FIXED BEAMS IN BENDING


 Candice McCormick
 2 years ago
 Views:
Transcription
1 FIXED BEAMS IN BENDING
2 INTRODUCTION Fixed or builtin beams are commonly used in building construction because they possess high rigidity in comparison to simply supported beams. When a simply supported beam of length, L, and flexural rigidity, EI, is subjected to a central concentrated load W, the maximum bending moment (BM) in beam is (WL/4), maximum deflection in beam is WI 3 /48EI, and maximum slope in beam is ±WL /16EI but if the ends of the same beam are fixed (i.e. built in walls), maximum bending moment is reduced to (WL/8), maximum deflection is reduced to WL 3 /19EI, and maximum slope is reduced to ±WL /3EI. If the allowable bending stress for the fixed beam is taken to be the same as for simply supported beam, then the load carrying capacity of the fixed beam is greatly improved. The overall reduction in bending moment, deflection, and slope in the fixed beam is due to the effect of fixing couples provided by the wall in keeping the slope and deflection at the ends to be zero. During building construction, the beams are cast along with the columns using reinforced cement concrete.
3 y x W simply supported beam x d y M ( x) EI W dx x W/ W/ W/ WL/4 the maximum bending moment (BM) in beam is (WL/4), maximum deflection in beam is WI 3 /48EI, and maximum slope in beam is ±WL /16EI W/ dy slope EI W x C1 dx x L, slope 0 0 W L C 4 WL dy C WL 1 16 dx 16EI 3 deflection W x WL xc 6 16 x 0, y C at y L 3 3 EIy W L WL L WL 3WL y WL 48EI 3 1 WL 48 3
4 FIXED BEAM BENDING MOMENT DIAGRAM A fixed beam can be considered as equivalent to a simply supported beam plus a beam of the same length, same material, same section subjected to end moments as shown in Figure 1 (a), (b) and (c). M D Figure 1 Fixed beam
5 A beam ABCD, of length L, fixed at ends A and D, carries loads W 1 and W at points B and C, respectively, is equivalent to a simply supported beam of length L, simply supported at ends A and D, carrying loads W 1 and W at B and C, respectively, plus a beam AD of length L, subjected to end moments M A and M D. Considering the beam as simply supported (SS) beam, bending moment diagram can be drawn as shown in Figure (a). Figure (b) shows the bending moment diagram of beam due to end moments. Due to end couples, there is convexity in beam, producing negative bending moment throughout. M D Figure (a) BM diagram of SS beam (b) BM diagram of end couples (c) BM diagram of fixed beam P 1, P are point of contraflexure
6 Due to loads and simple supports at ends, there is concavity in the beam producing positive bending moment throughout in the beam. Superimposing the two bending moment diagrams, we get the bending moment diagram for fixed beam as shown in Figure (c), in which P 1 and P are points of contraflexure. In the combined bending moment diagram there are positive bending moments and negative bending moments. At P 1 and P bending moment changes sign from negative to positive and from positive to negative as shown. M B M D Figure (a) BM diagram of SS beam (b) BM diagram of end couples (c) BM diagram of fixed beam P 1, P are point of contraflexure
7 Let us consider a section XX at a distance of x from end A of the beam. M, resulting bending moment at section = M x + M x X X M D B M x where M x = Bending moment at section when beam is considered as simply supported, and M x = Bending moment at section due to end moments M A and M D. M M x M A x L M A M D M A M D x M x Lx
8 FIXED BEAM SUPPORT MOMENTS If we refer to Figures (a), (b), (c), resultant bending moment at any section, M = M x + M x or M M x M x d y EI M x M x dx = Bending moment at section when beam is SS + Bending moment at section due to support moments. Multiplying both sides of by dx, we get EI d y dx dx M x dx M x dx
9 Integrating this equation over the length of the beam or L 0 EI d y dx EI dx dy dx L 0 L 0 M x dx aa L 0 M x dx = Area of the BM diagram of SS beam + area of BM diagram due to support moments where i D = Slope at end D, length is L, and i A = Slope at end A, length is zero But beam is fixed for both the ends, so slope at ends The equation becomes where EI 0= a +a a M A M D L
10 or M L A M D a This equation shows that the area of BM diagram due to support moments is numerically equal to the area of the BM diagram when beam is SS. d y Taking EI M x M x again and multiplying both the sides by x dx, and integrating dx both the sides over length of beam: L 0 EI d y dx EI x dx dy x y dx x dx ax ax Putting the values of slope and deflection at ends EI L 0 M L 0 x x dx But in a fixed beam, at the ends, slope and deflection both are zero, therefore, L 0 M Li y 0i y ax ax D D 0ax ax A A x
11 or But So, ax ax a = a i.e. distance of centroid of BM diagram when SS is equal to the distance of the centroid of BM diagram due to support moments from origin A, or from end of the beam. From the a diagram, area am A M D L ax M L L A M 3 L M A M D 6 D (origin at A), L L 3 Dividing the diagram into two triangles, we can determine that centroid of area A AD lies at from A and centroid of triangle ADD lies at from A (Figure 3). With the help of equations and, support moments M A, M D can be worked out. Figure 3 a diagram
12 GOVERNING DIFFERENTIAL EQUATION FOR DEFLECTION In pure bending of a prismatic beam, for an element dx bent by an infinitesimal angle dq, we can write y dq dx dx dq dq dx 1 The fiber length on a radius y can be found similarly. If the difference between the two fiber lenghts is du: du ydq dq ydq By dividing both sides by dx, the last term becomes K. The normal strain will be equal to: du dx dq 1 y y y dx y y From Hooke s law, x E x Ey In this equation, the variable y can assume both positive and negative values. Since, x M I y Ey M EI
13 Curvature In Cartesian coordinates the curvature of a line is defined as 1 d y dx dy 1 dx 3/ Where x locates a point on the elastic curve of a deflected beam and y gives the deflection of the same point from its initial position. dy Since the deflections tolerated in the majority of engineering structures are very small, slope curve is also very small. Therefore, the square of the slope can be neglected compared to one. dx of the elastic So, the curvature can be taken as 1 d y dx This simplification eliminates the geometric nonlinearity from the problem. Now, the governing differential equation for small deflections of elastic beams becomes: d y dx M EI
14 There are three governing equations for determining the deflection of a loaded beam: M ( x) EI d y dx V ( x) EI 3 d y 3 dx q( x) EI 4 d y 4 dx The choice of one of these equations for determining y depends on the case with which an expression for load, shear or moment can be formulated.
15 Macaulay s method: y Single moment: q( x) M O x a M O x a y Concentrated force: q( x) F O x a 1 F O y a x Uniformly distributed load: In order to keep distributed loads within specified boundaries, same magnitude but q( x) wo x a 0 a w O x negative direction distributed loads must be applied for the remaining length. Linearly increasing distributed load: dw q( x) x a dx 1 y dw dx x Quadratically increasing distributed load: c q( x) x a y a d w c dx x a
16 Macaulay s method:
17 FIXED BEAM WITH A CONCENTRATED LOAD AT CENTER Consider a fixed beam AB of length L, fixed at both ends A and B, carrying a point load W at its center as shown in Figure 4. It is equivalent to a simply supported beam AB with central load W and a beam AB subjected to end moments M A and M B. Since the beam is symmetrically loaded, end moments will be equal, i.e. M A = M B Figure 4 Equivalence of a fixed beam
18 In SS beam, BM at centre area of a BM diagram, Figure 4 a BM diagram
19 area of a diagram, a = M A L = M B L (M A = M B, because of symmetrical loading) or a = a or end moment, Resultant bending moment diagram is shown in Figure 8 (d), in which maximum positive BM maximum negative BM
20 The maximum bending moment in SS diagram is (WL/4) but when it is fixed at both the ends, maximum bending moment is reduced to 50%, i.e. (WL/8). Now let us calculate the maximum deflection in a fixed beam. Again consider a fixed beam AB of length L fixed at both the ends, load at centre is W as shown in Figure 5, reactions at ends. (due to symmetry) End couples Figure 5 Fixed beam with central point load Take a section X X, at a distance of x from end A, in the portion CB. Bending moment, (as obtained earlier)
21 or integrating it, we get where C 1 is constant of integration. (Term in bracket is to be omitted) At end A, because end is fixed. So, integration constant, C 1 = 0 Now integrating the above equation again, we get where C is another constant of integration. At end A, x = 0, deflection, y = 0, because end is fixed. So, 0 = omitted term + C, using Macaulay s method, constant C = 0.
22 Finally, the equation for deflection is Maximum deflection will occur at centre, i.e. at Putting this value of x, or
23 Example A beam of length L = 6 m, fixed at both the ends, carries a concentrated load of 30 kn at its centre. If EI = 7,000 knm for the beam, determine the fixing couple at ends and maximum deflection in the beam.
24 FIXED BEAM WITH UNIFORMLY DISTRIBUTED LOAD THROUGHOUT ITS LENGTH Consider a beam AB, of length L fixed at both the ends A and B, subjected to a uniformly distributed load of intensity w per unit length, throughout the length of the beam as shown in Figure 6 (a). It is equivalent to a SS beam of length L, subjected to udl, w throughout its length and a beam AB, subjected to end moments M A and M B as shown in Figure 6 (b) and (c). Since the beam is symmetrically loaded, end moments M A = M B Figure 6 (a) Fixed beam; (b) a BM diagram; (c) a BM diagram
25 When the beam is S.S, bending moment diagram is a parabola with, maximum bending moment at the centre. Area of BM diagram, area of BM diagram, a = a a = M A L = M B L (due to end moments) Reactions at supports = (due to symmetrical loading).
26 The resultant bending moment diagram for fixed beam is shown in Figure 30 (d), in which P 1 and P are points of contraflexure. The locations of P 1 and P can be determined. Figure 6 (d) BM diagram for fixed beam Now let s consider any section X X at a distance of x from end A as shown in Figure 7. Bending moment at section XX is or Integrating this, we get Figure 7 Fixed beam with uniform distributed loading
27 at x = 0, end A, slope, so constant C 1 = 0., Now Integrating this equation again, we get At end A, x = 0, deflection y = 0, so constant, C = 0 Finally, we have Maximum deflection in the beam will occur at the centre, putting the value of x =
28 Note that the deflection is reduced to only 0% of the deflection when the beam is simply supported at ends and maximum bending moment in beam is also reduced to from (in SS beam).
29 Example A beam with a 6 m span has its ends built in and carries a uniformly distributed load of 4 kn/m throughout its length. Find the (i) end moments, (ii) bending moment at the centre, and (iii) maximum deflection in the beam for E = 10 GPa, I = 4,800 cm 4.
30 FIXED BEAM WITH AN ECCENTRIC LOAD For this problem it is very much time consuming to draw a and a BM diagrams and then to determine support reactions and moments. Let us take a fixed beam AB, of length L fixed at both the ends A and B and carrying an eccentric load W at C, at a distance of a from end A, Figure 3. Say reactions at A and B are R A and R B and support moments at A and B are M A and M B respectively. There are four boundary conditions at two ends of fixed beam, and we can determine four unknowns R A, R B, M A, and M B. Consider a section X X at a distance of x from end A, in portion CB of the beam, bending moment at section XX is M = M A + R A x W (x a) or Integrating this equation, we get C (constant of integration) At x = 0, end A, slope,, because of fixed end 0 = omitted term + C 1 Figure 8 Fixed beam with eccentric load
31 Constant C l = 0 Integrating this equation we get At end A, x = 0, deflection, y = 0, because of fixed end 0 = omitted term + C Constant C = 0. So, At the end B, x = L, both deflection and slope are zero. Putting x = L in and
32 or or From these equations, or Reaction R A
33 Reaction R B Putting the value of R A in, we get
34 Similarly we can find out that support moment at B, In Figure 3, we have taken a C Bending moment at point C, when beam is in BM diagram. Bending moment diagram for fixed beam is A A P 1 C P BB as shown in Figure 9, where P 1 and P are points of contraflexure. End moments, Figure 9 BM diagram of fixed beam
35 Example Let us take numerical values of a and b, say a = m, b = 4 m, L = 6 m, W = 8 kn. What is the equation of deflection? (SS) BM diagram of fixed beam as shown in the figure. In a fixed beam with eccentric load, we know that the reaction R A, Moment M A, Deflection equation will become
36 Let us calculate deflection under the load W, i.e. at x = a a L b b or deflection,
37 Example A beam 8 m long, fixed at both the ends carries a vertical load of 4 kn at a distance of 3 m from left hand end. Determine: (i) support reactions, (ii) support moments, and (iii) location of points of contraflexure.
38 Example A fixed beam AB of length 6 m carries a uniformly distributed load (udl) of intensity w=6 kn/m over a length AC = m and a point load of 6 kn at point D at a distance of 4 m from A. Using double integration method, determine support reactions, support moments, and draw BM diagram for the beam. What is the deflection at point D of the beam if EI = 1,400 knm.
39 There are unknown reactions R A, R B at ends and unknown moments M A, M B as shown. In this problem it is convenient if we take origin at B and x positive towards left, and section XX in portion CA (because udl is in portion CA). Equation of BM, or
40 Integrating this equation, we get (constant of integration C 1 ) (at end B, x = 0, slope, because of fixed end.) So, 0 = omitted term + C 1 Constant C 1 = 0 Integrating this equation, we get (constant of integration C ) At end B, x = 0, deflection y = 0, because end is fixed.
41 0 = omitted terms + C Constant C = 0 Finally, the equation of deflection is To determine R B, M B, let us consider end A where x = 6 m, both slope and deflection are zero. Putting this value of x in and 0 = 6M B + 18R B or 0 = 6M B + 18R B 56 0 = 18M B + 36R B 68
42 or M B + 3R B = M B + R B = From these equations, reaction R B is obtained as, R B = 5.556kN Total load on beam = = 18 kn Reaction, R A = = kn Moment, M B = R B = = = knm To know bending moment M A, let us put x = 6 in and putting values of M A, R B also M A = x 6 6 (6 ) 3(6 4) = = knm.
43 Deflection at D: Putting x = m in Deflection,
44 BM Diagram of Fixed Beam Let us first draw BM diagram of an SS beam with same loads. Taking moments about A, let us determine reaction R B 6 R B = = 36 Reaction, R B = 6 kn Reaction, R A = = 1 kn Bending Moment
45 M D = 6 = + 1 KNm, M B = 0 In the figure AC D B shows a bending moment diagram of SS beam. Let us draw a diagram on this, taking M A = KNm M B = KN The resultant bending moment diagram for fixed beam is obtained as seen in the figure.
46 EFFECT OF SINKING OF A SUPPORT IN A FIXED BEAM During the construction of a building or a structure, due to the defect in material or in workmanship, one end of a fixed beam may sink by some amount. This type of sinking is common and may cause unintentional bending moment on the beam. Let us consider that one end of the fixed beam, i.e. support B, sinks by an amount δ, as shown in Figure 10 (a), say length of beam is L. Level of support B is below the level of support A by δ. Obviously there is no rate of loading on beam, but due to vertical pressure at B, sinking has occurred, therefore, Integrating it, we get Figure 10 (a) Sinking of a support
47 Say at x = 0, Reaction = R A, then Integrating this equation, we get (another constant of integration) Say at A, x = 0, bending moment is M A, so M A = 0 + C and Constant, C = M A Integrating this we get (constant of integration) But slope is zero at fixed end, A, i.e. at x = 0 0 = C 3 so Constant of integration, C 3 = 0
48 So, Integrating this we get (constant of integration) At x = 0, deflection, y = 0, 0 = C 4 so Constant of integration, C 4 = 0 Finally, we have At end B, x = L, slope,, deflection, y = δ
49 Putting these values in and From these two equations For equilibrium, R A = R B or Putting the values of M A, R A in bending moment at end B
50 Figures 10 (b) and (c) show the SF and BM diagrams of a fixed beam, whose one support has sunk. If a fixed beam of length L, carrying concentrated central load W, sinks by an amount δ at right hand support, then support moments will 6EI L 6EI L be now at one end to at the other end. Figure 10 (a) Sinking of a support, (b) Shear Force diagram; (c) Bending Moment diagram 6EI L 6EI L
51 EFFECT OF ROTATION OF A SUPPORT IN A FIXED BEAM Due to defective materials or defective construction, if one end of the fixed beam rotates (instead of keeping the slope unchanged), it causes development of support moments. Consider a beam AB, fixed at both ends but support B rotates by an angle i B as shown in Figure 11. In this case: Integrating this (constant of integration) Figure 11 (a) Beam in consideration at end A, x = 0, shear force R A
52 Integrating further (another constant of integration) At end A, x = 0, moment = M A Therefore, M A = 0 + C or constant of integration = M A Integrating this equation, we get (constant of integration) At x = 0, So, 0 = C 3 or constant C 3 = 0
53 Integrating it, we get (constant of integration) at x = 0, end A, deflection, y = 0. So, 0 = C 4 or constant, C 4 = 0 Finally we have,
54 Let us utilize end condition at B, where x = L, Using these values in and From the above equation, Putting this value of M A in, we get Reaction, Moment,
55 To maintain equilibrium, R B = R A (but in opposite direction) Using, bending moment at B, Figure 11 (b) shows BM diagram, and Figure 11 (c) shows SF diagram for this fixed beam. Figure 11 (a) Beam in consideration (b) BM diagram (c) SF diagram
56 Exercise A fixed beam AB of length 7 m rotates at end B by 0.1 in counterclockwise direction. If the beam carries a central load of 8 kn, determine support reactions and support moments. Given EI = 6,500 knm.
57 Example A beam AB of uniform section throughout and of span L is fixed at end A, and carries a central point load P. During loading, support B sinks by δ and rotates by an angle δ/l in anticlockwise direction. Determine the fixing moments at ends if EI is the flexural rigidity of the beam. The figure shows a fixed beam AB of span length L, carrying a load P at its centre, support B is below the level of support B by δ, and support B rotates by angle in counterclockwise direction. Let s say moments and reaction at A and B are M A, R A ; M B and R B, respectively. Now consider a section X X at a distance x from A in the portion CB of the beam. δ/l
58 Bending moment at the section or Integrating this, we get At end A, x = 0, slope So, 0 = omitted term + C 1, Constant, C 1 = 0
59 Integrating this again At end A, x = 0, deflection, y = 0, so 0 = omitted term + C Constant, C = 0. Finally the equation is, Putting x = L, at end B, where In and or
60 Adding the last two equations, we get or Substituting the value of M A Reaction,
61 Now putting the value of R A in Substituting the values of M A, R A in at x = L, this equation gives BM at end B
62 Example A fixed beam of length L carries a linearly increasing distributed load of intensity zero at the left hand end to w up to a distance a from the same end. Determine: (i) support reactions, (ii) support moments, if EI is the flexural rigidity of the beam. The figure shows a fixed beam of length L, carrying a linearly increasing distributed load of intensity zero at end A and w at a distance a from A, at section c. Consider a section XX, at a distance x from A, Rate of loading at section XX, Elementary load for small length of d x
63 Considering as an eccentric load, Support moments, dm A For eccentric loading Support moment M A, Similarly support moment M B,
64 Support Reactions: Total load on beam Taking moments, or Reaction,
65 Example A small beam in a bridge deck has to be propped temporarily at a particular point so that the prop can carry half the concentrated load occurring at that point. The beam is 5 m long and has both the ends built in at the same level as shown in the figure. Concentrated load occurs at 3 m from the left wall. The prop or column is a circular bar. Calculate its diameter so that as stated, the beam and column carry half the applied load. Second moment of area of beam is 36,000 cm 4. Young s modulus of the beam material is double the Young s modulus of the column material.
66 At the section under consideration for the beam, Load is distance b = 3 m, distance a = m, L = 5 m E b = Young s modulus of beam, then I b = Moment of inertia of beam section Deflection in beam at point C (from page 35) (y positive downwards, a and b shift) Putting the values,
67 Deflection in column, length of column = 1 m But both deflections are the same, so Area of crosssection of column, A c, area of column section Diameter of column,
68 Example A horizontal steel bar 60 mm in diameter is rigidly fixed at each end where the fixings are 1. m (L) apart. A rigid bracket is welded in the middle of the bar at right angles to the axis and in the same horizontal plane. Determine the radius of the arm of the bracket (R) at which a vertical load of 100 N can be suspended if the deflection of the load is not to exceed 0.5 mm. Given E = 00 GPa, shear modulus, G = 80 GPa.
69 Bar diameter, d = 60 mm f = 60 mm Moment of inertia, Polar moment of inertia, J = I = mm 4 E = kn/m EI = ( ) ( ) = 17.4 knm Deflection in the centre of the bar as fixed beam (from page 1), (y positive downwards)
70 δ deflection in the bar due to twisting = δ = mm T Twisting moment, T = WR = 1.R knm Torque will be equally divided in two portions AC and CB Torque on AC, Using Torsion formula (L'=L/)
71 Putting these values or
72 Example A beam of span L is fixed at both the ends. A couple M is applied to the beam at its centre about a horizontal axis at right angle to the beam. Determine the fixing couples at each support and slope at the centre of the beam. EI is the flexural rigidity of the beam.
FIXED BEAMS CONTINUOUS BEAMS
FIXED BEAMS CONTINUOUS BEAMS INTRODUCTION A beam carried over more than two supports is known as a continuous beam. Railway bridges are common examples of continuous beams. But the beams in railway bridges
More informationProcedure for drawing shear force and bending moment diagram:
Procedure for drawing shear force and bending moment diagram: Preamble: The advantage of plotting a variation of shear force F and bending moment M in a beam as a function of x' measured from one end of
More information2. Determine the deflection at C of the beam given in fig below. Use principal of virtual work. W L/2 B A L C
CE1259, Strength of Materials UNIT I STRESS, STRAIN DEFORMATION OF SOLIDS Part A 1. Define strain energy density. 2. State Maxwell s reciprocal theorem. 3. Define proof resilience. 4. State Castigliano
More informationSub. Code:
Important Instructions to examiners: ) The answers should be examined by key words and not as wordtoword as given in the model answer scheme. ) The model answer and the answer written by candidate may
More informationShear force and bending moment of beams 2.1 Beams 2.2 Classification of beams 1. Cantilever Beam Builtin encastre' Cantilever
CHAPTER TWO Shear force and bending moment of beams 2.1 Beams A beam is a structural member resting on supports to carry vertical loads. Beams are generally placed horizontally; the amount and extent of
More informationUNIT III DEFLECTION OF BEAMS 1. What are the methods for finding out the slope and deflection at a section? The important methods used for finding out the slope and deflection at a section in a loaded
More informationPURE BENDING. If a simply supported beam carries two point loads of 10 kn as shown in the following figure, pure bending occurs at segment BC.
BENDING STRESS The effect of a bending moment applied to a crosssection of a beam is to induce a state of stress across that section. These stresses are known as bending stresses and they act normally
More informationQUESTION BANK DEPARTMENT: CIVIL SEMESTER: III SUBJECT CODE: CE2201 SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1 STRESS AND STRAIN PART A
DEPARTMENT: CIVIL SUBJECT CODE: CE2201 QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1 STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State
More informationMAHALAKSHMI ENGINEERING COLLEGE
CE840STRENGTH OF TERIS  II PGE 1 HKSHI ENGINEERING COEGE TIRUCHIRPI  611. QUESTION WITH NSWERS DEPRTENT : CIVI SEESTER: IV SU.CODE/ NE: CE 840 / Strength of aterials II UNIT INDETERINTE ES 1. Define
More informationQUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS
QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1 STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State Hooke s law. 3. Define modular ratio,
More informationMAHALAKSHMI ENGINEERING COLLEGE
AHAAKSHI ENGINEERING COEGE TIRUCHIRAPAI  611. QUESTION WITH ANSWERS DEPARTENT : CIVI SEESTER: V SU.CODE/ NAE: CE 5 / Strength of aterials UNIT INDETERINATE EAS 1. Define statically indeterminate beams.
More informationPERIYAR CENTENARY POLYTECHNIC COLLEGE PERIYAR NAGAR  VALLAM THANJAVUR. DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK
PERIYAR CENTENARY POLYTECHNIC COLLEGE PERIYAR NAGAR  VALLAM  613 403  THANJAVUR. DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK Sub : Strength of Materials Year / Sem: II / III Sub Code : MEB 310
More informationOUTCOME 1  TUTORIAL 3 BENDING MOMENTS. You should judge your progress by completing the self assessment exercises. CONTENTS
Unit 2: Unit code: QCF Level: 4 Credit value: 15 Engineering Science L/601/1404 OUTCOME 1  TUTORIAL 3 BENDING MOMENTS 1. Be able to determine the behavioural characteristics of elements of static engineering
More informationBUILTIN BEAMS. The maximum bending moments and maximum deflections for builtin beams with standard loading cases are as follows: Summary CHAPTER 6
~ or CHAPTER 6 BUILTIN BEAMS Summary The maximum bending moments and maximum deflections for builtin beams with standard loading cases are as follows: MAXIMUM B.M. AND DEFLECTION FOR BUILTIN BEAMS Loading
More informationAdvanced Structural Analysis EGF Section Properties and Bending
Advanced Structural Analysis EGF316 3. Section Properties and Bending 3.1 Loads in beams When we analyse beams, we need to consider various types of loads acting on them, for example, axial forces, shear
More informationBEAM A horizontal or inclined structural member that is designed to resist forces acting to its axis is called a beam
BEM horizontal or inclined structural member that is designed to resist forces acting to its axis is called a beam INTERNL FORCES IN BEM Whether or not a beam will break, depend on the internal resistances
More informationBeams. Beams are structural members that offer resistance to bending due to applied load
Beams Beams are structural members that offer resistance to bending due to applied load 1 Beams Long prismatic members Nonprismatic sections also possible Each crosssection dimension Length of member
More information2 marks Questions and Answers
1. Define the term strain energy. A: Strain Energy of the elastic body is defined as the internal work done by the external load in deforming or straining the body. 2. Define the terms: Resilience and
More informationBE Semester I ( ) Question Bank (MECHANICS OF SOLIDS)
BE Semester I ( ) Question Bank (MECHANICS OF SOLIDS) All questions carry equal marks(10 marks) Q.1 (a) Write the SI units of following quantities and also mention whether it is scalar or vector: (i)
More informationSSCJE MAINS ONLINE TEST SERIES / CIVIL ENGINEERING SOM + TOS
SSCJE MAINS ONLINE TEST SERIES / CIVIL ENGINEERING SOM + TOS Time Allowed:2 Hours Maximum Marks: 300 Attention: 1. Paper consists of Part A (Civil & Structural) Part B (Electrical) and Part C (Mechanical)
More information18.Define the term modulus of resilience. May/June Define Principal Stress. 20. Define Hydrostatic Pressure.
CE6306 STREGNTH OF MATERIALS Question Bank UnitI STRESS, STRAIN, DEFORMATION OF SOLIDS PARTA 1. Define Poison s Ratio May/June 2009 2. What is thermal stress? May/June 2009 3. Estimate the load carried
More informationMechanics of Structure
S.Y. Diploma : Sem. III [CE/CS/CR/CV] Mechanics of Structure Time: Hrs.] Prelim Question Paper Solution [Marks : 70 Q.1(a) Attempt any SIX of the following. [1] Q.1(a) Define moment of Inertia. State MI
More informationCE6306 STRENGTH OF MATERIALS TWO MARK QUESTIONS WITH ANSWERS ACADEMIC YEAR
CE6306 STRENGTH OF MATERIALS TWO MARK QUESTIONS WITH ANSWERS ACADEMIC YEAR 20142015 UNIT  1 STRESS, STRAIN AND DEFORMATION OF SOLIDS PART A 1. Define tensile stress and tensile strain. The stress induced
More informationBEAM DEFLECTION THE ELASTIC CURVE
BEAM DEFLECTION Samantha Ramirez THE ELASTIC CURVE The deflection diagram of the longitudinal axis that passes through the centroid of each crosssectional area of a beam. Supports that apply a moment
More informationPDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [ ] Introduction, Fundamentals of Statics
Page1 PDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [2910601] Introduction, Fundamentals of Statics 1. Differentiate between Scalar and Vector quantity. Write S.I.
More informationUNITV MOMENT DISTRIBUTION METHOD
UNITV MOMENT DISTRIBUTION METHOD Distribution and carryover of moments Stiffness and carry over factors Analysis of continuous beams Plane rigid frames with and without sway Neylor s simplification. Hardy
More informationStructural Analysis I Chapter 4  Torsion TORSION
ORSION orsional stress results from the action of torsional or twisting moments acting about the longitudinal axis of a shaft. he effect of the application of a torsional moment, combined with appropriate
More informationUNIT I ENERGY PRINCIPLES
UNIT I ENERGY PRINCIPLES Strain energy and strain energy density strain energy in traction, shear in flexure and torsion Castigliano s theorem Principle of virtual work application of energy theorems
More informationConsider an elastic spring as shown in the Fig.2.4. When the spring is slowly
.3 Strain Energy Consider an elastic spring as shown in the Fig..4. When the spring is slowly pulled, it deflects by a small amount u 1. When the load is removed from the spring, it goes back to the original
More informationKINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK. Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV UNIT I STRESS, STRAIN DEFORMATION OF SOLIDS PART A (2 MARKS)
More informationSample Question Paper
Scheme I Sample Question Paper Program Name : Mechanical Engineering Program Group Program Code : AE/ME/PG/PT/FG Semester : Third Course Title : Strength of Materials Marks : 70 Time: 3 Hrs. Instructions:
More informationBending Stress. Sign convention. Centroid of an area
Bending Stress Sign convention The positive shear force and bending moments are as shown in the figure. Centroid of an area Figure 40: Sign convention followed. If the area can be divided into n parts
More informationR13. II B. Tech I Semester Regular Examinations, Jan MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) PARTA
SET  1 II B. Tech I Semester Regular Examinations, Jan  2015 MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) Time: 3 hours Max. Marks: 70 Note: 1. Question Paper consists of two parts (PartA and PartB)
More informationDEPARTMENT OF CIVIL ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF CIVIL ENGINEERING SUBJECT: CE 2252 STRENGTH OF MATERIALS UNIT: I ENERGY METHODS 1. Define: Strain Energy When an elastic body is under the action of external
More informationUNITI STRESS, STRAIN. 1. A Member A B C D is subjected to loading as shown in fig determine the total elongation. Take E= 2 x10 5 N/mm 2
UNITI STRESS, STRAIN 1. A Member A B C D is subjected to loading as shown in fig determine the total elongation. Take E= 2 x10 5 N/mm 2 Young s modulus E= 2 x10 5 N/mm 2 Area1=900mm 2 Area2=400mm 2 Area3=625mm
More information2012 MECHANICS OF SOLIDS
R10 SET  1 II B.Tech II Semester, Regular Examinations, April 2012 MECHANICS OF SOLIDS (Com. to ME, AME, MM) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry Equal Marks ~~~~~~~~~~~~~~~~~~~~~~
More informationUNIT II SLOPE DEFLECION AND MOMENT DISTRIBUTION METHOD
SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR Siddharth Nagar, Narayanavanam Road 517583 QUESTION BANK (DESCRIPTIVE) Subject with Code : SAII (13A01505) Year & Sem: IIIB.Tech & ISem Course & Branch: B.Tech
More information7.4 The Elementary Beam Theory
7.4 The Elementary Beam Theory In this section, problems involving long and slender beams are addressed. s with pressure vessels, the geometry of the beam, and the specific type of loading which will be
More informationtechietouch.blogspot.com DEPARTMENT OF CIVIL ENGINEERING ANNA UNIVERSITY QUESTION BANK CE 2302 STRUCTURAL ANALYSISI TWO MARK QUESTIONS UNIT I DEFLECTION OF DETERMINATE STRUCTURES 1. Write any two important
More informationSTRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS
1 UNIT I STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define: Stress When an external force acts on a body, it undergoes deformation. At the same time the body resists deformation. The
More informationStrength of Materials Prof. Dr. Suraj Prakash Harsha Mechanical and Industrial Engineering Department Indian Institute of Technology, Roorkee
Strength of Materials Prof. Dr. Suraj Prakash Harsha Mechanical and Industrial Engineering Department Indian Institute of Technology, Roorkee Lecture  28 Hi, this is Dr. S. P. Harsha from Mechanical and
More informationSRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDHYALAYA
SRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDHYALAYA (Declared as Deemedtobe University under Section 3 of the UGC Act, 1956, Vide notification No.F.9.9/92U3 dated 26 th May 1993 of the Govt. of
More informationPES Institute of Technology
PES Institute of Technology Bangalore south campus, Bangalore5460100 Department of Mechanical Engineering Faculty name : Madhu M Date: 29/06/2012 SEM : 3 rd A SEC Subject : MECHANICS OF MATERIALS Subject
More informationUNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation.
UNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation. At the same time the body resists deformation. The magnitude
More informationCHAPTER OBJECTIVES Use various methods to determine the deflection and slope at specific pts on beams and shafts: 2. Discontinuity functions
1. Deflections of Beams and Shafts CHAPTER OBJECTIVES Use various methods to determine the deflection and slope at specific pts on beams and shafts: 1. Integration method. Discontinuity functions 3. Method
More informationDeflection of Flexural Members  Macaulay s Method 3rd Year Structural Engineering
Deflection of Flexural Members  Macaulay s Method 3rd Year Structural Engineering 008/9 Dr. Colin Caprani 1 Contents 1. Introduction... 3 1.1 General... 3 1. Background... 4 1.3 Discontinuity Functions...
More informationFinite Element Analysis Prof. Dr. B. N. Rao Department of Civil Engineering Indian Institute of Technology, Madras. Module  01 Lecture  11
Finite Element Analysis Prof. Dr. B. N. Rao Department of Civil Engineering Indian Institute of Technology, Madras Module  01 Lecture  11 Last class, what we did is, we looked at a method called superposition
More informationDeflection of Flexural Members  Macaulay s Method 3rd Year Structural Engineering
Deflection of Flexural Members  Macaulay s Method 3rd Year Structural Engineering 009/10 Dr. Colin Caprani 1 Contents 1. Introduction... 4 1.1 General... 4 1. Background... 5 1.3 Discontinuity Functions...
More informationSemester: BE 3 rd Subject :Mechanics of Solids ( ) Year: Faculty: Mr. Rohan S. Kariya. Tutorial 1
Semester: BE 3 rd Subject :Mechanics of Solids (2130003) Year: 201819 Faculty: Mr. Rohan S. Kariya Class: MA Tutorial 1 1 Define force and explain different type of force system with figures. 2 Explain
More informationProblem 1: Calculating deflection by integration uniform load. Problem 2: Calculating deflection by integration  triangular load pattern
Problem 1: Calculating deflection by integration uniform load Problem 2: Calculating deflection by integration  triangular load pattern Problem 3: Deflections  by differential equations, concentrated
More informationChapter Objectives. Copyright 2011 Pearson Education South Asia Pte Ltd
Chapter Objectives To generalize the procedure by formulating equations that can be plotted so that they describe the internal shear and moment throughout a member. To use the relations between distributed
More informationChapter 8 Supplement: Deflection in Beams Double Integration Method
Chapter 8 Supplement: Deflection in Beams Double Integration Method 8.5 Beam Deflection Double Integration Method In this supplement, we describe the methods for determining the equation of the deflection
More informationCHAPTER 4: BENDING OF BEAMS
(74) CHAPTER 4: BENDING OF BEAMS This chapter will be devoted to the analysis of prismatic members subjected to equal and opposite couples M and M' acting in the same longitudinal plane. Such members are
More informationQUESTION BANK. SEMESTER: V SUBJECT CODE / Name: CE 6501 / STRUCTURAL ANALYSISI
QUESTION BANK DEPARTMENT: CIVIL SEMESTER: V SUBJECT CODE / Name: CE 6501 / STRUCTURAL ANALYSISI Unit 5 MOMENT DISTRIBUTION METHOD PART A (2 marks) 1. Differentiate between distribution factors and carry
More informationSN QUESTION YEAR MARK 1. State and prove the relationship between shearing stress and rate of change of bending moment at a section in a loaded beam.
ALPHA COLLEGE OF ENGINEERING AND TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING MECHANICS OF SOLIDS (21000) ASSIGNMENT 1 SIMPLE STRESSES AND STRAINS SN QUESTION YEAR MARK 1 State and prove the relationship
More informationMechanics of Solids notes
Mechanics of Solids notes 1 UNIT II Pure Bending Loading restrictions: As we are aware of the fact internal reactions developed on any crosssection of a beam may consists of a resultant normal force,
More informationOnly for Reference Page 1 of 18
Only for Reference www.civilpddc2013.weebly.com Page 1 of 18 Seat No.: Enrolment No. GUJARAT TECHNOLOGICAL UNIVERSITY PDDC  SEMESTER II EXAMINATION WINTER 2013 Subject Code: X20603 Date: 26122013 Subject
More information[8] Bending and Shear Loading of Beams
[8] Bending and Shear Loading of Beams Page 1 of 28 [8] Bending and Shear Loading of Beams [8.1] Bending of Beams (will not be covered in class) [8.2] Bending Strain and Stress [8.3] Shear in Straight
More informationSymmetric Bending of Beams
Symmetric Bending of Beams beam is any long structural member on which loads act perpendicular to the longitudinal axis. Learning objectives Understand the theory, its limitations and its applications
More informationtwenty one concrete construction: shear & deflection ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SUMMER 2014 lecture
ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SUMMER 2014 lecture twenty one concrete construction: Copyright Kirk Martini shear & deflection Concrete Shear 1 Shear in Concrete
More informationCHENDU COLLEGE OF ENGINEERING &TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING SUB CODE & SUB NAME : CE2351STRUCTURAL ANALYSISII UNIT1 FLEXIBILITY
CHENDU COLLEGE OF ENGINEERING &TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING SUB CODE & SUB NAME : CE2351STRUCTURAL ANALYSISII UNIT1 FLEXIBILITY METHOD FOR INDETERMINATE FRAMES PARTA(2MARKS) 1. What is
More informationModule 4 : Deflection of Structures Lecture 4 : Strain Energy Method
Module 4 : Deflection of Structures Lecture 4 : Strain Energy Method Objectives In this course you will learn the following Deflection by strain energy method. Evaluation of strain energy in member under
More informationUNIT I Thin plate theory, Structural Instability:
UNIT I Thin plate theory, Structural Instability: Analysis of thin rectangular plates subject to bending, twisting, distributed transverse load, combined bending and inplane loading Thin plates having
More information(Refer Slide Time: 01:00 01:01)
Strength of Materials Prof: S.K.Bhattacharya Department of Civil Engineering Indian institute of Technology Kharagpur Lecture no 27 Lecture Title: Stresses in Beams II Welcome to the second lesson of
More informationUNIVERSITY OF BOLTON SCHOOL OF ENGINEERING. BEng (HONS) CIVIL ENGINEERING SEMESTER 1 EXAMINATION 2016/2017 MATHEMATICS & STRUCTURAL ANALYSIS
TW21 UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BEng (HONS) CIVIL ENGINEERING SEMESTER 1 EXAMINATION 2016/2017 MATHEMATICS & STRUCTURAL ANALYSIS MODULE NO: CIE4011 Date: Wednesday 11 th January 2017 Time:
More informationStress Analysis Lecture 3 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy
Stress Analysis Lecture 3 ME 276 Spring 20172018 Dr./ Ahmed Mohamed Nagib Elmekawy Axial Stress 2 Beam under the action of two tensile forces 3 Beam under the action of two tensile forces 4 Shear Stress
More information6. Bending CHAPTER OBJECTIVES
CHAPTER OBJECTIVES Determine stress in members caused by bending Discuss how to establish shear and moment diagrams for a beam or shaft Determine largest shear and moment in a member, and specify where
More information3 Hours/100 Marks Seat No.
*17304* 17304 14115 3 Hours/100 Marks Seat No. Instructions : (1) All questions are compulsory. (2) Illustrate your answers with neat sketches wherever necessary. (3) Figures to the right indicate full
More informationBy Dr. Mohammed Ramidh
Engineering Materials Design Lecture.6 the design of beams By Dr. Mohammed Ramidh 6.1 INTRODUCTION Finding the shear forces and bending moments is an essential step in the design of any beam. we usually
More informationSTRENGTH OF MATERIALSI. Unit1. Simple stresses and strains
STRENGTH OF MATERIALSI Unit1 Simple stresses and strains 1. What is the Principle of surveying 2. Define Magnetic, True & Arbitrary Meridians. 3. Mention different types of chains 4. Differentiate between
More informationName :. Roll No. :... Invigilator s Signature :.. CS/B.TECH (CENEW)/SEM3/CE301/ SOLID MECHANICS
Name :. Roll No. :..... Invigilator s Signature :.. 2011 SOLID MECHANICS Time Allotted : 3 Hours Full Marks : 70 The figures in the margin indicate full marks. Candidates are required to give their answers
More informationChapter 7: Bending and Shear in Simple Beams
Chapter 7: Bending and Shear in Simple Beams Introduction A beam is a long, slender structural member that resists loads that are generally applied transverse (perpendicular) to its longitudinal axis.
More informationstructural analysis Excessive beam deflection can be seen as a mode of failure.
Structure Analysis I Chapter 8 Deflections Introduction Calculation of deflections is an important part of structural analysis Excessive beam deflection can be seen as a mode of failure. Extensive glass
More information3. BEAMS: STRAIN, STRESS, DEFLECTIONS
3. BEAMS: STRAIN, STRESS, DEFLECTIONS The beam, or flexural member, is frequently encountered in structures and machines, and its elementary stress analysis constitutes one of the more interesting facets
More informationAssumptions: beam is initially straight, is elastically deformed by the loads, such that the slope and deflection of the elastic curve are
*12.4 SLOPE & DISPLACEMENT BY THE MOMENTAREA METHOD Assumptions: beam is initially straight, is elastically deformed by the loads, such that the slope and deflection of the elastic curve are very small,
More informationCHAPTER 6: Shearing Stresses in Beams
(130) CHAPTER 6: Shearing Stresses in Beams When a beam is in pure bending, the only stress resultants are the bending moments and the only stresses are the normal stresses acting on the cross sections.
More informationUnit Workbook 1 Level 4 ENG U8 Mechanical Principles 2018 UniCourse Ltd. All Rights Reserved. Sample
Pearson BTEC Levels 4 Higher Nationals in Engineering (RQF) Unit 8: Mechanical Principles Unit Workbook 1 in a series of 4 for this unit Learning Outcome 1 Static Mechanical Systems Page 1 of 23 1.1 Shafts
More informationNAME: Given Formulae: Law of Cosines: Law of Sines:
NME: Given Formulae: Law of Cosines: EXM 3 PST PROBLEMS (LESSONS 21 TO 28) 100 points Thursday, November 16, 2017, 7pm to 9:30, Room 200 You are allowed to use a calculator and drawing equipment, only.
More informationCIVIL DEPARTMENT MECHANICS OF STRUCTURES ASSIGNMENT NO 1. Brach: CE YEAR:
MECHANICS OF STRUCTURES ASSIGNMENT NO 1 SEMESTER: V 1) Find the least moment of Inertia about the centroidal axes XX and YY of an unequal angle section 125 mm 75 mm 10 mm as shown in figure 2) Determine
More informationExternal Work. When a force F undergoes a displacement dx in the same direction i as the force, the work done is
Structure Analysis I Chapter 9 Deflection Energy Method External Work Energy Method When a force F undergoes a displacement dx in the same direction i as the force, the work done is du e = F dx If the
More informationShear Force V: Positive shear tends to rotate the segment clockwise.
INTERNL FORCES IN EM efore a structural element can be designed, it is necessary to determine the internal forces that act within the element. The internal forces for a beam section will consist of a shear
More informationChapter 2 Basis for Indeterminate Structures
Chapter  Basis for the Analysis of Indeterminate Structures.1 Introduction... 3.1.1 Background... 3.1. Basis of Structural Analysis... 4. Small Displacements... 6..1 Introduction... 6.. Derivation...
More informationTHE AREAMOMENT / MOMENTAREA METHODS:
THE AREAMOMENT / MOMENTAREA METHODS: The area moment method is a semi graphical method of dealing with problems of deflection of beams subjected to bending. The method is based on a geometrical interpretation
More informationENGINEERING COUNCIL DIPLOMA LEVEL MECHANICS OF SOLIDS D209 TUTORIAL 3  SHEAR FORCE AND BENDING MOMENTS IN BEAMS
ENGINEERING COUNCIL DIPLOMA LEVEL MECHANICS OF SOLIDS D209 TUTORIAL 3  SHEAR FORCE AND BENDING MOMENTS IN BEAMS You should judge your progress by completing the self assessment exercises. On completion
More informationModule 3. Analysis of Statically Indeterminate Structures by the Displacement Method
odule 3 Analysis of Statically Indeterminate Structures by the Displacement ethod Lesson 16 The SlopeDeflection ethod: rames Without Sidesway Instructional Objectives After reading this chapter the student
More informationChapter 3. Load and Stress Analysis
Chapter 3 Load and Stress Analysis 2 Shear Force and Bending Moments in Beams Internal shear force V & bending moment M must ensure equilibrium Fig. 3 2 Sign Conventions for Bending and Shear Fig. 3 3
More information[7] Torsion. [7.1] Torsion. [7.2] Statically Indeterminate Torsion. [7] Torsion Page 1 of 21
[7] Torsion Page 1 of 21 [7] Torsion [7.1] Torsion [7.2] Statically Indeterminate Torsion [7] Torsion Page 2 of 21 [7.1] Torsion SHEAR STRAIN DUE TO TORSION 1) A shaft with a circular cross section is
More informationmportant nstructions to examiners: ) The answers should be examined by key words and not as wordtoword as given in the model answer scheme. ) The model answer and the answer written by candidate may
More informationStrength of Materials Prof. S.K.Bhattacharya Dept. of Civil Engineering, I.I.T., Kharagpur Lecture No.26 Stresses in BeamsI
Strength of Materials Prof. S.K.Bhattacharya Dept. of Civil Engineering, I.I.T., Kharagpur Lecture No.26 Stresses in BeamsI Welcome to the first lesson of the 6th module which is on Stresses in Beams
More informationUNIVERSITY OF SASKATCHEWAN ME MECHANICS OF MATERIALS I FINAL EXAM DECEMBER 13, 2008 Professor A. Dolovich
UNIVERSITY OF SASKATCHEWAN ME 313.3 MECHANICS OF MATERIALS I FINAL EXAM DECEMBER 13, 2008 Professor A. Dolovich A CLOSED BOOK EXAMINATION TIME: 3 HOURS For Marker s Use Only LAST NAME (printed): FIRST
More informationChapter 4 Deflection and Stiffness
Chapter 4 Deflection and Stiffness Asst. Prof. Dr. Supakit Rooppakhun Chapter Outline Deflection and Stiffness 41 Spring Rates 42 Tension, Compression, and Torsion 43 Deflection Due to Bending 44 Beam
More informationMTE 119 STATICS FINAL HELP SESSION REVIEW PROBLEMS PAGE 1 9 NAME & ID DATE. Example Problem P.1
MTE STATICS Example Problem P. Beer & Johnston, 004 by Mc GrawHill Companies, Inc. The structure shown consists of a beam of rectangular cross section (4in width, 8in height. (a Draw the shear and bending
More informationCHAPTER 6 BENDING Part 1
Ishik University / Sulaimani Civil Engineering Department Mechanics of Materials CE 211 CHAPTER 6 BENDING Part 11 CHAPTER 6 Bending Outlines of this chapter: 6.1. Chapter Objectives 6.2. Shear and
More informationModule 3. Analysis of Statically Indeterminate Structures by the Displacement Method
odule 3 Analysis of Statically Indeterminate Structures by the Displacement ethod Lesson 21 The oment Distribution ethod: rames with Sidesway Instructional Objectives After reading this chapter the student
More informationDownloaded from Downloaded from / 1
PURWANCHAL UNIVERSITY III SEMESTER FINAL EXAMINATION2002 LEVEL : B. E. (Civil) SUBJECT: BEG256CI, Strength of Material Full Marks: 80 TIME: 03:00 hrs Pass marks: 32 Candidates are required to give their
More information= 50 ksi. The maximum beam deflection Δ max is not = R B. = 30 kips. Notes for Strength of Materials, ET 200
Notes for Strength of Materials, ET 00 Steel Six Easy Steps Steel beam design is about selecting the lightest steel beam that will support the load without exceeding the bending strength or shear strength
More informationMECHANICS OF MATERIALS. Analysis of Beams for Bending
MECHANICS OF MATERIALS Analysis of Beams for Bending By NUR FARHAYU ARIFFIN Faculty of Civil Engineering & Earth Resources Chapter Description Expected Outcomes Define the elastic deformation of an axially
More informationMechanics of Materials II. Chapter III. A review of the fundamental formulation of stress, strain, and deflection
Mechanics of Materials II Chapter III A review of the fundamental formulation of stress, strain, and deflection Outline Introduction Assumtions and limitations Axial loading Torsion of circular shafts
More informationENG2000 Chapter 7 Beams. ENG2000: R.I. Hornsey Beam: 1
ENG2000 Chapter 7 Beams ENG2000: R.I. Hornsey Beam: 1 Overview In this chapter, we consider the stresses and moments present in loaded beams shear stress and bending moment diagrams We will also look at
More informationCIV100 Mechanics. Module 5: Internal Forces and Design. by: Jinyue Zhang. By the end of this Module you should be able to:
CIV100 Mechanics Module 5: Internal Forces and Design by: Jinyue Zhang Module Objective By the end of this Module you should be able to: Find internal forces of any structural members Understand how Shear
More information