# S1 - CIVE207 Fall 2008 Assignment#1(Due Sept 16 Tuesday...

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CIVE207, Fall 2008, Assignment #1 (Due Sept. 16, Tuesday 12pm, Box 2 in Room MD475) CEl P1.34 A rectangular bar having width w = 6.00 in. an~ thi~imes~--- t = 1.50 in. is subjected to a tension load P. The normal and shear s.tresses on plane AB must not exceed 16 ksi and 8 ksi, respec- tIvely. Det~rmi~e the maximum load P that can be applied with- out exceedmg eIther stress limit. FIGURE P1.34/35 "'I' '-' . . . ,~r P4.11 The simple pin-connected structure carries a concentrated P3.23 Rigid bar BCD in Fig. P3.23 is supported by a pin at C and load P as shown in Fig. P4.11. The rigid bar is supported by strut by aluminum rod (I). A concentrated load P is applied to the lower AB and by a pin support at C. The steel strut AB has a cross- end of aluminum rod (2), which is attached to the rigid bar at D. The sectional area of 0.75 in. 2 and a yield strength 0f60 ksi. The diameter cross-sectional area of each rod is A = 0.20 in? and the elastic mod- ulus of the aluminum material is E = 10,000 ksi. After the load P is applied at E, the strain in rod (I) is measured as 900 JlE (tension). of the steel pin at Cis 0.5 in., and the ultimate shear strength is 54 Determine: ksi. If a factor of safety of 2.0 is required in both the strut and the pin at C, determine the maximum load P that can be supported by (a) the magnitude of load P. the structure. (b) the total deflection of point E relative to its initial position. I' 'I 50 in. ~(l)-l Rigid bar _._.m".,D 20 in. Connection detail FIGURE P3.23 FIGURE P4.11

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1.34 A rectangular bar having width w = 6.00 in. and thickness t = 1.50 in. is subjected to a tension load P. The normal and shear stresses on plane AB must not exceed 16 ksi and 8 ksi, respectively. Determine the maximum load P that can be applied without exceeding either stress limit. Fig. Pl.34 Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle B are P an = 2A (l+cos2B) (a) and Tnt = ~sin2B (b) 2A The angle B for inclined plane AB is calculated from 3 tanB = - = 3 :. B = 71.5651 ° 1 The cross-sectional area of the bar is A = wxt = (6.00 in.)(1.50 in.) = 9.0 in?
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## This note was uploaded on 10/06/2009 for the course CIVE 207 taught by Professor Shao during the Fall '09 term at McGill.

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S1 - CIVE207 Fall 2008 Assignment#1(Due Sept 16 Tuesday...

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