the ridge as shown in Figure 5.6. For the loading indicated i) determine the support reactions and ii) sketch the axial load, shear force and bending moment Apply the three equations of static equilibrium to the force system in addition to the VA 5.0 - (12.0 x 4,0) - 25.0 - 35.0 -20,0 + VU 0 E J A+ 12,0 + 8,0 + 5,0 + 8,0 + H 0 +vc J) - 0

## X

IMslilonaftwIvriMiinaLrtri - Hi only Ilf ' ItiiiiiixiancrUfrwniaLiod - i- orlv Consider ij 1.0 i < J Aj 0 -S- roiiiiinjfUrDirirmnlwn - ii oflf e.g. Fi (ki,i 8i)+(ki,2 S2)+(ki,3 53)+(kM 64) The bending moment at any position 'x' along the element can be expressed as Boundary Conditions wfoert Jf**(h (kflcstion < 0 nnd stopc Substitute for x and 0 in equation (2) (x 0, 0 -i.O) Substitute for x and 5 in equation (3) (x 0, 5 0) Substitute for x and 5 in equation (3) (x 0, 5 0) Substitute for x...

## Example 32 Two Dimensional Plane Truss

Consider the pin-jointed, plane-frame ABC loaded as shown in Figure 3.9. Construct a table in terms of tension coefficients and an X Y co-ordinate system as shown in Table 3.1. The equilibrium equations are solved in terms of the 't' values and hence the member forces and support reactions are evaluated and entered in the table as shown in Table 3.1. Consider joint B There are only two unknowns and two equations, hence Adding both equations substitute for in tlic first equation rc -4.76 Force...

## Example 85 Frame

An asymmetric uniform, frame is pinned at supports A and G and is subjected to a system of factored loads as shown in Figure 8.15. Assuming the Xverticai.ioad 1.7 and -horizontal loads 1.4 determine the required plastic moment of resistance Mp of the section. Number of diienLes-of-iniliiwrtmnasy It, (3w + r) - 3 1(3 * 3) + < l) - (3 x 4) 1 Number ofpoi& iMe hinjie positions ft 5 (B, C. D, E and F) (i i. 3 beam ntcchnnisrns an< 11 sway mcchanism) Consider each independent mechanism...

## Image Of Simply Support Beam With Central Point Load Hinge Mechanism

The load deflects zero at the supports and 5 at the centre Average displacement of the load 2 4 The Internal Work Done in developing the hinges is found from the product of the moment induced (i.e. Mp) and the amount of rotation (e.g. 0) for each hinge. Internal Work Done - Moment* Rotation Tor each hinge position The External Work Done by the applied load system is found from the product of the load and the displacement for each load. KMcrctnl Work Done (Load x Displacement) > internal Woffc...

## What Is Rn In Structural Analysis

This equation represents the average value of stress in the cross-section which will induce the yield stress at mid-height of the column for any given value of n. Experimental evidence obtained by Perry and Robertson indicated that the hypothetical initial curvature of the column could be represented by which was combined with a load factor of 1.7 and used for many years in design codes to determine the critical value of average compressive stress below which overall buckling would not occur....

## Bending Rotational Stiffness

A fundamental relationship which exists in the elastic behaviour of structures and structural elements is that between an applied force system and the displacements which P is the applied force, k is the stiffness, 5 is the displacement. A definition of stiffness can be derived from this equation by rearranging it such that when 5 1.0 i.e. unit displacement the stiffness is 'the force necessary to maintain a UNIT displacement, all other displacements being equal to zero.' The displacement can...

## Example 33 Three Dimensional Space Truss

The space frame shown in Figure 3.10 has three pinned supports at A, B and C, all of which lie on the same level as indicated. Member DE is horizontal and at a height of 10 m above the plane of the supports. The planar dimensions z-x, x-y and z-y of the frame are indicated in Figure 3.11. Determine the forces in the members when the frame carries loads of 80 kN and 40 kN acting in a horizontal plane at joints E and D respectively as shown. Lcnjjili aTmcnbergt .nJ r t-t1 J it - IM nt ijut io.O'...

## Free and Fixed Bending Moments

When a beam is free to rotate at both ends as shown in Figures 4.70 a and b such that no bending moment can develop at the supports, then the bending moment diagram resulting from the applied loads on the beam is known as the Free Bending Moment Figure 4.70 Free Bending Moment Diagrams Figure 4.70 Free Bending Moment Diagrams When a beam is fixed at the ends encastre such that it cannot rotate, i.e. zero slope at the supports, as shown in Figure 4.71, then bending moments are induced at the...

## Equivalent Uniformly Distributed Load Method for the Deflection of Beams

In a simply supported beam, the maximum deflection induced by the applied loading always approximates the mid-span value if it is not equal to it. A number of standard frequently used load cases for which the elastic deformation is required are given in In many cases beams support complex load arrangements which do not lend themselves either to an individual load case or to a combination of the load cases given in Appendix 2. Provided that deflection is not the governing design criterion, a...

## Plastic Neutral Axis

Obviously at all stages of loading, the compression force FC induced by the applied moment must equal the tension force FT . This being so, then at the formation of the plastic hinge where all the material is subjected to the same stress i.e. py, the plastic neutral axis must be that axis which equally divides the area into two separate parts, i.e. Fi Compression Force Ac py Tens ion Force A j x py Ac Area in compression, Ay Area in tension Py - yield slress and Force in compression Force in...

## Static Method for Continuous Beams

In the static method of analysis the 'Free Bending Moment' diagrams for the structure are drawn and the 'Fixed Bending Moment' diagrams are then added algebraically. The magnitude and 'sense' ve or -ve of the moments must be such that sufficient plastic hinges occur to cause the collapse of the whole or a part of the structure. In addition, for collapse to occur, adjacent plastic hinges must be alternatively 'opening' and 'closing'. For uniform beams the plastic moment of resistance of each...

## Method of Tension Coefficients

The method of tension coefficients is a tabular technique of carrying out joint resolution in either two or three dimensions. It is ideally suited to the analysis of pin-jointed space-frames. Consider an individual member from a pin-jointed plane-frame, e.g. member AB shown in Figure 3.8 with reference to a particular X-Y co-ordinate system. If AB is a member of length LAB having a tensile force in it of TAB, then the components of this force in the X and Y directions are TAB Cos0 and TAB Sin0...