## Member DEF

U wif'(2.0 4.0) - 26.565 Cos 0,894 Sin a-0.447 A 55um if,ndsl Liiinpres5i(in to lio positive. Axial force +< 24.30 Cost ) 4 (J9.07* S in ) + 43.66 kN Shoor force -(24,30m Sin ) + (49.o > Chr) + 33-DI kN Axial ft - 4 CM JO x Cteff) 4 I .QTx & bi ) - 4- 22 JO Hi Shear fbrec - (24.30 X Sintf) + (49.07.x CoStf) - - 9.91 kN an'(2.0 6.0)- IfcOS* Coi Q* 0.947 Sin 0 0316 Assume axial eompicssion to beposilivA, Aljoinl D Axial forac + (24.3 0 x Cos-0 ) + (23.93 x Sin O ) + 30.57 kN Shear force -...

## Problem

2.0 m j.Om 3.0 in j.O m m 2.0 m j.Om 3.0 in j.O m m I 8 ftN a 5.1.4 Solutions Statically Determinate Rigid-Jointed Frames T< i ic Sl (itsilty IJtlcrminiHi Kigid-JuinU'il f'nimcs tVu IiU-im N u in 6k r 5.2 No, I Assuming pOiilivt Ixuidinj immicnLs induM linsinii ins id i' Lhu frjilli 3J)+ J3i45 * 1 jO) - + kNm Mo-+(13.45* 5.0) - (1.34 * 6.33) - (5.0 * 2.S3) + (5.0 x 0.67) t 47.97 kNm Mgm +06,55 X 5.0) - 11.34 K 4.67) - S.U x 4. )- + 47.7*i kNm .14-* - (K.O K 1.0) - ( L.J1 k 4.0) 4- (,. J5 *...

## The equation for the slope at x

S.S j 5.0, 10.0. 8.0 t , + .ir - - f.v 2T --- 61 - - .i - 91 - The equation for the deflection at x e.g. the deflection at the mid-span point can be determined from equation (5) by substituting the value of x 5.0 and ignoring the when their contents are -ve, i.e. The maximum deflection can be determined by calculating the value of x when the slope, i.e. equation (4) is equal to zero and substituting the calculated value of x into equation In most simply supported spans the maximum deflection...

## Info

(6.0 K -1,0X2.0) + ( 16.0 -( s 8,0) 0 5.0X2.5) + ( 13-0 i-0) (16.0x4.0) From equation (2) 40.0 + Hn 0 . Ho - J0 0 kN Frttdequation(3) 37101 0 ir +-16,5 kN t Fromaquation(1) 104.0 + 46.5 0 ., + 57.5 kN Assuming positive bending moments induce tension inside the frame Mb -(6.0x4.0 (2.0) -48.0 kNm Mc +(46.5x3.0)-(40.0x4.0) -20.50 kNm The values of the end-forces Fj to F8 can be determined by considering the equilibrium of Fi 57.50 kN 1 .-, fi 24,0 fcN +ve f LFj 0 +45.5 -(li.O * 5,0) *Ft-0 +vc * 0...

## J fi 7 J

Applw.'J i.ii il load P< jd + 5 S.33 fcN (tvusion) yfPL > _ 3823x1500x0 33 ABa ) f- ffr + jj > **- 2.41 - 1.S7 - - 4.2S mm t El y AV. At, Topic Drtcrmimilc rtrinis I'Vaiiifs. - Drfkclinn 11 sir Unil I.(Hid I'rolikni Numbii- 4,13 Pupi No, J (SlmlEI) is not a cunliuuous funaicm and Ute product LiHi'g.rjt ninsl Ik1 CYJituak-d between cagli cf lite di wonlini ilics, i.e. C, lg B and B Co A, The valine Of Ef tH wisilicm 'jr alnni lhe hcnm bttWCn laud A fcgivei hy Ef+EI (j- 4.oy4j 0.2 SF.ix...

## A

When the slenderness of an element is high, the element fails by excessive lateral deflection (i.e. buckling) at a value of stress considerably less than the yield or crushing values as shown in Figures 6.3(a) and (b). The failure of an element which is neither short nor slender occurs by a combination of buckling and yielding crushing as shown in Figures 6.4(a) and (b).

## Pinned

Consider the beam shown in Figure 4.68 in which a unit rotation is imposed at end A as before but the remote end B is pinned. The force (MA) necessary to maintain this displacement can be shown (e.g. using McCaulay's Method) to be equal to (3EI) L, which represents the reduced absolute stiffness of a pin-ended beam. It can therefore be stated that 'the stiffness of a pin-ended beam is equal to 34xthe stiffness of a fixed-end beam.' In addition it can be shown that there is no carry-over moment...

## Statically Indeterminate Beams

In many instances multi-span beams are used in design, and consequently it is necessary to consider the effects of the continuity on the support reactions and member forces. Such structures are indeterminate (see Chapter 1) and there are more unknown variables than can be solved using only the three equations of equilibrium. A few examples of such beams are shown in Figure 4.57(a) to (d).

## Secant ModulusEs

The 'secant modulus' is equal to the slope of a line drawn from the origin of the stressstrain graph to a point of interest beyond the elastic limit as shown in Figure 2.4. The secant modulus is used to describe the material resistance to deformation in the inelastic region of a stress strain curve and is often expressed as a percentage of Young's Modulus, e.g. 75 -0.75E.

## The equation for the bending moment at x is

The equation for the deflection at x is 16.33 -3.0 - 2.0 -6.0 +Ajt+B where A and B are constants of integration related to the boundary conditions. Boundary Conditions In this problem, assuming no settlement occurs at the supports then the deflection is equal Substituting for x and y in equation (3) B 0 Substituting for x and y in equation (3) B 0 The general equations for the slope and bending moment at any point along the length of

## Equilibrium

All structural analyses are based on satisfying one of the fundamental laws of physics, i.e. is the force system acting on a body a is the acceleration of the body Structural analyses carried out on the basis of a force system inducing a dynamic response, for example structural vibration induced by wind loading, earthquake loading, moving machinery, vehicular traffic etc., have a non-zero value for 'a' the acceleration. In the case of analyses carried out on the basis of a static response, for...

## Ji LN X19

PL 35 0> 4 SjwAC - * - + VV - + 3S.0 kNm 4 1.0 Span FC Jfaj + + + 3.0 fcN'm & K + (35J0* 2.0) -(4.0x 0 t'cAfa + 17-5 kN CoiuMcr ihi vertical equilibrium of ihe twain -+ve y Hfy 0 + 8.0 x 6.0 k 3,0) - (6.0 x Fxfcj-O ffcc iw + 24-0 tN Consider iht vertical equilibrium of the beam -+ve f - 0 + J'ffl +i'nc -CS-OKi-O O +24.0 kN Consider pun DE- 0 +(20.0 * 3.0) - (4.0 x I j0 ntty Q l'n lKt + 15-0 liN Consider ilw nieitl equilibrium ofiht beam vet 0 + + iW - 0 r,K - + i.0 hN + (6.0 x 2.0 * 1.0)...

## Pa rcd Yc fki 409 kN

Vca fl d +Fbfi*d (+4.09- 5.61) - 1.S2 kN + Vve (+ 5.61 + 13,72) + 19.33 kN -13.72 kN Consider the vertical equilibrium of the beam + (L0 x 8.0x 4.0) - (8-0 x Fbc frw) 0 Equation Ik , .1lJ.iik T (1) Consider the vertical equilibrium of the beam Consider the vertical equilibrium of the beam Consider the vertical equilibrium of the beam The final vertical support reactions are given by (i) + (ii) Check tht vsrtiicsl cquilibrtum Total vertical fort + 2r5S + 41.& I + 10933 + 36. The final...

## Plastic Cross Section Properties

When using elastic theory in design, the acceptance criterion are based on permissible or working stresses. These are obtained by dividing the yield stress py of the material by a suitable Factor of Safety. The loads adopted to evaluate an actual working stress are working loads. In a structure fabricated from linearly elastic material, the Factor of Safety (F. of S.) can also be expressed in terms of the load required to produce yield stress and the working load. This is known as the Load...

## Method ofSections

The method of sections involves the application of the three equations of static equilibrium to two-dimensional plane frames. The sign convention adopted to indicate ties (i.e. tension members) and struts (i.e. compression members) in frames is as shown in Figure 3.1. Joint Strut - compression member Joint The method involves considering an imaginary section line which cuts the frame under consideration into two parts A and B as shown in Figure 3.4. Since only three independent equations of...

## R7N

COinple i lln Unill jad llblelO determine llle value of f vjn Topic Urail Load Method fur lit' 11 lilion of Pin-Join led Frames ProMem Ni nfacn J, Fn& i Nft 2 Zf W + 10.35 mm j Hori unmi dtiliiliimlljolal H A tp v a Dull I.-lulI in llieltLiri unUil direclitHi aL joint Band dWnriiK ihe values of (lie Jf-fdrKS usinjt joinl resolution 4S before. tM5 -ml 1 -0,47 fi- fort jf M Vu, X ft T TOJJ Complete ik Unii Lok Eublv m determbis ik vahie of in.it Topic Unit Load Prklltad for Dcilrclion...

## Mathematical Modelling

The purpose of mathematical modelling is to predict structural behaviour in terms of loads, stresses and deformations under any specified, externally applied force system. Since actual structures are physical, three-dimensional entities it is necessary to create an idealized model which is representative of the materials used, the geometry of the structure and the physical constraints e.g. the support conditions and the externally applied force system. The precise idealisation adopted in a...

## Figure 821

Consider the equilibrium of he frame on the Itfl-hiind side ill 13 D Mn - 0 - 20,0 + x 2,0) 0 .-, Yx - + 10 Ji kN f Consider ihu equilibrium of (hi Irsme on (he right-hand side it C Elfe 0 - 20. + (20.0 xJ, )-( rE. x 2.0) 0 - - 0JJ kN +V61 + 10.32 - 30.0 20.0 - 0.32 + f 0 . . Kb - + 4fc0 Mi f EP, 0 - IJjQ-fl llr. - -1-15.0 kN BeadingmomentBfCL a- + (IMl* 4.0)-(30.0* 2,0) - tS.72 kNmS.V, tk'ntlins moment st Ci M& - + ( 15j0 h 3-0)- ( 1 i-0 K 1.5) - 20.63 +1.87 kNin s H EJcnttiiig moment al L...

## The joint at C can rotate either in a clockwise direction or an anticlockwise direction

The independent mechanisms can be entered into a table as before and the possible combinations investigated. In this example ID 2 and consequently a minimum of three hinges is required to induce total collapse. Since mechanisms (i) and (iv) have a significantly higher associated Mp value these have been selected to combine with the joint mechanism to produce a possible combination Mechanism (vi) the addition of mechanisms (i)+(iv)+(v)(a) Independent and Combined Mechanisms for Example 8.6

## Mechni i m il in 19 ljiiil CIO

Nate nu inlcma ivork i done l i upp > r1 12 hiiertinl Wofk Dona Externa I Work Done My (2 + J-(30J K lhe faings it joint t is assumed to tJcvelop in DKlltKt CM ni Ci- Internal Wrtk DoilC OHQ I Vork l iwe Mri +20+ ) U5jOK I.SiJ) J MjB- 22.50 H 5.63 fcNm The hinge at joint C is assumed 10 develop in member CFG a C . luteniid W-nrk Dont1 t Mejniil Work I June 1.5(3) 2. tfO- 22.50 Mf 1 JS kNm The liinge at joint C is assumed la de veto ) in member CFG atCj.

## Example 86 Joint Mechanism

In framed structures where there are more than two members meeting at a joint there is the possibility of a joint mechanism developing within a collapse mechanism. Consider the frame shown in Figure 8.20 with the collapse loads indicated. At joint C individual hinges can develop in members CBA, CDE and CFG giving three possible hinge positions at the joint in addition to positions B, D F and G.

## St020

Intimai Work J oabc < Exlcmal Woifc Done .un< 0+ HJ )- (io il)+< 4t> * Mf fy-RIO* 5 + 40 xitjH y tr0n ii Wp-70,0 ftNm intimai Work J oabc < Exlcmal Woifc Done Topic h lit iKihiis - Riuii Juinlnl Fnnici I Problem Number 3,7- Sialic Method Assume the vertical component of HMCtioii il support D to bu tlm rcJniicLint reliction. (1) iitatieaLly determinate feu ce system (II) Porc system dae lu redundjut reaction Consider tysfcm (I) Apply the three equations of Sialic equilibrium the faree...

## DW

U is the total strain energy of the structure due to the applied load system, W is the force acting at the point where the displacement is required, A is the linear displacement in the direction of the line of action of W. This form of the theorem is very useful in obtaining the deflection at joints in pin-jointed structures. Consider the pin-jointed frame shown in Figure 3.14 in which it is required to determine the vertical deflection of joint B. The member forces induced by the applied load...

## Strain Energy Axial Load Effects

Consider an axially loaded structural member of length 'L', cross-sectional area 'A', and of material with modulus of elasticity 'E' as shown in Figure 3.12(a) When an axial load 'P' is applied as indicated, the member will increase in length by '5L' as shown in Figure 3.12(b). Assuming linear elastic behaviour 5L this relationship is represented graphically in Figure 3.13. The work-done by the externally applied load 'P' is equal to (average value of the forcexdistance through which the force...

## SL term

(Note under normal applied loading the AH) The effects of temperature change in members can also be accommodated in a similar manner in this case the 5L term is related to the coefficient of thermal expansion for the material, the change in temperature and the original length, a is the coefficient of thermal expansion, At is the change in temperature a reduction being considered negative and an increase being positive. Since this is an elastic analysis the principle of superposition can be used...

## Fabrication Errors Lackoffit

During fabrication it is not unusual for a member length to be slightly too short or too long and assembly is achieved by forcing members in to place. The effect of this can be accommodated very easily in this method of analysis by adding additional terms relating to each member for which lack-of-fit applies. The 5L term for the relevant members is equal to the magnitude of the error in length, i.e. AL where negative values relate to members which are too short and positive values to members...

## Problem 814

8.11 Solutions Plastic Analysis Rigid-Jointed Frames 2 < i ) c I1 l.iMi- Anulji - Hijjicl JuinlctL rnmcs 2 'rol km Number it, 10 - Kill un 1k Methud l'i LL Nd, J CIhx 1l coIIjjihi iiiLtlianisin iV willi hinges .i It , C, and I1', (ij , 'I liinecs) The value of Mf nbliiiiiL-d (Ll > kNrn slum Id btL checked by ensuring that the hcudirji nipmfiil in the fi imc dncs not eieecrf fhe i lcvjiiH .l .inlue at any location. Topic Plttlfc AiihIvsif - Hi itt Juinlrd Fnuum 1 Pruhkin Number 8.11...

## Example 419 Singlespan Encastre Beam

Determine the support reactions and draw the bending moment diagram for the encastre beam loaded as shown in Figure 4.75. Consider the beam in two parts. (i) Fixed Support Reactions The values of the fixed-end moments are given in Appendix 2. Consider the rotational equilibrium of the beam Consider the rotational equilibrium of the beam Consider the vertical equilibrium of the beam Consider the vertical equilibrium of the beam Beams 311 Consider the rotational equilibrium of the beam

## Structural Loading

All structures are subjected to loading from various sources. The main categories of loading are dead, imposed and wind loads. In some circumstances there may be other loading types which should be considered, such as settlement, fatigue, temperature effects, dynamic loading, or impact effects (e.g. when designing bridge decks, crane-gantry girders or maritime structures). In the majority of cases design considering combinations of dead, imposed and wind loads is the most appropriate. Most...

## Introduction

The use of beams plate-girders does not always provide the most economic or suitable structural solution when spanning large openings. In buildings which have lightly loaded, long span roofs, when large voids are required within the depth of roof structures for services, when plated structures are impractical, or for aesthetic architectural reasons, the use of roof trusses, lattice girders or space-frames may be more appropriate. Such trusses girders frames, generally, transfer their loads by...

## Kinematic Method for Continuous Beams

In this method, a displacement is imposed upon each possible collapse mechanism and an equation between external work done and internal work absorbed in forming the hinges is developed. The collapse mechanism involving the greatest plastic moment, Mp, is the Consider the previous Example 8.1 of an encastre beam with a uniformly distributed load. The hinge positions were identified as occurring at A, B and the mid-span point (since the beam and loading are symmetrical). Assuming rigid links...

## Stress Strain Relationship

The plastic analysis and design of structures is based on collapse loads. A typical stress-strain curve for a ductile material having the characteristic of providing a large increase in strain beyond the yield point without any increase in stress, (e.g. steel) is given in Figure 2.39. When adopting this curve for the theory of plasticity (see Chapter 8) it is idealised as indicated in Figure 2.40 If a beam manufactured from material with a characteristic stress strain curve as shown in Figure...

## Fixed End Moments

The horizontal deflection at the rafter level 5DH 0.85AB (0.8x20) 16 mm Final values of support reactions * The maximum value along the length of members BCD can be found by identifying the point of zero shear as follows Examples in structural analysis 444 tciwim fliMMe 5.3.2 Problems Moment Distribution - Rigid-Jointed Frames with A series of rigid-jointed frames are indicated in Problems 5.13 to 5.18 in which the relative EI values and the applied loading are given. In each case i) sketch the...

## Q P fa

P is the force due to the applied load system u is the force due to the applied imaginary Unit load applied at B P is a multiplying factor to reflect the value of the load applied at B (since the unit load is an imaginary force the value of P zero and is used here as a mathematical convenience.) The total strain energy in the structure is equal to the sum of the energy stored in all the members Using Castigliano's 1st Theorem the deflection of joint B is given by Examples in structural analysis...

## Figure 320

The horizontal deflection i E The horizontal deflection i E The resultant deflection at joint B can be determined from the horizontal and vertical components evaluated above, i.e. R 101.013+121.0s) 157.6 2 AE i-TaD''(L2lJMl l0lj01)-i 50,15 A similar calculation can be carried out to determine the vertical deflection at joint D. The reader should complete this calculation to determine the member forces as indicated in Figure 3.21. Tlie member u forces for vertical dcflcction At joint I> Tlie...

## Example 49 Superposition Beam

Using superposition this beam can be represented as the sum of Using superposition this beam can be represented as the sum of (32,0 + 30.0) 62.0 kN Ma (- 64.0 - 90,0) 154.0 kN Shear Force at B len-iiAd (-S.0 - 30.0) - 38,0 kN Shear Force at B ht-h i side - 8.0 kN Bending Moment at B - 4,0 kNm 4.4.5 Example 4.10 Superposition -Beam 5 Using superposition this beam can be represented as the sum of 10.0+7.5 - 1.5) 16.0 kN Vn (- 2,0 + 7.5 + 7.5) - 13.0 kN Shear Force at B lef baodtidi - 8.0 kN S h...

## 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...

## Stratural Analysis By 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...

## 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...

## 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

## 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'...

## 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...

## 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...