applied to the beam cross section. Calculate (a) the tension normal stress in the wood at point K. (b) the compression normal stress in the steel at point H. Steel (2) Wood (1) H K N.A.

The wood column is ixed at its base and free at its top. Determine the load P that can be applied to the edge of the column without causing the column to fail either by buckling or by yielding. Ew = 12 GPa, sY = 55 MPa.

6:53 PM M ..39 and 300 mm long solid copper bar of 20 mm diameter (See Figure 1). The composite bar is heated 35°C above room temperature, clamped at both ends and then allowed to cool to room temperature. If the distance between the clamps is unchanged find the stress in each bar. For steel, use a = 12.5 x 100% E = 200 GPa and for Aluminum use a = 17 x × 10**; E = 130 GPa 500 m Figure Q1 Question 2 A narrow steel strip, 15 mm thick, is clad by two magnesium plates, one is 2.5 mm thick and the other is 3.5 mm thick as show in Figure 2. Calculate the change in stress in each strip for a centigrade degree rise in temperature of the compound strip. Assume perfect bonding of the strips along their length. For steel, a = 12 x 10°/°C; E = 200 GN/m². For magnesium, a = 27 x 10°/°C; E = 45 GN/m² 3.5mm 15 mm 2.5 mm Figure Q2 Question 3 A horizontal beam AD, 10 m long, carries a uniformly distributed load of 25kN/m, together with a concentrated load of 36 kN at the left-hand end A (see Figure…

The steel channel is used to reinforce the wood beam. Determine the maximum stress in the steel and the wood if the beam is subjected to a moment of M=1.2kN.m Est = 200GPa, Ew = 12GPa. 12mm A 3.'S. IO. **** SI TEMININING FEMININ... *** 12mm INI MINI SEUR 5.1.3. TE 2277 *** ....... ---- 175mm 100mm 12mm 1.2kN.m

How to solve for the centroid of the composite body using the principle of moments and treat each part as a finite element of the whole figure (Show a step by step solution)

The rectangular doubly reinforcement stress concrete block with the arrangement of reinforcement of 2N28 bars on top and 3N28 bars on bottom. The modulus of elasticity are Ec =23,500MPa and Es = 200,000MPa. Dead load is 18KN/m and live load is 12KN/m. f'c = 25MPa 800 730 N12 ligs (fx.t-500MPa) Ast = 3N28 350 Span = 10m Figure 3 Use the Transformed section approach considering the presence of steel. Calculate the concrete area and transformed steel area in both tension and compression shown in Figure 3. n=Es/Ec Ast=1232mm^2 Use n-1 for the transformed area to do the calculate. Unit: mm^2 with two decimal

The column above is subjected to axial loading only. Assume the following: • E= 29000 ksi G = 10000 ksi • = 986 in4 A = 60 in? • L= 20 ft • Grade 60 steel (60 ksi yield stress) • The column is fixed at the top and pinned at the bottom Which of the following is closest to the maximum normal stress the column can support (before yielding or buckling)? 167 ksi 24 ksi 42 ksi 60 ksi

The column is built up by gluing the two boards together. If the wood has an allowable normal stress of sallow = 6 MPa, determine the maximum allowable eccentric force P that can be applied to the column.

What is the manufacturing tensional yield strength (lbf) of grade S135 drill pipe? a. This is a trick question because drill pipe size is not provided b. The answer can’t be calculated because the length of the pipe is not given c. The yield strength is 135000 psi and is a property of the steel d. The yield strength is 135000 psi and depends on the cross-sectional area

2- The Tensile stress of AISI 1035 Cold-drawn steel is; (a) 550 MPa. (b) 500 MPa. (c) 450 MPa.