{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

15_Mechanics Homework Mechanics of Materials Solution

# 15_Mechanics Homework Mechanics of Materials Solution - a...

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: a. ‘k l l [”7 C‘Ml m m ,,,A,4,,,,,, ,Hi .M ., _, ,...._ ... ......__.__.... _.r ..___. n,.... , .- __, we at .e. :1 W, 771..” 7'“. , - .- —- -- 1-.-_.- A--_.-—.—-——-.— .-.......-_.... t PROBLEM 1.17 [.17 Two wooden planks, each Tls: - in. thick and 6 in. wide, arejoined by the glued mortise joint shown. Knowing that the joint will fail when the average shearing stress in the glue reaches 120 psi, determine the smallest allowable length d ofthe cuts if the joint is to withstand an axial load of magnitude P = 1200 lb. ' r—d—al ‘ H 7 ' SOLUTION Seven suriaces .31er +113 +9.}.p 30w; ?= 1200 )5. Area. A= (7303-5) at: 3833 P _ P 49 .. Moo . “I A“ *2 nd‘ﬁa d='-633m *" i l PROBLEM 1.18 1.18 A load P is applied to a steel rod supported as shown by an aluminum plate into which a Odin—diameter hole has been drilled. Knowing that the shearing stress must not exceed 18 ksi in the steel rod and 10 ksi in the aluminum plate, determine the largest load P which maybe applied to the rod. SOLUTION Four s‘i'eef A, = “trait = '|T(o.G\(o.‘~ll : 0.7540 in‘ Z’l‘ E ‘5 P: AITI " (OJWYI-ﬂd '4: H.357 kips For aapuminun A1 2 Walt =- n(i.c)(o.253 - l.2.‘>’66 m1 l 11 = -E 13-.- A175; = 0.9.54.6)(10‘) : 12.577 taps Liw'J-i‘nj waive 0‘? P Is ‘I'hc .srnatﬂ’er-VaJue -'- P=12.\$7 Rip: 4 / 4 / ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online