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1.Stress and Strain 1.1��Internal loadings 1-1��The curved rods of radius R lie in (a)the horizontal plane xy and (b)the vertical plane yz�� as shown in the figure of Prob. 1-1. Determine the resultant internal loadings acting on any cross section �� for the given loads�� dimension and angle. SOLUTION (a)Shear force��Torque��Bending moment . (b)Normal force��Shear force���� Bending moment . 1-2��Members AB and BC are pin-supported at A and C and jointed at B�� as shown in the figure of Prob. 1-2. Determine the resultant internal loadings acting on the section through point D on member AB. SOLUTION Member BC is a two-force member. Its two perpendicular components are FBCx and FBCy. The internal loadings on section D:. 1-3��Member BC is supported by member AB�� as shown in the figure of Prob. 1-3. If F=30kN�� determine the resultant internal loadings acting on the section at the fixed end A. SOLUTION The support reactions: RC=F/3�� RB=2F/3. The internal loadings on section A: FN= 2F/3=20kN�� V=0�� M=2F/3×2m=40kNm. 1-4��Determine resultant internal loadings acting on section a-a and vertical section b-b�� as shown in the figure of Prob. 1-4. Each section passes through the centerline at point C. SOLUTION The support reactions: The internal loadings on section a-a: The internal loadings on section b-b: 1-5��Determine the resultant internal loadings in the beam on sections through points C and D�� as shown in the figure of Prob. 1-5. Point D is located to the left of the load 10kN. SOLUTION The support reactions: From the right segment of the beam sectioned through C�� the internal loadings on section C: From the right segment of the beam sectioned through D�� the internal loadings on section D: Prob. 1-6 1-6��The rod AB of mass m rotates about the vertical axis with angular velocity��as shown in the figure of Prob. 1-6. If the rod has sectional area S and length l�� write the expression for the axial force FN(x)during rotation. SOLUTION The inertial force of the rod during rotation: Based on the method of the section�� the axial force expression is 1.2��Average normal stress in an axially loaded bar 1-7��A long retaining wall AD is supported by concrete thrust blocks and braced by wood shores set BC at a 30 angle. The simplified analysis model is depicted in the figure of Prob. 1-7�� where the base of the wall and both ends of the shore are assumed to be pinned�� and the pressure of the soil against the wall is assumed to be triangularly distributed with a maximum intensity of 80kN/m. For the given cross section of the shore 100mm×100mm�� determine the compressive stress in the shore. SOLUTION The internal loading: Resultant force acting on AD�� F=1/2×(80kN/m×4.5m)=180kN. The average normal stress: 1-8��A container of weight G=500N is suspended by a system of steel wire as shown in the figure of Prob. 1-8. The side length of the container is a=400mm�� and the maximum permissible load for the wire is 290N. Examine if the wire is safe to carry the weight when it is 1.7m long. If angle �� is adjustable�� determine the minimum length of the wire for safely carrying the weight. SOLUTION The internal loading in the wire: The wire is unsafe to carry the weight when it is 1.7m long. The maximum loading in the wire: Fu= 290N. The minimum length of the wire for safely carrying the weight is 2m. 1-9��The rigid beam BC has a length L and is suspended by two rods BD and CE with an equal cross-sectional area A�� as shown in the figure of Prob. 1-9. If the allowable stresses of two rods are [?1] and [?2]�� where [?1]=2[?2]�� determine the distance x and the maximum load F that the structure can support when the load F moves from B to C. SOLUTION The internal maximum loadings of rod 1 and rod 2:. At the critical state�� the following equation of equilibrium can be established: The maximum load F that the structure can support is 1-10��If the maximum average normal stress in any bar does not exceed [?]=125MPa�� determine the cross-sectional area of the truss members AB and CD to support the external load F=150kN�� as shown in the figure of Prob. 1-10. SOLUTION The reaction at support D: From joint D�� the internal loadings of members BD and CD can be determined. Member CB is a zero-force member. From the equations of equilibrium at joint B�� we have Cross-sectional areas of members AB and CD: 1.3��Average shear stress in connections 1-11��Two plates are fastened by four 19mm diameter rivets�� as shown in the figure of Prob. 1-11. Determine the largest load F that can be applied to the plates if the maximum shear and bearing stresses in each rivet do not exceed the allowable shear stress a�� and the maximum tensile stress in each plate does not exceed the allowable tensile stress. (Unit: mm) SOLUTION To