Strength of materials - Axial force, shear force and bending moment

STRENGTH OF MATERIALS TRAN MINH TU - University of Civil Engineering, Giai Phong Str. 55, Hai Ba Trung Dist. Hanoi, Vietnam 1/10/2013 1 2 CHAPTER 1/10/2013 2 Axial Force, Shear Force and Bending Moment 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 Contents 1/10/2013

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3 2.1. Introduction 1/10/2013 4 - 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 5 - Types of supports 2.1. Introduction Idealized A A V A H - Pin support: prevents the translation at the end of a beam but does not prevent the rotation. 1/10/2013 6 - Roller support: prevents the translation in the vertical direction but not in the horizontal direction, and does not prevent the rotation. 2.1. Introduction A A V 1/10/2013 7 2.1. Introduction - Fixed (clamped) support: the bar can neither translate nor rotate. A A V A H AM 1/10/2013 8 2.1. Introduction 1/10/2013 9 2.1.2. Types of beams 2.1. Introduction 1/10/2013 10 2.2. Internal Stress Resultants y z xMx My Mz Qx NZ Qy In general, internal stress resultants (internal forces) consist of 6 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 3 internal stress resultants exert on this plane (zoy) . y z xMx NZ Qy • Nz – axial force (N); • Qy – shear force (Q); • Mx - bending moment (M) 1/10/2013 11 2.2. Internal force Resultants  To determine the internal force resultants => Using the method of sections. Q N Q N NN N 1/10/2013 12  Sign convention: 2.2. Internal force Resultants N N NN • Axial force: positive when outward of an element, negative when inward of an element • Shear force: positive when acts clockwise against an element, negative when acts counterclockwise against an element • Bending moment: positive when compresses the upper part of the beam and negative when compresses the lower part of the beam 1/10/2013 13 2.2. Axial, Shear and Moment diagram • Because of the applied loadings, the beams develop an internal shear force and bending moment that, in general, vary from point to point along the axis of the beam. In order to properly design a beam it therefore becomes necessary to determine the maximum shear and moment in the beam. • One way to do this is expressing N, Q and M as the functions of their arbitrary position z along the beam’s axis. These axial, shear and moment functions can then be plotted and represented by graph called the axial, shear and moment diagram 2.1.3. Axial, Shear and Moment diagram 1/10/2013 14 2.2. Axial, Shear and Moment diagram N, Q z M z Moment diagram to be plotted by side, which it is in tension 1/10/2013 15 Example 2.1: Draw the shear and moment diagram for the beam shown in the figure. Solution: 1. Support reactions VA VB F a b C   0A BM V a b Fa      0B AM V a b Fb     B Fa V a b     A Fb V a b    Recheck: 0Y  2.3. Example 1/10/2013 16 F a b VA VB C 1 1 Section 1 – 1: VA z1 Q M N 0N  10 z a    0A A Fb Y Q V Q V a b         Section 2 – 2:   1 0 1 10A A Fbz M M V z M V z a b         0N  20 z b    0B B Fa Y Q V Q V a b             2 0 2 20B B Faz M M V z M V z a b         2 2 VB z2 Q M N Segment AC Segment BC A B 2.3. Example 1/10/2013 17 2.3. Example Comment 1: F a b VA VB Fb a+b a+b Fa + N M Q Fab a+b F C   : Fb AC Q a b     : Fa BC Q a b      1: Fbz AC M a b     2: Faz BC M a b   The section on which the concentrated force acts, the shear force diagram has “jump” 1/10/2013 18 2.3. Example L q VA VB Example 2.2: Draw the shear and moment diagram for the beam shown in the figure. Solution: 1. Support reactions 2 . 0 2 A B ql M V l   . 2 A q l V  2 . 0 2 B A ql M V l   . 2 B q l V  Symmetrically: . 2 A B q l V V   Or: 2. Internal force resultant’s functions: Segment 1-1 (0 ≤ z  L) . 2 ql Q q z   2. . 2 2 ql q M z z   1 1 Q zVA M N q 0AY Q qz V    2 0 0 2 A qz M M V z    1/10/2013 19 2.3. Example  Comment 2 L q VA VBqL/2 qL/2 + Q L/2 qL2/8 M . 2 ql Q q z  2. . 2 2 ql q M z z  0 2 A qL z Q   2 B qL z L Q    0 0Az M   0Bz L M   2 qL M ' qz  0 2 L M ' z   0M'' q     2 2 8 max z L/ qL M M     The section on which the shear force is equal to zero then the bending moment is maximum. 1/10/2013 20 2.3. Example – example 2.3 1. Support reactions: .( ) 0A BM V a b M    .( ) 0B AM V a b M    B M V a b    A M V a b    2. Stress resultants: AC: Section 1-1 ( 0 ≤ z1  a) y A M Q V a b      VA VB a b C M .x AM V z  Q VA M z1 VB M Q z2 1 1 2 2 Section 2-2 ( 0 ≤ z2  b) y A M Q V a b      2.x BM V z 1/10/2013 21 2.3. Example – example 2.3 a b VA VB M (a+b) M (a+b)Ma (a+b) Mb (a+b) M Q M C M y A M Q V a b      1.x AM V z  AC: ( 0 ≤ z1  a) y A M Q V a b      BC: ( 0 ≤ z2  b) 2.x BM V z Comment 3 The section on which the concentrated moment acts, the bending moment diagram has “jump” 1/10/2013 22 2.4. Relationships between transverse loads, shear forces, and bending moments - Consider the beam shown in the figure, which is subjected to an arbitrary loading. The free-body diagram for a small segment dz of the beam: ( ) 0Y Q dQ Q q z dz     ( ) dQ q z dz   ( ) 0 2 2 dz dz M M dM M Q dQ Q       dM Q dz   2 2 ( ) d M dQ q z dz dz   - Positive distributed load q(z): acts upward on the beam q(z) > 0 1/10/2013 23 2.4. Relationships between loads, shear forces, and bending moments Application: - Recognizing the type of Q and M diagrams when the distributed load’s function is known, i. e if the distributed load’s function is n- degree, then the shear force’s function will be (n+1)-degree and the bending moment function will be (n+2)-degree. - The section, on which the shear force is equal to zero then the bending moment is maximum. - Determining Q, M on the arbitrary section, when knowing the value of Q and M on the specific section. •Qright = Qleft + Sq ( Sq – Area of distributed load diagram q) •Mright = Mleft + SQ ( SQ – Area of shear force diagram Q) 1/10/2013 24 2.4. Relationships between loads, shear forces, and bending moments q z q(z) A B ( ) B B A A dQ q z dz  B A qQ Q S  Sq Q z Q(z) A B SQ ( ) B B A A dM Q z dz  B A QM M S  1/10/2013 25 2.4. Relationships between loads, shear forces, and bending moments 1/10/2013 26 2.5. Graphical Method for Constructing Shear and Moment Diagrams - Base on the relationships between loads, shear force, and bending moments - Knows the distributed load q(z) => Predict the types of shear force and bending moment diagram => Indentify the necessary number of points to construct the diagram. • q=0 => Q=const => QA=? (or QB) M linear => MA=? and MB=? • q=const => Q linear => QA=? QB=? M quadratic => MA=?; MB=?; maximum? convex, concave,..? 1/10/2013 27 • Support reactions q F=qa VA VB .3 2 .2 . 0B AM V a qa a F a    5 3 AV qa  .3 2 . .2 0A BM V a qa a F a    4 3 BV qa  Segment AC: 2a a C q=const Q linear QA=VA QC=VA+Sq=5qa/3-2qa=-qa/3 M quadratic: MA=0 MC=MA+SQ=4qa 2/3; Mmax=25qa2/18 5 3 qa 1 3 qa + 5a/3 Mmax=25qa 2/18 4qa2/3 2.5. Graphical Method for Constructing Shear and Moment Diagrams -Example 1/10/2013 28 2.5. Graphical Method for Constructing Shear and Moment Diagrams -Example 2a a VA VB C 5 3 qa 4 3 qa 1 3 qa + 5a/3 Mmax=25qa 2/18 4qa2/3 Segment BC: q F=qa q= 0 Q = const QB= - VB M linear: MB=0 MC = MB - SQ=4qa 2/3 Q M 1/10/2013 29 F a aa A B C D 2.5. Graphical Method for Constructing Shear and Moment Diagrams - Example a aa A R B B C D VD F R 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: 2 D F V R  1/10/2013 30 2.5. Graphical Method for Constructing Shear and Moment Diagrams -Example 0 ( ) 2 2 C B Q F Fa M M S a      (Q)2 F 2 F (M) 2 Fa 0 ( ) 2 2 C D Q Fa Fa M M S       F a aa A B C D B C D VD F R a. Segment BC: q(z)=0 => Q=const => QB= R = F/2 => M linear 0BM  b. Segment CD: q(z)=0 => Q=const => QD= -VD = - F/2 => M linear 0DM  1/10/2013 31 2.) Mean beam AB: 2 F (Q) 2 Fa (M) F a aa A B C D A R B 2.5. Graphical Method for Constructing Shear and Moment Diagrams -Example 1/10/2013 32 2.5. Graphical Method for Constructing Shear and Moment Diagrams -Example a aa A B C D 2 Fa 2 Fa 2 F 2 F 2 F (Q) (M) 3.) The Shear force and bending moment diagram of a system of the beams 1/10/2013 33 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 34 2.6. Normal, Shear force and bending moment diagram of frame VD VA HA a a a F q A D B C Draw the axial force, shear force and bending moment diagram of the frame: with q=8kN/m, F=5kN, a=1m Solution: 1. Support reactions: x y 0 5( )AX H F kN    1 .1 .1. .1 0 2 A DM V q F    9( )DV kN  1 .1 .1. .1 .2 0 2 D A AM V q F H     1( )AV kN  2. Axial force diagram N AB: 1AB AN V kN  BC: 1BC AN V kN  CD: 0CDN  1/10/2013 35 2.6. Normal, Shear force and bending moment diagram of frame 1 1 1 + + N kN 3. Shear force and bending moment diagram AB: q=0 Q const  5A AQ H kN   5 5 +  M linear: 0AM  0 5.1 5B A QM M S kNm      5 Q kNM kNm 1/10/2013 36 2.6. Normal, Shear force and bending moment diagram of frame BC: q=0 Q const  0BQ   M linear: 5BM kNm  5 0 5C B QM M S kNm      1 1 1 + + N kN 5 5 + 5 Q kNM kNm 5 CD: q=const 9D DQ V kN      M quadratic: 0DM  0 ( 1 9).1/ 2 5C D QM M S kNm         Q linear: 9 ( 8.1) 1C D qQ Q S kN         9 1 - 5 1kN Equilibrium of joint 5kNm 1kN 5kNm 1/10/2013 37 Draw the shear and moment diagram for the beam shown in the figure. 2.7. Home works 1/10/2013 38 2.7. Homework Draw the shear and moment diagram for the beam shown in the figures. 1/10/2013 39 Draw the shear and moment diagram for the compound beam shown in the figures. 2.7. Homework 1/10/2013 40 THANK YOU FOR YOUR ATTENTION ! 1/10/2013 41

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