Normal Stress
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Transcript of Normal Stress
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Bautista, Eldin E. Mec 32/ C5
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Rod AB:
AB=
=175 MPa
A= d1
4
P= 40kN + 30kN = 70kN = 70000N
175MPa= 70000N
d1
4
d1=22.57mm answer
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Rod BC:
BC=
=150 MPa
A= d2
4
P= 30kN = 30000N
150MPa= 30000N
d2
4
d2=15.96mm answer
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A rod is composed of an aluminum section rigidly attached between steel and bronze sections, as shown in Fig. P-107. Axial loads are applied at the positions indicated. If P = 3000 lb and the cross sectional area of the rod is 0.5 in2, determine the stress in each section.
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For steel: stAst=Pst st(0.5)=12 st=24 ksi answer
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For aluminum: alAal=Pal al(0.5)=12 al=24 ksi answer For bronze: brAbr=Pbr br(0.5)=9 br=18 ksi answer
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An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown in Fig. P-108. Axial loads are applied at the positions indicated. Find the maximum value of P that will not exceed a stress in steel of 140 MPa, in aluminum of 90 MPa, or in bronze of 100 MPa.
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For bronze: brAbr=2P100(200)=2P P=10000N
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For aluminum: alAal=P 90(400)=P P=36000N For Steel: stAst=5P P=14000N For safe value of P, use the smallest above. Thus, P=10 000N answer
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The homogeneous bar shown in Fig. P-106 is supported by a smooth pin at C and a cable that runs from A to B around the smooth peg at D. Find the stress in the cable if its diameter is 0.6 inch and the bar weighs 6000 lb.
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MC=0 5T+10(334T)=5(6000) T=2957.13lb
T=A 2957.13=[14(0.62)] =10458.72 psi answer
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Determine the largest weight W that can be supported by two wires shown in Fig. P-109. The stress in either wire is not to exceed 30 ksi. The cross-sectional areas of wires AB and AC are 0.4 in2 and 0.5 in2, respectively.
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Free body diagram of Joint A
or wire AB: By sine law (from the force polygon): TABsin40=Wsin80 TAB=0.6527W ABAAB=0.6527W 30(0.4)=0.6527W W=18.4kips answer For wire AC: TACsin60=Wsin80 TAC=0.8794W