Question

P8.052 (Multistep)

A tubular steel column CD supports horizontal cantilever arm ABC, as shown in the figure. Column CD has an outside diameter of 8.000 in. and a wall thickness of 0.400 in. The loads are PA=350 lb and PB=560 lb. Dimensions of the structure are a=5.2 ft, b=8.4 ft, and c=12.4 ft. Determine the maximum compression stress at the base of column CD.

b a PB PA с B A с D

Part 1

Determine the cross-sectional area and the area moment of inertia for column CD.

A=	9.5456	in.2
Iz=	69.11	in.4

Part 2

Determine the internal axial force in column CD. Use the sign convention for internal axial forces.

F=

- 910

lb

Part 3

Determine the magnitude of the internal moment in column CD. In this case, since you’re entering the magnitude, enter a positive value.

M=

9464

lb⋅ft

Part 4

Determine the axial stress due to the internal force in column CD. This is the component of the normal stress produced by the force calculated in Step 2. Use the sign convention for normal stresses.

σaxial=

- 95.332

psi

Part 5

Determine the maximum compression bending stress in column CD. This is the largest magnitude compressive stress due to bending. Use the sign convention for normal stresses, so enter your answer with the correct sign for a compressive stress.

σbend,max=

psi

Part 6

Determine the maximum compression stress at the base of column CD. Use the sign convention for normal stresses.

σmax=

psi

Show transcribed image text

Answer 1

Free bedy viagram Of AC. P3 560/6 by PA350 lb - 5.04 A Rcr qidlo 3910 M NO 91016 6264 .6 free bedy cinqram Arial 12o1ce Diagram olio Bending HOME OF Diagram of Re z Reaction force on Ac by co on at clor) at ir internal Axial force on co MC a Reaction MO1691 Acat c by or internal moment inco CO RYOM R.8.0 af AL. cavuatting vertical korres

Pyoni P.B.0 of ac ovuaring vertical rovees. ft 2 PAtP3 = 560 +350 Rez aloeb Equating MONont MIC PB Y6 + PA*19+5) Mc * 560 48•4+ 350x (5•2484) Mc - 4404 + 4760 Mc = 9464 lb ft out side dia nefer do 8.ooo PN ! ܘ ܕ Thicke ness to.von in inside dion ciex of (o dizdo-at zdo-20-490 2 8.00 - 20.00 di =+72iN cross OHO Sectional AYO Az • II (do-div) A? (8.000) (72)

1 -0.485 (64 - 5184) А ? 9.54-56 Y of interia. of col Aler Honent . T zu (dou-di") (18.000)*- (7-2)) I 7T tu :. I = 0-5965 int 6.9.11 7 Part -> ☺ Internal Arial force ColUHN PRC zglo fb compressive P-900 db, part- E- 2 Moment in co Magnitude of internal M z Mc z 9464 26 ft Mart - < . Axial stress due to Pinternal force inco

4) Tayil F A 2-919 954 5 6 1-95:33 PS part - > MaxiMUM COMPressive bending Shessisco. Ebending a lex de 2.-9464x8.000 2 25460 69.11 = 547.76 Part- z 2-4 Mayimom Compressive stess at the base of co column tmax a Faxil & obend. ending 2-95:33 - 63452.899 :.Cmax :-643.09.
ANSWER