ending Internal orces mi@seu.edu.cn
Contents Introduction( 弯曲变形简介 ) The orms of Internal orces in eams( 梁中内力的形式 ) Symmetric ending( 对称弯曲 ) Types of eams( 静定梁的分类 ) Types of Supports and Reactions( 梁的支撑种类及相应的支座反力形式 ) Internal Releases( 内力释放器 ) Determinacy of eams( 梁的静定性 ) Sign Convention of Shearing orces and ending Moments( 剪力和弯矩的符号规则 ) Direct Calculation of Shearing orces( 剪力简易计算法 )
Direct Calculation of ending Moments( 弯矩简易计算法 ) Diagram of Shearing orces & ending Moments( 剪力图与弯矩图 ) Differential Relations among Loads, Shearing orces and ending Moments( 弯矩 剪力和荷载之间的微分关系 ) Diagram of Internal orces for Plane rames( 平面刚架的内力图 ) Diagram of Internal orces for Curved eams( 曲梁的内力图 ) 3
Introduction 4
Introduction 5
Introduction 6
Introduction 7
Introduction 8
Introduction 9
The orm of Internal orces in eams Internal forces on cross-sections of bending beams m m M m M m m m S S The internal forces on cross-section m-m can be simplified as shearing forces and bending moments at cross-sectional centroid. 10
Symmetric ending eam: one type of prismatic bars which takes bending as the main form of deformation. Symmetric longitudinal plane: formed by the centroidal axes of cross sections. Symmetric bending: all external forces & moments acting in the symmetric longitudinal plane (beam axis being bent from a straight line to a curve within the symmetric longitudinal plane). 11
Types of eams Simply supported beam Cantilever beam eam with an overhang Compound beam
Types of Supports and Reactions ixed support Hinged or pinned support Roller support M R Rx Rx Ry Ry Ry 13
Internal Releases xial release Shear release Moment release 14
Internal Releases
Determinacy of eams Statically determinate beams Statically indeterminate beams - Cantilever beam - Simply supported cantilever beam - Simply supported beam - Continuous beam - Overhanging beam - ixed beam 16
Sign Convention of Shearing orces and ending Moments Shearing forces - Positive: left up / right down (clockwise loop) S S S S ending moment - Positive: upper half under compression / lower half under tension (top concave / bottom convex) M M M M 17
Sample Problem ind the shearing force and bending moment at the left and right cross-section at point C. Solution: C a a 1. Reaction forces at the supports. Internal forces (assuming positive) at the right crosssection of C. S C M a C a SC M C 18
3. Internal forces (assuming positive) at the left crosssection of C. SC a MC a a M C C SC a Note: for a concentrated force acting at C, the change of bending moment is zero while that of shearing force is equal to the magnitude of the concentrated force. 19
Sample Problem ind the shearing force and bending moment at the left and right cross-section at point C. a C Solution: a a 1. Reaction forces at the supports. Internal force at the right cross-section of C. S C M a C a SC M C 0
3. Internal force at the left cross-section of C. C a SC a SC a M a C a M C Note: or a concentrated moment acting at C, the change of shearing force is zero while that of bending moment is equal to the magnitude of the concentrated moment. 1
Direct Calculation of Shearing orces The shearing forces at a beam cross-section is equal to the net of the external forces collected from either side of the cross-section. Upward net from left portion or downward net from right portion results in positive shearing force at the cross-section. ll forces acting on a beam, including the reaction forces at the supports, must be taken into consideration.
Sample Problem Directly calculate the shearing force at cross-section C without the explicit use of method of section. 1 3 4 q C Solution: SC 1 3 - ased on the left portion from cross-section C: - ased on the right portion from cross-section C: ql SC 4 Note: in practice, we pick the portion with simpler loadings. 3
Direct Calculation of ending Moments The bending moment at a beam cross-section is equal to the net of the external moments with respect to the cross-section centroid, collected from either side of the cross-section Clockwise net from left portion or counter clockwise net from right portion results in positive bending moment. ll forces acting on a beam, including the reaction forces at the supports, must be taken into consideration. 4
Sample Problem Directly calculate the bending moment at cross-section C without the explicit use of method of section. Solution: - ased on the left portion from cross-section C M C d d 1 1 M1 1 M 1 C - ased on the right portion from cross-section C M C d d 5
Diagram of Shearing orces & ending Moments Equation of shearing forces and bending moments - Equation of shearing forces: S = S (x) - Equation of bending moments: M = M(x) - x: denotes the position of cross-section. Diagram of shearing forces and bending moments - bscissa: cross-section position (x) - Ordinate: shearing forces ( S ) / bending moments (M) 6
Sample Problem Draw the diagram of shearing forces and bending moments x 1 x C Solution: 1. Reaction force at the supports:. Equation of internal forces S( x1) S( x) a a x1 [0, a) x [0, a) Positive: left & upward Negative: right & upward 7
S 0.5 x1 M( x ) x1 [0, a) 1 x1 Positive: left & clockwise 0.5 x x 1 C x M( x) x x [0, a) Positive: right & counter clockwise a a M 0.5 a 8
Sample Problem Draw the diagram of shearing forces and bending moments x 1 a x C Solution: a a 1. Reaction forces at the supports S 0.5. Equations of internal forces S ( x1 ) ; S ( x) x x M ( x1) ; M ( x) 1 M -0.5a 0.5a 9
Sample Problem ql S ql x / q l ql ql / Plot the diagram of shearing forces and bending moments Solution: 1. Reaction forces at the supports. Equation of internal forces ( ) ql S x qx ql x M ( x) x qx q l ql x 8 M ql /8 0 xl 3. Diagram of internal forces 30
Relations among Loads, Shearing orces and ending Moments q( x) q( x) M (x) M( x) dm( x) x dx x q x dx x d x 0 S S S ( x) ( ) d ( ) S S x S x 1 M x dm x qxd x M x S xdx 0 31
Relations among Loads, Shearing orces and ending Moments x q x dx x d x 0 S S S x d S x q x, S S1 q x dx dx x1 1 M x dm x qxd x M x S xdx 0 x dm x x, M M xdx S 1 S dx x1 q x dm dx x Sign convention of q(x): up positive / down negative. 3
Relations among Loads, Shearing orces and ending Moments Without distributed load: q(x) = 0 q x d x S 0 x const dx Note: the slope of the diagram of shearing forces is zero. S dm x dx S x const M x const x const Note: the slope of the diagram of bending moments is constant. 33
Relations among Loads, Shearing orces and ending Moments With uniformly distributed load: q(x) = const q x d S x const const const S x x dx Note: the slope of the diagram of shearing forces is constant. dm x dx S x M x const x const x const Note: M(x) is a second order polynomial of x. 34
Relations among Loads, Shearing orces and ending Moments Loads Diagram of shearing forces Diagram of bending moments No load (P=0 & q=0) S 0 S 0 0 0 S S Uniformly distributed q=const q 0 q 0 0 q q 0 Concentrated load (P) C Concentrated moment (m) C Discontinuous s Const C C Continuous but not differentiable M Discontinuous M C 35
Sample Problem Draw the diagram of shearing forces and bending moments. Solution: ql 8 5ql 8 S l ql / l ql /8 ql /8 C q M 36
Sample Problem Draw the diagram of shearing forces and bending moments. Solution: 3ql 8 ql 8 S 3ql 8 l x q 3l 8 x l l ql 8 M 9ql 18 ql 16 37
Sample Problem Draw the diagram of shearing forces and bending moments. Solution: qa qa qa S q qa qa / a a a qa qa / M qa / 38
Sample Problem Draw the diagram of shearing forces and bending moments. Solution: qa / q a q a qa qa S qa qa / qa / M qa / qa /8 39
Diagram of Internal orces for Plane rames Plane frames are composed of prismatic bars with different orientations in the same plane. Rigid joint ar Neighboring bars are connected via rigid joints ar Internal forces acting on cross-sections of component bars include axial force, shearing force, and bending moment. 40
Diagram of Internal orces for Plane rames Sign convention - xial forces: tension positive / compression negative - Shearing forces and bending moments: the same sign convention when the plane frame is observed from inside. Drawing convention - xial forces and shearing forces: draw at either side of the bar with sign labeled. - ending moments: draw at the side under tension without sign labeled. 41
q = 1 kn/m 4 m 3 m Sample Problem Draw the diagram of axial forces, shearing forces and bending moments for the plane frame shown below. Solution: 1. reaction forces at the supports. M 811 4 / 11 0 3 5 kn x 0 1 4 1 0 3 kn x x y 0 y 8 0 3 kn y D y 8 kn 1 m m E x C 1 kn 4
q = 1 kn/m 4 m 3 m. Diagram of internal forces 1) Diagram of axial forces C : N1 5kN DC : N 1kN D : 3kN N3 D E 5 kn C 1 kn D 8 kn 1 m m E 3 kn C 1 kn 5 kn 3 kn 3 kn 43
q =1 kn/m 4 m 3 m ) Diagram of shearing forces C:horizontal line DC:horizontal line D:oblique straight line D 8 kn 1 m m E C D 1kN E 3 kn C 1 kn 5 kn 1 kn 3 kn 3 kn 5 kn 3 kn 44
3 m q = 1 kn/m 4 m 3 m 3) Diagram of bending moments 8 kn C: oblique straight line (two-point) CD: oblique straight line (two-point) D 1 m m E C D: parabola (three-point) 3 knm D 4 knm C E 3 knm 1 kn 4 knm 4.5 knm 7 knm 3 kn 5 kn 0 3 kn 0 45
Diagram of Internal orces for Curved eams Draw the diagram of axial forces, shearing forces and bending moments for the curved beam shown below. Solution: Static equilibrium N sin cos M y Rsin S 46
Contents Introduction( 弯曲变形简介 ) The orms of Internal orces in eams( 梁中内力的形式 ) Symmetric ending( 对称弯曲 ) Types of eams( 静定梁的分类 ) Types of Supports and Reactions( 梁的支撑种类及相应的支座反力形式 ) Internal Releases( 内力释放器 ) Determinacy of eams( 梁的静定性 ) Sign Convention of Shearing orces and ending Moments( 剪力和弯矩的符号规则 ) Direct Calculation of Shearing orces( 剪力简易计算法 ) 47
Contents Direct Calculation of ending Moments( 弯矩简易计算法 ) Diagram of Shearing orces & ending Moments( 剪力图与弯矩图 ) Differential Relations among Loads, Shearing orces and ending Moments( 弯矩 剪力和荷载之间的微分关系 ) Diagram of Internal orces for Plane rames( 平面刚架的内力图 ) Diagram of Internal orces for Curved eams( 曲梁的内力图 ) 48