ME-313 STRENGTH OF MATERIALS October 11, 2000 TEST I OPEN TEXT BOOK

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1 ME-313 STRENGTH OF MATERIALS October 11, 000 TEST I OEN TEXT BOOK Name: 1. Three plates of are bolted together as shown in the figure below. Design the bolted connection by choosing both plate thickness and bolt diameter. The connection must carry a load of 175 kn, while the allowable stresses are σ a 10 Ma allowable tensile stress on the plates σ b 60 Ma allowable bearing stress on the plates τ a 100 Ma allowable shear stress on the bolts (5 pts.) 60 mm 180 mm 60 mm 80 mm 80 mm 80 mm 80 mm roblem 1. A meter for measuring torque uses the angle of twist of a shaft to indicate torque. The shaft is to be 150 mm long and made of 6061-T6 aluminum alloy. Determine the required diameter of the shaft if it is desired to have a twist of 10.0 when a torque of 5.0 N.m is applied to the meter. For the shaft thus designed, compute the torsional shearing stress and the factor of safety with respect to the yield stress in shear. Is this satisfactory? If not, what would you do? (5 pts.) 3. A built-up beam is made by fastening four L x x 1 / angles to a 1 / in. x 16 in. plate as shown in the figure below. Compute the moment of inertia for the beam. (5 pts.)

2 . A solid rod (5 mm diameter) is surrounded by an aluminum tube (inner diameter 0 mm, outer diameter 5 mm) as shown in the figure below. A rigid bearing block is supported on the aluminum tube and a gap of 0. mm exists between the block and the top of the rod. Determine the permissible load if the maximum allowable stress is 10 Ma in the and 100 Ma in the aluminum. (5 pts.) 0. mm Aluminum Tube OD 5 mm ID 0mm Steel 5 mm diameter 500 mm roblem 3 roblem

3 ME-313 STRENGTH OF MATERIALS October 11, 000 TEST I Solutions 1. Three plates of are bolted together as shown in the figure below. Design the bolted connection by choosing both plate thickness and bolt diameter. The connection must carry a load of 175 kn, while the allowable stresses are σ a 10 Ma allowable tensile stress on the plates σ b 60 Ma allowable bearing stress on the plates τ a 100 Ma allowable shear stress on the bolts (5 pts.) 60 mm 180 mm 60 mm 80 mm 80 mm 80 mm 80 mm Solution roblem 1 Ignore stress concentration The bolts: Each bolt carries a load of /. The diameter of the bolts is determined as below. τ all πd d πτ all π N (N / m ) m 33.38mm The plate thickness is determined by two factors, the allowable bearing stress and the allowable tensile stress.

4 o Bearing stress / σb td N t 6 σ d (N / m ) b 3 m m 10.08mm o Tensile stress σ a Area (0.18m.0338m) t N t m (N / m ) m 11.0mm o Select the larger of the two thickness values, i.e mm The design therefore requires o Bolt diameter d33.38 mm o late thickness t 11.0 mm

5 . A meter for measuring torque uses the angle of twist of a shaft to indicate torque. The shaft is to be 150 mm long and made of 6061-T6 aluminum alloy. Determine the required diameter of the shaft if it is desired to have a twist of 10.0 when a torque of 5.0 N.m is applied to the meter. For the shaft thus designed, compute the torsional shearing stress and the factor of safety with respect to the yield stress in shear. Is this satisfactory? If not, what would you do? (5 pts.) Solution π φ rad 180 G 6 10 TL φ JG TL 5Nm 0.15m J φg 0.175rad 6 10 (N / m ) πd J 3 9 N / m d π d m τ FS τ (from table at the back of the book) 10 10Ma 96.88Ma m Tc Td 5Nm ( m) 6 τ (N / m ) 96.88Ma 10 J J m τ for AA 6061 T6 is 10Ma from the back of the book. yield yield actual 1.5 This factor of safety appears to be sufficient. To increase the factor of safety, one could select a different aluminum alloy with a higher yield strength. Changing the diameter or changing to a material other than aluminum will change the angle of twist for the applied torque, and may not be acceptable. 10 m

6 3. A built-up beam is made by fastening four L x x 1 / angles to a 1 / in. x 16 in. plate as shown in the figure below. Compute the moment of inertia for the beam. (5 pts.) Solution The rectangular portion is being bent about its centroid. Therefore 3 bh I rect in 1 The moment of inertia I x of the L shapes about their centroid can be obtained from the tables at the back of the book. In the built up beam, the centroid of the L shapes is displaced (8-1.18)in. from the axis of bending Therefore, the moment of inertia of the L sections about the axis of bending can be determined using the parallel axis theorem, i.e. x I ' I + Ay 5.56in in (6.8in) in The moment of inertia of the entire cross-section is therefore total I 170.7in in 890.6in

7 . A solid rod (5 mm diameter) is surrounded by an aluminum tube (inner diameter 0 mm, outer diameter 5 mm) as shown in the figure below. A rigid bearing block is supported on the aluminum tube and a gap of 0. mm exists between the block and the top of the rod. Determine the permissible load if the maximum allowable stress is 10 Ma in the and 100 Ma in the aluminum. (5 pts.) 0. mm Aluminum Tube OD 5 mm ID 0mm Steel 5 mm diameter 500 mm Solution Assume the is loaded to reach its allowable stress value, then ε L 10Ma E ε L 10Ma 00Ga mm 0.35mm Since the aluminum tube is 0.mm longer L ε σ alumin um alumiunum alumiunum mm 0.55mm 500mm ε alumiunum E alumiunum Ga 77Ma A stress of 77 Ma is less that the allowable stress for aluminum. So this is OK Total load is determined from the load supported by the rod and the aluminum tube.

8 total total total + alu min um σ 6 π (N / m ) A + σ alumin um A 687N + 570N 9N 9.kN alumin um [ 0 10 m ] 3 6 π ( 5 10 m) (N / m ) ( 5 10 m) ( )