Normal Stress

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    13-Sep-2015
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presentation for mec32

Transcript of Normal Stress

  • Bautista, Eldin E. Mec 32/ C5

  • Rod AB:

    AB=

    =175 MPa

    A= d1

    4

    P= 40kN + 30kN = 70kN = 70000N

    175MPa= 70000N

    d1

    4

    d1=22.57mm answer

  • Rod BC:

    BC=

    =150 MPa

    A= d2

    4

    P= 30kN = 30000N

    150MPa= 30000N

    d2

    4

    d2=15.96mm answer

  • 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.

  • For steel: stAst=Pst st(0.5)=12 st=24 ksi answer

  • For aluminum: alAal=Pal al(0.5)=12 al=24 ksi answer For bronze: brAbr=Pbr br(0.5)=9 br=18 ksi answer

  • 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.

  • For bronze: brAbr=2P100(200)=2P P=10000N

  • 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

  • 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.

  • MC=0 5T+10(334T)=5(6000) T=2957.13lb

    T=A 2957.13=[14(0.62)] =10458.72 psi answer

  • 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.

  • 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