Page 1 Shear Force and Bending Moments Consider a section x-x at a distance 6m from left hand support A 5kN 10kN 8kN 4m 5m 5m 1m A C D B R A = 8.2 kN R B =14.8kN E x x 6 m Imagine the beam is cut into two pieces at section x-x and is separated, as shown in figure Page 2 Shear Force and Bending Moments Consider a section x-x at a distance 6m from left hand support A 5kN 10kN 8kN 4m 5m 5m 1m A C D B R A = 8.2 kN R B =14.8kN E x x 6 m Imagine the beam is cut into two pieces at section x-x and is separated, as shown in figure To find the forces experienced by the section, consider any one portion of the beam. Taking left hand portion Transverse force experienced = 8.2 – 5 = 3.2 kN (upward) Moment experienced = 8.2 × 6 – 5 × 2 = 39.2 kN-m (clockwise) If we consider the right hand portion, we get Transverse force experienced = 14.8 – 10 – 8 =-3.2 kN = 3.2 kN (downward) Moment experienced = - 14.8 × 9 +8 × 8 + 10 × 3 = -39.2 kN-m = 39.2 kN-m (anticlockwise) 5kN A 8.2 kN 10kN 8kN B 14.8 kN 4 m 6 m 9 m 1 m 5 m Page 3 Shear Force and Bending Moments Consider a section x-x at a distance 6m from left hand support A 5kN 10kN 8kN 4m 5m 5m 1m A C D B R A = 8.2 kN R B =14.8kN E x x 6 m Imagine the beam is cut into two pieces at section x-x and is separated, as shown in figure To find the forces experienced by the section, consider any one portion of the beam. Taking left hand portion Transverse force experienced = 8.2 – 5 = 3.2 kN (upward) Moment experienced = 8.2 × 6 – 5 × 2 = 39.2 kN-m (clockwise) If we consider the right hand portion, we get Transverse force experienced = 14.8 – 10 – 8 =-3.2 kN = 3.2 kN (downward) Moment experienced = - 14.8 × 9 +8 × 8 + 10 × 3 = -39.2 kN-m = 39.2 kN-m (anticlockwise) 5kN A 8.2 kN 10kN 8kN B 14.8 kN 4 m 6 m 9 m 1 m 5 m 5kN A 8.2 kN 10kN 8kN B 14.8 kN 3.2 kN 3.2 kN 39.2 kN-m 39.2 kN-m Thus the section x-x considered is subjected to forces 3.2 kN and moment 39.2 kN-m as shown in figure. The force is trying to shear off the section and hence is called shear force. The moment bends the section and hence, called bending moment. Page 4 Shear Force and Bending Moments Consider a section x-x at a distance 6m from left hand support A 5kN 10kN 8kN 4m 5m 5m 1m A C D B R A = 8.2 kN R B =14.8kN E x x 6 m Imagine the beam is cut into two pieces at section x-x and is separated, as shown in figure To find the forces experienced by the section, consider any one portion of the beam. Taking left hand portion Transverse force experienced = 8.2 – 5 = 3.2 kN (upward) Moment experienced = 8.2 × 6 – 5 × 2 = 39.2 kN-m (clockwise) If we consider the right hand portion, we get Transverse force experienced = 14.8 – 10 – 8 =-3.2 kN = 3.2 kN (downward) Moment experienced = - 14.8 × 9 +8 × 8 + 10 × 3 = -39.2 kN-m = 39.2 kN-m (anticlockwise) 5kN A 8.2 kN 10kN 8kN B 14.8 kN 4 m 6 m 9 m 1 m 5 m 5kN A 8.2 kN 10kN 8kN B 14.8 kN 3.2 kN 3.2 kN 39.2 kN-m 39.2 kN-m Thus the section x-x considered is subjected to forces 3.2 kN and moment 39.2 kN-m as shown in figure. The force is trying to shear off the section and hence is called shear force. The moment bends the section and hence, called bending moment. Shear force at a section: The algebraic sum of the vertical forces acting on the beam either to the left or right of the section is known as the shear force at a section. Bending moment (BM) at section: The algebraic sum of the moments of all forces acting on the beam either to the left or right of the section is known as the bending moment at a section 3.2 kN 3.2 kN F F Shear force at x-x M Bending moment at x-x 39.2 kN Page 5 Shear Force and Bending Moments Consider a section x-x at a distance 6m from left hand support A 5kN 10kN 8kN 4m 5m 5m 1m A C D B R A = 8.2 kN R B =14.8kN E x x 6 m Imagine the beam is cut into two pieces at section x-x and is separated, as shown in figure To find the forces experienced by the section, consider any one portion of the beam. Taking left hand portion Transverse force experienced = 8.2 – 5 = 3.2 kN (upward) Moment experienced = 8.2 × 6 – 5 × 2 = 39.2 kN-m (clockwise) If we consider the right hand portion, we get Transverse force experienced = 14.8 – 10 – 8 =-3.2 kN = 3.2 kN (downward) Moment experienced = - 14.8 × 9 +8 × 8 + 10 × 3 = -39.2 kN-m = 39.2 kN-m (anticlockwise) 5kN A 8.2 kN 10kN 8kN B 14.8 kN 4 m 6 m 9 m 1 m 5 m 5kN A 8.2 kN 10kN 8kN B 14.8 kN 3.2 kN 3.2 kN 39.2 kN-m 39.2 kN-m Thus the section x-x considered is subjected to forces 3.2 kN and moment 39.2 kN-m as shown in figure. The force is trying to shear off the section and hence is called shear force. The moment bends the section and hence, called bending moment. Shear force at a section: The algebraic sum of the vertical forces acting on the beam either to the left or right of the section is known as the shear force at a section. Bending moment (BM) at section: The algebraic sum of the moments of all forces acting on the beam either to the left or right of the section is known as the bending moment at a section 3.2 kN 3.2 kN F F Shear force at x-x M Bending moment at x-x 39.2 kN Moment and Bending moment Bending Moment (BM): The moment which causes the bending effect on the beam is called Bending Moment. It is generally denoted by ‘M’ or ‘BM’. Moment: It is the product of force and perpendicular distance between line of action of the force and the point about which moment is required to be calculated.Read More

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