Lecture Strength of Materials I: Chapter 2 - PhD. Tran Minh Tu

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Lecture Strength of Materials I: Chapter 2 - PhD. Tran Minh Tu

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Chapter 2 - Axial force, shear force and bending moment. The following will be discussed in this chapter: Internal stress resultants; relationships between loads, shear forces, and bending moments; graphical method for constructing shear and moment diagrams; normal, shear force and bending moment diagram of frame.

STRENGTH OF MATERIALS 1/10/2013 TRAN MINH TU - University of Civil Engineering, Giai Phong Str 55, Hai Ba Trung Dist Hanoi, Vietnam CHAPTER Axial Force, Shear Force and Bending Moment 1/10/2013 Contents 2.1 Introduction 2.2 Internal Stress Resultants 2.3 Example 2.4 Relationships between loads, shear forces, and bending moments 2.5 Graphical Method for Constructing Shear and Moment Diagrams 2.6 Normal, Shear force and bending moment diagram of frame 1/10/2013 2.1 Introduction - Structural members are usually classified according to the types of loads that they support - Planar structures: if they lie in a single plane and all loads act in that same plane 2.1.1 Support connections - Structural members are joined together in various ways depending on the intent of the designer The three types of joint most often specified are the pin connection, the roller support, and the fixed joint 1/10/2013 2.1 Introduction - Types of supports - Pin support: prevents the translation at the end of a beam but does not prevent the rotation A Idealized HA 1/10/2013 V A 2.1 Introduction - Roller support: prevents the translation in the vertical direction but not in the horizontal direction, and does not prevent the rotation A V A 1/10/2013 2.1 Introduction - Fixed (clamped) support: the bar can neither translate nor rotate A HA MA V A 1/10/2013 2.1 Introduction 1/10/2013 2.1 Introduction 2.1.2 Types of beams 1/10/2013 2.2 Internal Stress Resultants In general, internal stress resultants (internal forces) consist of components • Nz – Normal force • Qx, Qy – Shear forces • Mx, My – Bending moments • Mz – Torsional moment  Planar structures: if they lie in a single plane and all loads act in that same plane => Only internal stress resultants exert on this plane (zoy) • Nz – axial force (N); • Qy – shear force (Q); • Mx - bending moment (M) 1/10/2013 Mz Mx x Qx z NZ My Qy y Mx x z NZ Qy y 10 2.5 Graphical Method for Constructing Shear and Moment Diagrams -Example F=qa • Support reactions M B q  VA 3a  2qa.2a  F a   VA  qa  M A  VB 3a  2qa.a  F.2a  2a a VA VB qa +  VB  qa Segment AC: q=const QA=VA C 5a/3 Q linear qa QC=VA+Sq=5qa/3-2qa=-qa/3 M quadratic: MA=0 1/10/2013 MC=MA+SQ=4qa2/3; Mmax=25qa2/18 4qa2/3 Mmax=25qa2/18 27 2.5 Graphical Method for Constructing Shear and Moment Diagrams -Example F=qa Segment BC:q= q Q = const QB= - VB M linear: MB=0 MC = MB - SQ=4qa2/3 C 2a VA a VB qa + Q 5a/3 qa qa M 4qa2/3 1/10/2013 Mmax=25qa2/18 28 2.5 Graphical Method for Constructing Shear and Moment Diagrams - Example F Example 2.5: Draw the shear force and bending moment diagram for the compound beam shown in the figure: Solution: System of beams ABCD consists of: + Secondary beam BCD + Main beam AB 1) Secondary beam BCD: - Support reactions: F VD  R  A B D C a a F D B C R VD R A B a 1/10/2013 a a a 29 2.5 Graphical Method for Constructing Shear and Moment Diagrams -Example F a Segment BC: q(z)=0 A B D => Q=const => QB= R = F/2 => M linear C  MB   M C  M B  SQ   ( a F Fa a)  2 b Segment CD: q(z)=0 => Q=const => QD= -VD = - F/2 => M linear  MD  a a F D B C R VD F (Q) F Fa Fa  M C  M D  SQ   (  )  2 1/10/2013 (M) Fa 30 2.5 Graphical Method for Constructing Shear and Moment Diagrams -Example F A B D 2.) Mean beam AB: C a A a a R B F Fa 1/10/2013 (Q) (M) 31 2.5 Graphical Method for Constructing Shear and Moment Diagrams -Example 3.) The Shear force and bending moment diagram of a system of the beams A B D C a F a a F (Q) F Fa (M) Fa 1/10/2013 32 2.6 Normal, Shear force and bending moment diagram of frame - The frame is composed of several connected members that are fixed connection The design of these structures often requires drawing the shear and moment diagram for each of the members - Using a method of section, we determine the axial force, the shear force, and the bending moment acting on each members - Always draw the moment diagram on the tensile side of the member 1/10/2013 33 2.6 Normal, Shear force and bending moment diagram of frame Draw the axial force, shear force and bending moment diagram of the frame: with q=8kN/m, F=5kN, a=1m q Solution: C D Support reactions: X 0 H A  F  5(kN ) M  V  q  F   VD  9(kN )  A D M D  VA  q.1  F  H A   VA  1(kN ) 2 Axial force diagram N a a F y a A AB: N AB  VA  1kN BC: N BC  VA  1kN CD: NCD  1/10/2013 VD B HA x VA 34 2.6 Normal, Shear force and bending moment diagram of frame Shear force and bending moment diagram AB: q=0  Q  const  QA  H A  5kN  M linear:  M A   M B  M A  SQ   5.1  5kNm + 5 Q kN M kNm + N kN + 1/10/2013 1 35 2.6 Normal, Shear force and bending moment diagram of frame BC: q=0  Q  const  QB   M linear:  M B  5kNm  M C  M B  SQ    5kNm CD: q=const Q linear:  QD  VD  9kN  M quadratic:  M D   QC  QD  Sq  9  (8.1)  1kN  M C  M D  SQ   (1  9).1/  5kNm 1 + M kNm Q kN N kN Equilibrium of joint 5kNm + + 1/10/2013 - 1kN 5kNm 1kN 36 2.7 Home works Draw the shear and moment diagram for the beam shown in the figure 1/10/2013 37 2.7 Homework Draw the shear and moment diagram for the beam shown in the figures 1/10/2013 38 2.7 Homework Draw the shear and moment diagram for the compound beam shown in the figures 1/10/2013 39 1/10/2013 40 THANK YOU FOR YOUR ATTENTION ! 1/10/2013 41 ... 1/10 /20 13 2. 1 Introduction - Fixed (clamped) support: the bar can neither translate nor rotate A HA MA V A 1/10 /20 13 2. 1 Introduction 1/10 /20 13 2. 1 Introduction 2. 1 .2 Types of beams 1/10 /20 13 2. 2... qa QC=VA+Sq=5qa/ 3-2 qa=-qa/3 M quadratic: MA=0 1/10 /20 13 MC=MA+SQ=4qa2/3; Mmax =25 qa2/18 4qa2/3 Mmax =25 qa2/18 27 2. 5 Graphical Method for Constructing Shear and Moment Diagrams -Example F=qa Segment... M VA Q Q z2 VB Section 2- 2 ( ≤ z2  b) Qy  VA   1/10 /20 13 M x  VB z2 M ab 20 2. 3 Example – example 2. 3 M AC: ( ≤ z1  a) M Qy  VA   ab M x  VA z1 C a b VA VB BC: ( ≤ z2  b) M Qy

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