2 internal force theory 2015

CHAPTER 2: INTERNAL FORCES 2.1 Free Body Diagram 2.2 Internal Forces & Method of Section 2.3 Sign Conventions 2.4 Procedure 2.5 Diagrams of internal forces 2.6 Relationships between loads, shear force and bending moment 2.1 FREE BODY DIAGRAM Useful definition Free body diagram: A sketch of the outlines shape of the body isolated from its surrounding On this sketch, all forces and couple moments that the surrounding exert on the body together with any support reactions must be shown correctly Only then applying equilibrium equations will be useful 2.2 INTERNAL FORCES & METHOD OF SECTIONS Internal loadings: These internal loading acting on a specific region within the body can be attained by the Method of Section Method of Section: Imaginary cut is made through the body in the region where the internal loading is to be determined The two parts are separated and a free body diagram of one of the parts is drawn Only then applying equilibrium would enable us to relate the resultant internal force and moment to the external forces 2.2 INTERNAL FORCES & METHOD OF SECTIONS CROSS SECTION METHOD 2.2 INTERNAL FORCES & METHOD OF SECTIONS Point O is often chosen as the centroid of the sectioned area CROSS SECTION METHOD 2.2 INTERNAL FORCES & METHOD OF SECTIONS CROSS SECTION METHOD 2.2 INTERNAL FORCES & METHOD OF SECTIONS Four types of internal forces can be defined: Normal force, N This force acts perpendicular to the area Shear Force, V This force lies in the plane of the area (parallel) Torsional Moment, T This torque is developed when the external loads tend to twist one segment of the body with respect to the other Bending Moment, M This moment is developed when the external loads tend to bend the body CROSS SECTION METHOD 2.2 INTERNAL FORCES & METHOD OF SECTIONS • If the body is subjected to a co-planar force system then only normal force N, 01 shear force V, and 01 bending moment M component exist on the section CROSS SECTION METHOD 2.2 INTERNAL FORCES & METHOD OF SECTIONS Consider a bar at “balance” state (ie free body diagram) P4 P1 F P5 P2 i 1 0 iy F i 1 Fi : P1, P2…, P6 0 iz P3 P6 An imaginary cross section P1 P2 A O Mx > Nz > z Qy > P3  m F   i 1 x i Qy > Nz > O z Mx > P4 P5 B P6 y y Fi : P1, P2, P3, Nz, Qy, Mx Fi : P4, P5, P6, Nz, Qy, Mx CROSS SECTION METHOD 2.3 SIGN CONVENTION CROSS SECTION METHOD 2.3 SIGN CONVENTION Shear force: clockwise Bending moment: compresses the upper part of the bar or elongates the lower part Normal force: elongates CROSS SECTION METHOD 2.3 SIGN CONVENTION • If the internal shear force causes a clockwise rotation of the beam segment Then it is considered to positive • If the internal moment causes compression in the top fibers then it is considered to be positive CROSS SECTION METHOD 2.3 SIGN CONVENTION Stress Under General Loadings • A member subjected to a general combination of loads is cut into two segments by a plane passing through Q • The distribution of internal stress components may be defined as, F x  x  lim A0 A  xy  lim A0 V yx A Vzx  xz  lim A0 A • For equilibrium, an equal and opposite internal force and stress distribution must be exerted on the other segment of the member CROSS SECTION METHOD 2.4 PROCEDURE Determination of support reactions by studying the equilibrium of the whole structure (Xác định phản lực liên kết cách xét cân tòan hệ) Imagine a section passing through the body (Tưởng tượng mặt cắt qua vật thể) Equilibrium of one divided part (Xét cân phần bị chia) CROSS SECTION METHOD 2.4 PROCEDURE After sectioning, decide which segment of the body will be studied If this segment has a support or connection than a free body diagram for the entire body must be done first to calculate the reactions of these supports Pass an imaginary section through the body at the point where the resultant internal loadings are to be determined and put the three unknowns (V, Mo, N) at the cut section Then apply equilibrium Suggestion: take the summation of moment around a point on the cut section (V and N will not appear in this equation) and solve directly for Mo) CROSS SECTION METHOD 2.4 PROCEDURE Determine the reactions using the equilibrium conditions of the overall structure Cut the beam at the cross section at which shear force and bending moment are to be determined Draw a free-body diagram Set up equilibrium equations of the F.B.D to determine shear force and bending moment at the cross section CROSS SECTION METHOD 2.4 PROCEDURE Example: Determine internal forces on the cross section at C (Xác định nội lực tiết diện C)  Fx  0  NC   Fy  0  VC  58,8N  MC  0  MC  5,69N.m CROSS SECTION METHOD 2.5 DIAGRAMS OF INTERNAL FORCES Beams are long straight bars having constant cross section area and support loads that are applied perpendicular to its longitudinal axis CROSS SECTION METHOD 2.5 DIAGRAMS OF INTERNAL FORCES In order to properly design a beam, the maximum values for V and M in the beam have to be found This could be done through the shear force and bending moment At each location z, values of V(z) and M(z) are obtained by using the procedure of determining internal forces on the cross section at z V and M vary throughout the length of the beam This means that V = V(z) and M = M(z) Graphs are plotted as values of V or M versus distance z along the axis of the beam Graphs are called shear force and bending moment diagrams 2.5 DIAGRAMS OF INTERNAL FORCES EXAMPLE 1: Cantilevered beam and concentrate load 2.5 DIAGRAMS OF INTERNAL FORCES EXAMPLE 2: Simply supported beam and concentrate load Remind: Given: beam AB, length L, concentrate load P at distance L1 (from A) Problem: plot V and M diagram? EXAMPLE 3: Cantilevered beam and uniformed distribution load EXAMPLE 4: Simply supported beam and uniformed distribution load 2.5 DIAGRAMS OF INTERNAL FORCES REMARKS If we let the cross section to move from left end to right end of the beam and always consider the left-hand side segment then: Whenever we see a external concentrate force or concentrate moment, there will be a sudden change of the shear diagram or moment diagram Value of the change in the diagram is equal to that of force or moment Direction of the change in the diagram follows that of the change of the force or moment Whenever we see a change of external force or moment (including reaction force), it is necessary to add one more time of considering the internal force formulation i.e the internal force diagrams will have one more segment 2.6 RELATIONSHIPS BETWEEN LOADS, SHEAR FORCE DIAGRAM AND BENDING MOMENT DIAGRAM 2.6 RELATIONSHIPS BETWEEN LOADS, SHEAR FORCE DIAGRAM AND BENDING MOMENT DIAGRAM The concentrated loads cause abrupt changes in the shear force wherever they are located As the differentials are small, the bending moment does not change as we pass through the point of application of a concentrated load