CHPER TWO STRIN, HOOK S W, XI OD PROMS Introduction: In engineering the deformation of a body is specified using the concepts of norm and shear rain. Whenever a force is applied to a body, it will tend to change the body s shape. These changes are referred to as deformation, and they may be highly visible or practicly unnoticeable. Deformation of a body can so occur when the temperature of the body is changed. The figure shows the before and after positions of three different line segments on this rubber membrane which is subjected to tension. The vertic line is lengthened, the horizont line is shortened, and the inclined line changes its length and rotates. Norm Strain (i Strain): If an ai load P is applied to the bar in the Figure, it will change the bar s length o to a length. We will define the average norm rain e (epsilon) of the bar as the change in its length d (delta) = - o divided by its origin length, that is: o o (dimensionless quantity) d 1
Ductile and Brittle Materis Engineering materis are commonly classified as either ductile or brittle materis. ductile materi lies one having a relatively large tensile rain up to the point of rupture (for eample, ructur eel or uminum) whereas a brittle materi has a relatively sml rain up to this same point. Ca iron and concrete are eamples of brittle materis. Very Ductile Moderately Ductile Brittle arge Moderate Sml Ductile: warning before fracture Brittle: No warning
Hooke s aw From the origin O to the point cled proportion limit, the ress-rain curve is a raight line. This linear relation between elongation and the ai force causing was fir noticed by Sir Robert Hooke in 1678 and is cled Hooke's law that within the proportion limit, the ress is directly proportion to rain. E Where: E denotes the slope of the raight-line portion O. The quantity E is equ to the slope of the ress-rain diagram from O to it is cled the modulus of elaicity of the materi in tension, or, as it is often cled, Young s modulus of elaicity.
EXMPE -1 Determine the average norm rains in the two wires in the Figure shown if the ring at moves to. The origin length of each wire is B B C The norm rains: B B 5.01004 5 B.0110 5 C C ( 0.01) ( 0.01) C B C C 4 The fin lengths are: 5m (4 0.0) (4 0.0) 5.01004m 5.0m 5.0 5 4.4 10 5 m/ m m/ m EXMPE - When force P is applied to the rigid lever arm BC shown in the Figure, the arm rotates counterclockwise about pin through an angle of 0.05. Determine the norm rain in wire BD. BB tan(0.05) 400 BB 400 tan(0.05) 0.49mm DB DB 00 0.49 00.49mm DB DB DB 00.49 00 00 0.00116m/ m 4
Deflection of ily oaded Rods Using Hooke s law and the definitions of ress and rain, we will now develop an equation that can be used to determine the elaic displacement of a member subjected to ai loads. The free-body diagram of the aily loaded element is shown in Figure. The resultant intern ai force will be a function of since the etern diributed loading will cause it to vary ong the length of the bar. Provided the ress does not eceed the proportion limit, we can apply Hooke s law P d d P d P d E( ) d d E 0 P d E For conant load P, cross-section area and Modulus of elaicity E, the above equation becomes: P E If the ructur member has some segments with different load, area, and E in each segment, the tot deflection equs the sum of the deflection of each segment. P E P P 5
EXMPE - The uniform eel bar shown in the Figure has a diameter of 50 mm and is subjected to the loading shown. Determine the displacement at D. ssume E=00 GPa. D D D P E 70 1000 1500 0 1000 1000 50 1000 000 5 00000 5 00000 5 00000 0.0891mm P (kn) This negative result indicates that point D moves to the left. EXMPE -4 member is made of a materi that has a specific weight of γ = 6 kn/m and modulus of elaicity of E=9 GPa. Determine how far of the cone s end is displaced due to gravity when it is suspended in the vertic position. y W V y 0. 1 0.1y Vol. Fy 0 P 0 Pdy E 110 0.014y 6 m y 6 1000 0.01047y 6.8y dy 0.014y 9 10 0 9 (0.1y) y 6.8y. 10 9 y 0.01047y 0 ydy. 10 6 y 0 Py very sml amount 6
EXMPE -5 bronze bar is faened between a eel bar and an uminum bar as shown in the Figure. i loads are applied at the positions indicated. Find the large vue of P that will not eceed an overl deformation of.0 mm, or the following resses: 140 MPa in the eel, 10 MPa in the bronze, and 80 MPa in the uminum. Use E = 00 GPa, E = 70 GPa, and Ebr = 8 GPa. P P P bro bro bro 140 480 6700N bro 10 650 P 80 0 P 9000N 1800N P 1000 P 000 P 1500 480 00000 650 70000 480 8000 1 1 P 96000 1175 6560 P 84611N bro P -P +P Use the smle P=1800 N 7
EXMPE -6 Rigid beam B res on the two short pos shown in Figure.. C is made of eel and has a diameter of 0 mm, and BD is made of uminum and has a diameter of 40 mm. Determine the displacement of point F on B if a vertic load of 90 kn is applied over this point. Take E = 00 GPa, E = 70 GPa. B B F PCC CE 60 1000 00 10 00000 PBDBD E BD 0.86mm 0.86mm 0 1000 00 0.10mm 0.10mm 0 70000 400 0.10 0.184 0.5mm 600 8
EXMPE -7 uniform concrete slab of tot weight W is to be attached, as shown in the Figure, to two rods whose lower ends are on the same level. Determine the ratio of the areas of the rods so that the slab will remain level. Take E = 00 GPa, E = 70 GPa. P P W M 0 6P W 0 6P P P ( ) ( ) E E W W 6 4 00 70 M W 00 W 4W 8W 70 00 8.809 70 6 9
EXMPE -8 The horizont rigid beam BCD is supported by vertic bars and and is loaded by vertic forces P1=400 kn and P=60 kn acting at points and D, respectively. Bars and are made of eel (E=00 GPa) and have cross-section areas =11,100 mm and =9,80 mm. The diances between various points on the bars are shown in the figure. Determine the vertic displacements of points and D. F F Shortening of bar : F 464.4 E 980 00 Shortening of bar : M M B 96kN F E 0 400 1.5 F 464kN C 0 400 F 96 0.4m 11100 00.1 D ( 1.5 1 7 D 5 5 0.6m 0.4 0.6 0.mm ) From displacement diagram: 1.5 60.6 1.5 60.1 1 7 0.6 0.4 0.88mm 5 5 10
EXMPE -9 For the composite column shown: 1. Determine the maimum load P if the tot deflection =0.5mm.. Draw the vertic deflection diagram. P P tot ( ) ( ) E E P 00 P 500 0.5 00 50 50 70 100 100 0.5 0.0006P 0.00071P P 190.8kN 190.8500 0.16mm 70 100 100 190.800 0.114mm 00 50 50 11