The 1.4-kip load P is supported by two wooden members of uniform cross section t

Question

The 1.4-kip load P is supported by two wooden members of uniform cross section that are joined by the simple glued scarf splice shown. Determine the normal and shearing stresses in the glued splice. Fig. P1.31 and P1.32 Rod AB is made of brass (Eb = 15E6 psi, alpha b = 11.6 E-6/degree F) and rod CD of aluminium (Ea = 10.1 E6 psi, alpha a = 13.1 E-6/degree F). Knowing that at 60degree F a 0.02" gap exists between the ends of the two rods, determine a) the normal stress in each rod after the temp has been raised to 180degree F, b) the deformation of rod AB at that time. A 4' concrete post reinforced by four steel bars, each of 3/4"Phi. Given; alphas = 6.5E-6/degree F Es = 29E6 psi alpha c = 5.5E-6/degree F Ec = 3.6E6 psi Determine the normal stress induced in the concrete (steel by a temperature rise of 80degreeF, Allow materials to deform freely due to DeltaT deltas = delta C deltaTS - delta PS = delta TC + delta PC The rigid bar AD is supported as shown by two steel wires of 1/16"Phi (E = 29E6 psi) and a pin and bracket at D. Knowing that the wires were initially tant, determine; the tension in each wire when a 120# load P is applied at B the deflection @ pt B A 250mm bar of 15Times30mm rectangular cross section consists of 2 aluminium layers, of the same thickness. If it subjected to centric forces of magnitude p = 30KN, and knowing that Ea = 70GPa and Eb = 105GPa determine the normal stress, in the aluminum layers in the brass layer

Explanation / Answer

P=1.4 kip=6227.51024 N

A=15 in2 = 9.6774*10-3 m2

area of splice=A'=A/sin60 =0.0111 m2

normal stress=Psin60/A' = 482.63 kPa

shear stress=Pcos60/A' = 278.64 kPa

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.