Problem " Â F y = 0. ) R A + 2R B + R C = 200 kn ) 2R A + 2R B = 200 kn [using symmetry R A = R C ] ) R A + R B = 100 kn

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1 Problem 0. Three cables are attached as shown. Determine the reactions in the supports. Assume R B as redundant. Also, L AD L CD cos 60 m m. uation of uilibrium: + " Â F y 0 ) R A cos 60 + R B + R C cos kn ) R A + R B + R C 00 kn ) R A + R B + R C 00 kn ) R A + R B 00 kn [using symmetry R A R C ] ) R A + R B 00 kn uation of Compatibility: To compute the downward (-ve) deformation (d L ) due to the external load (in this case we do not have any force in the member BD) + " Â F y 0 ) F AD cos kn 0 [using symmetry F AD F CD ] ) F AD 00 kn Figure : Single and double shear.

2 Hence, d L F AD L AD A cos 60 (00 kn) ( m) A 00 knm A Similarly, to compute the upward (+ve) deformation (d R ) due to the redundant reaction R B (in this case we have force R B in the member BD) d R R BL AD A cos 60 Using the euation of compatibility + R BL BD A R B ( m) + R B ( m) A A d d L + d R 0 ) 00 knm + R B ( m) + R B ( m) A A A ) 00 kn R B 80 kn 5 ) R A 00 kn 80 kn 0 kn R C 0 Isotropic Material The material properties are same in every direction. Homogeneous Material The material properties are same for every position. Poisson s Ratio For the axially loaded member s x A P shown in the figure, even if s y s z 0 here but e y, e z 6 0 due to the transverse contraction. The lateral strains are eual in this case for a homogeneous isotropic material and a material constant, known as Poisson s ratio (n), can be defined as n lateral strain axial strain e y e x e z e x Using Hooke s law (s x e x ) e y e z

3 Multiaxial Loading For multiaxial loading the generalized Hooke s law is given by e x + s x ns y e y + s y e z ns y + s z Figure 5: Multiaxial loading. Shearing Strain The shearing strain is defined as shown in the figure. Hooke s law for shearing stress and strain is t xy Gg xy t yz Gg yz t zx Gg zx where G is the modulus of rigidity or shear modulus. Figure 6: Shear stresses and strains. G ( + n) For a general stress condition in an isotropic linearly elastic material the generalized Hooke s law: e x + s x ns y e y + s y e z ns y + s z t xy Gg xy t yz Gg yz t zx Gg zx Problem. Figure 7: Single and double shear. A bolt of diameter 0 mm is tightened such that the decrease in its diameter is 0 µm. Using the property of steel, 00 G and G 77. G determine the internal force in the bolt. Given d y 0 µm m, d 0 mm 0.0 m. n G e y d y d e x e y n

4 Hence, the internal force in the bolt pd P sa (e x ) p(0.0) ( ).77 N m Problem. The plate shown in the figure is subjected to biaxial loading. Compute the change in length of the sides and the diagonal. Also, compute the change in the angle ACB. Assume 00 G, n 0.9. Given s x 00 M, s y 0, s z 0 M. Using generalized Hooke s law for multiaxial loading: Figure 8: Problem. e x + s x ns y e y + s y e z (0 0 6 ) (00 06 ) 0.9 ( ) ns y + s z 0.9 (00 06 ) Hence, the changes in lengths d AB l AB e x (0. m) ( ) m mm d BC l BC e z (0. m) ( ) m mm The change in thickness d t te y (0.0 m) ( ) mm

5 To estimate the change in length of the diagonal, first calculate the length of the diagonal before deformation: l AC l AB + l BC The length of the diagonal after deformation (l AB ( + e x )) +(l BC ( + e z )) Hence, the change in length of the diagonal d AC (l AB ( + e x )) +(l BC ( + e z )) l AB + l BC m 0.05 mm The change in angle ACB: D tan l AB( + e x ) l BC ( + e z ) + e x + e z.89 0 l AB l BC Relative change in the angle ACB D tan tan 5 00% 0.09%. The change in volume DV V V 0 (l AB ( + e x ) l BC ( + e z ) t( + e y )) (l AB l BC t) (l AB l BC t) (e x + e y + e z ) V 0 (e x + e y + e z ) m mm 3 Problem 3. Determine the average shear stress in the pin (dia 0 mm) at B. From the free-body diagram of ABC Â F y 0 B y (000 N) 0 B y 000 N Â M B 0 F CD (0. m) (000 N) (0.5 m) 0 F CD 5000 N Â F x 0 B x F CD 0 B x F CD 5000 N

6 Hence, the reaction in the pin R B B x + B y 5385 N. Since the pin is under double shear the shear stress in the pin is Figure 9: Problem 3. t R B pd N M p(0.0) m The bearing stress in member BCD s b R B dt 5385 N (0.0 m) (0.0 m) M The bearing stress in the support s b R B dt N (0.0 m) (0.005 m) M Stresses on Inclined Sections Consider the axially loaded bar as shown in the figure. Compute the stresses (s and t ) on an inclined plane a a 0. Sign Convention: Normal stress from tension is positive and shear stress producing counter-clockwise rotation is positive. Using the above sign convention and the free-body diagram, we can write s N P cos P A A A cos s x cos cos t V P sin P A A A cos sin s x cos sin cos Figure 30: Stresses on an inclined plane.

7 Hence, s s x cos s x ( + cos ) t s x cos sin s x sin Problem. Determine the stresses developed on the inclined plane a a 0. The axial stress developed in the bar Hence, s x P A N 0.00 m N/m 5 M s s x 5 M ( + cos ) ( + cos 60 ) 8.75 M t s x 5 M sin sin M For a block on the plane a a 0 the complete stress diagram is shown below. To obtain this use the following: side a a 0 : Substitute 30 to estimate s 30 and t 30. side b b 0 : Substitute to estimate s 0 and t 0. side a b: Substitute to estimate s 0 and t 0. side a 0 b 0 : Substitute to estimate s 60 and t 60. Figure 3: Problem.

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