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Preview text ISM for Problem Determine the resultant internal normal force acting on the cross section through point A in each column. In (a), segment BC weighs 300 and segment CD weighs In (b), the column has a mass of 200 ( a) Given: L BC 3m kg wBC 300 m kg wCA 400 m FB 5kN L CA 1.2m g 9.81 m 2 s FC 3kN Solution: Fy ( ) ( ) FA g L BC g L CA FB 2FC 0 ( ) ( ) FA g LBC g LCA FB 2FC FA 24.5 kN ( b) Given: g 9.81 L 3m m 2 s FB 8kN w 200 Ans kg m F1 6kN F2 4.5kN Solution: Fy FA ( L) g FB 2F1 2F2 0 FA ( L ) g FB 2F1 2F2 FA 34.89 kN Ans Problem Determine the resultant internal torque acting on the cross sections through points B and C. Given: T D m T BC m T AB m Solution: Equations of equilibrium: Mx T B T BC T D 0 T B TD T B 150 m Mx Ans TC TD 0 T C TD T C 500 m Ans Problem A force of 80 N is supported the bracket as shown. Determine the resultant internal loadings acting on the section through point A. P 80N Given: 30deg 45deg a 0.3m b 0.1m Solution: Equations of equilibrium: NA cos ( ) 0 NA cos ( ) NA 77.27 N Ans VA sin ( ) 0 VA sin ( ) VA 20.71 N Ans MA cos ( ) a cos ( ) sin ( ) ( b a sin ( ) ) 0 MA cos ( ) a cos ( ) sin ( ) ( b a sin ( ) ) MA m Ans Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD. Problem The beam AB is pin supported at A and supported a cable BC.
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Determine the resultant internal loadings acting on the cross section at point D. P Given: a 0.8m b 1.2m c 0.6m d 1.6m e 0.6m Solution: 36.87 deg a d 14.47 deg atan atan Member AB: sin ( ) ( a b) ( b) 0 ( b) FBC sin ( ) ( a b) FBC 12.01 kN Segment BD: cos ( ) cos ( ) 0 ND cos ( ) cos ( ) ND kN Ans VD sin ( ) sin ( ) 0 VD sin ( ) sin ( ) VD 0 kN Ans MD 0 ( ) MD sin ( ) sin ( ) MD 0 m sin ( ) Ans Note: Member AB is the member. Therefore the shear force and moment are zero. Problem Solve Prob. For the resultant internal loadings acting at point E.
P Given: a 0.8m b 1.2m c 0.6m d 1.6m e 0.6m Solution: 36.87 deg a d 14.47 deg atan atan Member AB: sin ( ) ( a b) ( b) 0 FBC ( b) sin ( ) ( a b) FBC 12.01 kN Segment BE: cos ( ) cos ( ) 0 NE cos ( ) cos ( ) NE kN Ans VE sin ( ) sin ( ) 0 VE sin ( ) sin ( ) VE 0 kN Ans e ME 0 ( ) ME sin ( ) sin ( ) e ME 0 m Ans Note: Member AB is the member. Therefore the shear force and moment are zero. Problem The force F 400 N acts on the gear tooth. Determine the resultant internal loadings on the root of the tooth, i.e., at the centroid point A of section P 400N Given: Solution: 30deg 45deg a 4mm b 5.75mm Equations of equilibrium: For section a VA cos ( ) 0 VA cos ( ) VA 386.37 N Ans NA sin ( ) 0 NA sin ( ) NA 103.53 N Ans sin ( ) a cos ( ) b 0 MA sin ( ) a cos ( ) b MA 1.808 m Ans Problem The beam supports the distributed load shown.
Determine the resultant internal loadings on the cross section through point C. Assume the reactions at the supports A and B are vertical. KN w1 4.5 m Given: kN w2 6.0 m a 1.8m b 1.8m d 1.35m e 1.35m Solution: c 2.4m L 1 a b c L 2 d e Support Reactions: L2 L1 L 1 0.5 L1 L2 L 1 0 3 L2 L 1 ( 0.5 ) L2 1 22.82 kN L1 ( ( )( ) ) ( ( ) ) Ay L1 L2 0 Ay L 1 L 2 Ay 12.29 kN Equations of Equilibrium: For point C NC 0 Ans ( a VC 0 VC ( a VC 3.92 kN Ans MC ( a 0.5 ( a b) ( a b) 0 MC ( a 0.5 ( a b) ( a b) MC 15.07 m Ans Note: Negative sign indicates that VC acts in the opposite direction to that shown on FBD. Problem Determine the resultant internal loadings acting on (a) section and (b) section Each section is located through the centroid, point C.