# HW 1 - 4 CHAPTER 1 Tension Compression and Shear Problem...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 4 CHAPTER 1 Tension, Compression, and Shear Problem 1.2-5 The cross section of a concrete pier that is loaded unifonnly in compression is shown in the figure. (a) Detennine the average compressive stress eTc in the concrete if the load is equal to 2500 k. (b) Detennine the coordinates x and y of the point where the resultant load must act in order to produce unifonn nonnal stress. Solution 1.2-5 Concrete pier in compression Iy (a) AVERAGE COMPRESSIVE STRESS eTc t 16 in. -+ 16 in. -+ 16 in. ~---x 48 in. P = 2500k P 2500 k ..... eT =-= 2=1.70ksi c A 1472 in. .•.•.•..••.••.•.,.2 '. <D.' ' . '(J). (b) COORDINATES OF CENTROID C 0·· • • .... :4 . . . :., 1 ~ From symmetry, y = 2(48 in.) = 24 in. x l: xA x = --'-' (see Chapter 12, Eq. 12-7a) .- USE THE FOLLOWING AREAS: A _ 1 _ _ _ A I = (48 in.)(20 in.) = 960 in,2 x = A (XI A) + 2i 2 A 2 + X3 A 3) I A 2 = A 4 = 2(16 in.)(l6 in.) = 128 in. 2 1 --'-2 [(10 in.)(960 in. 2 ) 1472m. A = (16 in.)(16 in,) = 256 in. 2 3 + 2(25.333 in.)(128 in. 2 ) A = Al + A 2 + A 3 + A 4 +(28 in.)(256 in. 2 )] = (960 + 128 + 256 + 128) in. 2 =15.8 in. .- = 1472 in. 2 Problem 1.2-6 A car weighing 130 kN when fully loaded is pulled slowly up a steep inclined track by a steel cable (see figure). The cable has an effective cross-sectional area of 490 mm 2 , and the angle a of the incline is 30°. Calculate the tensile stress eTt in the cable. 5 SECTION 1.2 Normal Stress and Strain Solution 1.2-6 Car on inclined track FREE-BODY DIAGRAM OF CAR w W = Weight of car T ~ Tensile force in ,cable a = Angle of incline A = Effective area of cable R l' R 2 = Wheel reactions (no friction force between wheels and rails) EQUILIBRJUM IN THE INCLINED DIRECTION I-FT=O ~l!'- T-Wsina=O T= Wsin a TENSILE STRESS lN THE CABLE T Wsin a U ------ ,- A - A SUBSTITUTE NUMER1CAL VALUES: W = 130 kN a = 30° A = 490mm 2 ( 130 kN) (sin 30°) U, = 490 mm2 = 133 MPa -- Problem 1.2-7 Two steel wires, AB and BC, support a lamp weighing 18 Ib (see figure). Wire AB is at an angle a = 34° to the horizontal and wire BC is at an angle f3 = 48°. Both wires have diameter 30 mils. (Wire diameters are often expressed in mils; one mil equals 0.001 in.) Determine the tensile stresses u AB and u BC in the two wires. Solution 1.2-7 Two steel wires supporting a lamp FREE-BODY DIAGRAM OF POINT B SUBSTITUTE NUMER1CAL VALUES: -T AB (0.82904) + T Bc (0.66913) = 0 T AB (0.55919) + T BC (0.743 14) - 18 = 0 f3 = 48° d = 30 mils = 0.030 in. SOLVE THE EQUATIONS: 7Td 2 TAB == 12.163 Ib T BC = 15.069 Ib A =-=706.9 X 10-6in. 2 4 TENS1LE STRESSES lN THE WlRES TAB . U AB = A = 17,200 pSI -- EQUAT10NS OF EQUILIBRIUM T Be . Use = - = 21,300 pSI -- 'i.F, = 0 - TAB cos a + T BC cos f3 = 0 A 'i.F y = 0 TAB sin a + T BC sin f3 - W = 0 22 CHAPTER 1 Tension, Compression, and Shear Problem 1.5-5 A bar of monel metal (length L = 8 in., diameter d = 0.25 in.) is loaded axially by a tensile force p = 1500 lb (see figure from Prob. 1.5-4). Using the data in Solution 1.5-5 Bar of monel metal in tension...
View Full Document

{[ snackBarMessage ]}

### Page1 / 13

HW 1 - 4 CHAPTER 1 Tension Compression and Shear Problem...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online