Chapter 5 Axial Force, Shear, and Bending Moment

Size: px

Start display at page:

Download "Chapter 5 Axial Force, Shear, and Bending Moment"

妥李
5 years ago
Views:

1 Chapter 4 Axial Force, Shear, and Bending-moment 4-1 Introduction 1. Beams In many instances in structural and machine design, members must resist forces applied laterally or transversely to their axes. Such members are called beams. Here only the cases that all loads act and the deflections occur in the same plane----plane of bending----are considered. 梁的轴线受载后将变成曲线 ----elastic curve. 此种变形称为弯曲变形.

2 2. Classification of beams Simply supported or simple beam Overhanging beam Cantilever or fixed beam

3 通常梁有各方面的简化 : (1) 构件的简化 : 有时不考虑横截面的具体形状, 并忽略构造上的枝节, 如键槽销孔阶梯等, 简化为一直杆. (2) 载荷的简化 : 载荷作用范围小时, 处理为集中力或集中力偶, 若连续作用, 则处理为分布载荷. (3) 支座的简化 :fixed support,or either pins or rollers. 所以, 除了上述的三种梁外, 还有 continuous beams, restrained beams, 属于静不定问题.

4 4-2 Direct approach: axial force, shear, bending moment 要求得梁上某处的内力, 依然可用截面法. 梁平衡, 局部也平衡. 1. Axial force in beams From F x = 0 one can get the axial force P The line of action of the axial force will always be directed through the centroid of the beam s crosssection to avoid bending.

5 2. Shear in beams From F y =0, one has internal force V which is called the shear or shear force acting at right angles to the axis of the beam. The shear is the resultant of stresses distributed over the cross section and is numerically equal to the algebraic sum of all the vertical components of the external forces acting on the isolated segment, but it is opposite in direction. Sense of the shear force: 左上右下为正, 反之为负. 如图.

6 3. Bending moment in beams From M z =0, one has bending moment M acting on the cross section of the beam. The bending moment is the resultant moment of stresses distributed over the cross section and is numerically equal to the external moment acting on the isolated segment, but it is opposite in direction. 上面关于 P, V, M 正负的规定, 使得在求解内力时, 无论研究左段还是研究右段都将得到一样的结果.

7 4-3 Shear and Bending-Moment Diagrams 剪力, 弯矩随轴截面位置的变化曲线分别称为剪力图与弯矩图. 本节以例题为主, 介绍如何求梁的内力以及梁的内力图的画法.

8 Example 4-1 Construct axial-force, shear, and bending-moment diagrams for the cantilever loaded with an inclined force at the end; see the figure. Solution: 1. Calculate reactions 2. Calculate P, V, M. Study left segment: F x =0 P(x)=P F y =0 V(x)=P M Cz =0, PL-Px+M(x)=0 M(x)=-PL+Px M(x) 3. Draw the P, V, M diagrams

9 Example 4-2 Plot shear and a bending-moment diagrams for a simple beam with a uniformly distributed load; see Fig.. Solution: 1 Calculate the reactions 2 Calculate V(x), M(x). Study the left segment F y =0, 0.5w 0 L - w 0 x - V(x)=0 M Cz =0, -0.5w 0 Lx+0.5w 0 x 2 +M(x)=0 C 注意 : 如果在截面处假设的未知内力沿正方向, 则计算结果为正, 内力图上即为正, 结果为负, 内力图上为负. 避免二次判断.

10 Example 4-3 Simple beam with load is shown in the figure. Plot shear and bending-moment diagrams. Solution: 1. Reactions R A =Pb/l, R B =Pa/l 2 V(x), M(x) for AC: study left segment V 1 (x)=r A =Pb/l (0 x a) M(x)=R A x=pbx/l (0 x a) for BC: study right segment V 2 (x)=-r B =-Pa/l (a x l) M(x)=R B (l-x)=pa(l-x)/l (a x l) 3. Shear, bending-moment diagrams M max =M(a)=Pab/l M(0)=M(l)=0 Bending-moment

11 Example 4-4 Simple beam with load is shown in the figure. Plot shear and bending-moment diagrams. Solution: 1. Reactions R A =R B =M 0 /l 2 V(x), M(x) AC: V 1 (x)=r A = M 0 /l (0 x a) M(x)=R A x= M 0 x/l (0 x a) BC: V 2 (x)=+r B =+ M 0 /l (a x l) M(x)=-R B (l-x)= -M 0 (l-x)/l (a x l) 3. Shear, bending-moment diagrams M max = M 0 a/l M(0)=M(l)=0

12 由以上几个最基本载荷形式的例题可以看出, 剪力图和弯矩图有如下一些规律 : (1) 在梁上没有分布载荷的地方, 剪力图为水平线, 弯矩图为直线, 一般为斜线 ; (2) 在有均布载荷的梁段上, 剪力图为斜线, 弯矩图则为抛物线 ; (3) 在集中力作用处, 剪力图有突变, 突变值即为该处集中力的大小, 弯矩图在此则为一折角 ; (4) 在集中力偶作用处, 剪力图没有变化, 弯矩图则有突变, 突变之值即为该处集中力偶之力偶矩. 掌握这些规律, 有助于正确而简捷的画出梁上的剪力图和弯矩图.

13 Example 4-5 For the beam loaded as shown in the figure, write explicit expression for V(x) s and M(x) s along the spans. Solution: Q 1. Reactions: Q= =9 kn R A =Q 2/9=2kN, R B =Q-R A =7kN 2. V(x), M(x) x x R A q(x) for AC: V(x)=R A =2kN (0 x 3) M(x)=R A x=2x knm (0 x 3) for BC: q(x)=3/6 (x-3)=0.5(x-3) V(x)=R A -0.5q(x-3)= (x-3) 2 =2-0.25(x-3) 2 (3 x 9) M(x)= R A x-0.5q(x-3) (x-3)/3 = 2x - (x-3) 3 /12 (3 x 9) R B

14 Example 4-6 Beam with load is shown in the figure. Plot bending-moment diagram and determine M max. P Solution: Using superposition principle PL M max =0.25Pl+0.125w 0 l 2 + P

15 4-4 Shear and Bending-moments by Integration The beam with loads is shown in the figure. Determine the shear and bending-moment. Consider a beam element dx long. For equilibrium, F y [ M ( x) = 0 V ( x) [ V ( x) + dv ( x)] + q( x) dx = 0 m A dv ( x) dx = 0 = q( x) + dm ( x)] V ( x) dx 略去二阶微量 q( x) dx dm( x) dx 1 dx 2 = V ( x) M ( x) [ q( x) dx]( 1 2 d dx) = 2 0 M( x) dx 2 d = q( x) d d d

16 By integrating, we have: x x = q( x) dx + C1 M ( x) = 0 0 V ( x) V ( x) dx + C 2 所以, 又可得到剪力与弯矩图的如下规律 : 1) 梁上某段无分布载荷时, 该段剪力图为水平线, 弯矩图为斜线. V>0, M 图递增, V<0, M 图递减. 2) 某段有向下分布载荷时, 该段剪力图递减, 弯矩图为凹向下曲线 (concave down); 反之, 则凹向上 (concave up). (Vice verse) 3) 在 V=0 处, M=M max or M=M min 弯矩取极值. 此外, 弯矩极值也可能发生在集中力或集中力偶作用处. 此时, 相应的曲线会有突变, 突变值等于集中力或集中力偶. 利用以上规律可以比较方便的作出剪力图与弯矩图, 并校核其正确性.

17 Example 4-7 Construct shear and moment diagrams for the symmetrically loaded beam shown in Fig. By the integration process. Solution: 1. Reactions 2. V-diagram 3. M-diagram The moment diagram is obtained by summing up the area of the shear diagram.

18 EXAMPLE 4-8 Consider a simple beam with a uniformly increasing load intensity from an end, as shown in Fig.(a). The total applied load is W. (a) Construct shear and moment diagrams with the aid of the integration process. (b)derive expressions for V and M using integration. Solution (a)since the total load W=kL 2 /2, k=2w/l 2. reactions are W/3 and 2W/3, downward. Since the rate of applied load is smaller on the left end than on the right, the shear diagram is concave upward. The point of zero shear occurs as

19 b. Boundary conditions: M(0)=0, M(L)=0 M(L)=0

20 Example 4-9 Construct the bending moment diagram fro the horizontal beam loaded as shown as following. Bending M Shear V

21 Example 4-10 Construct shear and bending moment diagrams for loaded beam shown as following with the aid of the integration process.

22 15(10 + x x 1 1 = 12 ) x 1 = 0 15(10 + x x 2 2 ) x = ( x 2 15) = 0

kNΠm 2 5 % GB [4 5 ] ( ) ; m 100 ; 60m m ( A 114 [3 ] B 113) GB

kNΠm 2 5 % GB [4 5 ] ( ) ; m 100 ; 60m m ( A 114 [3 ] B 113) GB 37 3 Vol137 No13 2 0 0 4 3 CHINA CIVIL ENGINEERINGJOURNAL Mar. 2 0 0 4 ( J GJ322002) ( ) : (J GJ322002) 56 : ; ; ; ; : TU973 + 112 : A : 10002131X (2004) 0320001211 EXPLANATION ON SOME QUESTIONS ABOUT