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IZMIR INSTITUTE OF TECHNOLOGY 4.0 16.07.2015 Department of Architecture AR232 Spring2006 INTERNAL FORCES AND MOMENTS IN MEMBERS 4.1 Introduction 4.2 Idealization of Structural Systems and Loads 4.3 Method of Analysis 4.4 Differential Equailibrium Relations 4.5 Examples Dr. Engin Aktaş 1 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR232 Spring2006 4.1. Introduction • Structural members are subjected to many different types of external forces. In the design of these members, it is essential to compute the stresses and deformations produced by the external loads The study of structural members √ 1. √ 2. 3. 4. 21.02.2005 Idealization of the structural system and the loads Computation of the external reactions Determination of the internal forces Calculation of the stresses and deformations caused by these internal forces Dr. Engin Aktaş 2 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR232 Spring2006 4.2. Idealization of Structural Systems and Loads 4.2.1 Types of Supports (1) Roller (Link) Supports B A A (a) Link (b) Rod (c) Roller 21.02.2005 B (d) Roller Dr. Engin Aktaş (a) and (b) can resist the force only in the direction of line AB (c) and (d) can only resist a force which ‘s perpendicular to the direction of movement of the roller. 3 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR232 Spring2006 (2) Pin or Hinge Supports (a) and (b) can resist a force acting in any direction in the plane. Therefore, in general, hinged or pinned supports have two unknowns (a) (b) (3) Fixed, clamped, built-in or encastre supports This type of support is capable of resisting a force in any direction and is also capable of resisting a moment or a couple. 21.02.2005 Dr. Engin Aktaş 4 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR232 Spring2006 4.2.2 Types of Loads A force is transmitted to a member through a post, hanger or another member. In such cases the load is applied over a very limited portion of the member. Such forces are idealized as concentrated forces and shown with a single vector acting at a point. In general, forces and loads are distributed over a finite length or area of the member. These are idealized as continuously distributed loads. dF y General Load Distribution on Member x dx If the total force on length Dx is denoted as DF, then the intensity of loading, q is defined as DF dF Dx 0 Dx dx q lim 21.02.2005 Dr. Engin Aktaş 5 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture y y q q(x)=constant AR232 Spring2006 q0 x xl l xl q0 x xl x l (b) Triangularly distributed (a) Uniformly distributed y q q q q1 2 1 x l y q q0 sin x l q0 q2 q1 x x l (d) Sinusoidal (c) Trapezoidal 21.02.2005 Dr. Engin Aktaş 6 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR232 Spring2006 4.3. Method of Analysis Internal forces in members are determined by passing an imaginary section across the member at the point concerned, and writing the equilibrium equations for the free body either to left or right of the cut. y My Fy Fx Mz z 21.02.2005 Fz Mx x Fx – Normal Force My, Mz – Bending Moments Fy, Fz – Shear Force Mx – Twisting moment of torque Dr. Engin Aktaş 7 IZMIR INSTITUTE OF TECHNOLOGY 2 kN/m Example Compute the internal forces at point D for the beam given. Department of Architecture 10 kN 1m 3m REy RBy 226 5RBy 10 4 43 1 0 2 F At D (use right portion) x N 1m RBy 14 kN 43 4 5R 0 M 0 2 2 1 10 1 B Ey 2 M=6 kNm REy 6 kN D Fx 0 REx 5 kN N=5 kN + M D C A 2m + 4 kN/m E REx B 5 kN ME 0 AR232 Spring2006 D E Q 6 kN 21.02.2005 E 5 kN 6 kN 0 5 N 0 N 5 kN (comp) 43 6Q 0 Q 0 5 kN 2 43 M 0 2 6 3 M 0 M 6 kNm D 2 Fy 0 Dr. Engin Aktaş 8 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR232 Spring2006 4.4. Differential Equilibrium Relations q(x) q y Q+DQ M x x O Q M+DM Dx Dx DQ Fy 0 Q DQ qDx Q 0 Dx q Dx Dx M O 0 Q DQ 2 Q 2 M M DM 0 Q Dx DQ Dx DM 0 2 Dx is very small compared to the other D M DQ Q 2 two terms and will be neglected Dx dM dQ Q q & In the limit, when Dx approaches zero dx dx These are basic differential equations relating the external load, q(x), the shear force, Q, and the bending moment, M. 21.02.2005 Dr. Engin Aktaş 9 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR232 Spring2006 Let’s write this relations in integral form as i 1 i 1 dQ qdx i i 1 or Qi 1 Qi qdx i i i 1 dM Qdx i i 1 i 1 or M i 1 M i Qdx i i i and i+1 are any two sections along the member First equation shows that the change in shear between any two sections is equal to minus the area of the load diagram between these two sections. Second equation shows that the change in moment between any two sections is equal to minus the area of the shear diagram between these sections. 21.02.2005 Dr. Engin Aktaş 10 IZMIR INSTITUTE OF TECHNOLOGY Example Department of Architecture Sketch the shearing force and bending moment diagrams 4 kN 0.5 kN/m for the beam given 0 RA RD 4 05 2 0 A RA RD 5 kN + M A 0 4RD 43 05 21 0 RD 3.25kN RA 1.75kN R F AR232 Spring2006 D y I B 2m C 1m 1m RD A II 0.5 kN/m MBA A x QBA 1.75 kN between A and B 0 x 2 QBA 1.75 0.5x 2 x M BA 1.75x 0.5 2 0.5 kN/m MCB A x QCB 1.75 kN between B and C 2 x 3 QCB 0.75kN M CB 1.75x 0.52x 1 @ x = 2m @ x = 3m QB 0.75kN M B 2.5kNm QC 0.75kN M C 3.25kNm It is to be noted that because of the 4 kN load at C, these values correct strictly just short of C. 21.02.2005 Dr. Engin Aktaş 11 IZMIR INSTITUTE OF TECHNOLOGY III Department of Architecture between C and D 4 kN MCD 0.5 kN/m A B x @ x = 4m 1.75 kN 3 x 4 QCD 3.25 M CD 1.75x 0.52x 1 4x 3 QCD C AR232 Spring2006 QD 3.25kN (just shortof D) MD 0 QB QA Area of theload diagram between A and B 4 kN 0.5 kN/m D A B C 1.75 kN QB 1.75 0.5 2 QB 0.75kN 3.25 kN M B M A Area of theshear diagram between A and B 3.25 kN Q 1.75 kN + ___ 0.75 kN 2.5 kNm M 21.02.2005 + 1.75 0.75 M B 0 2 M B 2.5 kNm 2 3.25 kNm M=0.75x+1 M=13-3.25x Dr. Engin Aktaş 12 IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture RA=4 kN Example y 4 kN/m AR232 Spring2006 RB=8 kN q can be written as q x Therefore Q 6m RA RB 8 kN 4 kN 3.46 m + __ Q x Q QA qdx 0 4 x 6 x 4 x2 Q 4 xdx Q 4 6 3 0 x3 0 4 x 12 3.46 m where Q 0 3 @x 6m Q 4 12 8 kN x M M A Qdx 9.24 kNm 0 x2 x3 M 4 dx 4 x 3 9 0 @x 6m M 0 x + M at Q 0 [email protected] x 12 M max 4 12 21.02.2005 Dr. Engin Aktaş 12 12 9.24 kNm 9 13 IZMIR INSTITUTE OF TECHNOLOGY Example Department of Architecture 3 kN B C A 1.9 kN 6.1 kN AR232 Spring2006 QB QA thearea of theload diagram A toB QB 1.9 5 QB 3.1kN 1m 5m The maximum moment occurs where the shearing force is zero, @ x=1.9 m, and is given by 3.1 kN 1.9 kN 1.9 m + __ __ 3.1 m Q 1.9 1.9 M max M A 2 3 kN M max 0 1.81 kNm 3.61 M max 1.81 kNm 2 + 3.8 m __ and for MB 3.1 3.1 M B M max 2 3 kNm M B 1.81 4.81 M B 3 kNm 21.02.2005 Dr. Engin Aktaş 14