EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 9780100257061
Author: BEER
Publisher: YUZU
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Chapter 1, Problem 63RP
A couple M of magnitude 1500 N • m is applied to the crank of an engine. For the position shown, determine (a) the force P required to hold the engine system in equilibrium, (b) the average normal stress in the connecting rod BC, which has a 450-mm2 uniform cross section.
Fig. P1.63
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Procedure:1- Cartesian system, 2D3D,type of support2- Free body diagram3 - Find the support reactions4- If you find a negativenumber then flip the force5- Find the internal force3D∑Fx=0∑Fy=0∑Fz=0∑Mx=0∑My=0\Sigma Mz=02D\Sigma Fx=0\Sigma Fy=0\Sigma Mz=05- Use method of sectionand cut the elementwhere you want to find
Procedure:1- Cartesian system, 2D3D,type of support2- Free body diagram3 - Find the support reactions4- If you find a negativenumber then flip the force5- Find the internal force3D∑Fx=0∑Fy=0∑Fz=0∑Mx=0∑My=0\Sigma Mz=02D\Sigma Fx=0\Sigma Fy=0\Sigma Mz=05- Use method of sectionand cut the elementwhere you want to findthe internal force andkeep either side of the
Procedure:
1- Cartesian system, 2D3D,
type of support
2- Free body diagram
3 - Find the support reactions
4- If you find a negative
number then flip the force
5- Find the internal force
3D
∑Fx=0
∑Fy=0
∑Fz=0
∑Mx=0
∑My=0
ΣMz=0
2D
ΣFx=0
ΣFy=0
ΣMz=0
5- Use method of section
and cut the element
where you want to find
the internal force and
keep either side of the
Chapter 1 Solutions
EBK MECHANICS OF MATERIALS
Ch. 1.2 - Two solid cylindrical rods AB and BC are welded...Ch. 1.2 - Two solid cylindrical rods AB and BC are welded...Ch. 1.2 - Two solid cylindrical rods AB and BC are welded...Ch. 1.2 - Two solid cylindrical rods AB and BC are welded...Ch. 1.2 - A strain gage located at C on the surface of bone...Ch. 1.2 - Two brass rods AB and BC, each of uniform...Ch. 1.2 - Each of the four vertical links has an 8 36-mm...Ch. 1.2 - Link AC has a uniform rectangular cross section 18...Ch. 1.2 - Three forces, each of magnitude P = 4 kN, are...Ch. 1.2 - Link BD consists of a single bar 1 in. wide and 12...
Ch. 1.2 - For the Pratt bridge truss and loading shown,...Ch. 1.2 - The frame shown consists of four wooden members,...Ch. 1.2 - An aircraft tow bar is positioned by means of a...Ch. 1.2 - Two hydraulic cylinders are used to control the...Ch. 1.2 - Determine the diameter of the largest circular...Ch. 1.2 - Two wooden planks, each 12 in. thick and 9 in....Ch. 1.2 - When the force P reached 1600 lb, the wooden...Ch. 1.2 - A load P is applied to a steel rod supported as...Ch. 1.2 - The axial force in the column supporting the...Ch. 1.2 - Three wooden planks are fastened together by a...Ch. 1.2 - A 40-kN axial load is applied to a short wooden...Ch. 1.2 - An axial load P is supported by a short W8 40...Ch. 1.2 - Link AB, of width b = 2 in. and thickness t=14...Ch. 1.2 - Determine the largest load P that can be applied...Ch. 1.2 - Knowing that = 40 and P = 9 kN, determine (a) the...Ch. 1.2 - The hydraulic cylinder CF, which partially...Ch. 1.2 - For the assembly and loading of Prob. 1.7,...Ch. 1.2 - Two identical linkage-and-hydraulic-cylinder...Ch. 1.5 - Two wooden members of uniform rectangular cross...Ch. 1.5 - Two wooden members of uniform rectangular cross...Ch. 1.5 - The 1.4-kip load P is supported by two wooden...Ch. 1.5 - Two wooden members of uniform cross section are...Ch. 1.5 - A centric load P is applied to the granite block...Ch. 1.5 - A 240-kip load P is applied to the granite block...Ch. 1.5 - A steel pipe of 400-mm outer diameter is...Ch. 1.5 - A steel pipe of 400-mm outer diameter is...Ch. 1.5 - A steel loop ABCD of length 5 ft and of 38-in....Ch. 1.5 - Link BC is 6 mm thick, has a width w = 25 mm, and...Ch. 1.5 - Link BC is 6 mm thick and is made of a steel with...Ch. 1.5 - Members AB and BC of the truss shown are made of...Ch. 1.5 - Members AB and BC of the truss shown are made of...Ch. 1.5 - Link AB is to be made of a steel for which the...Ch. 1.5 - Two wooden members are joined by plywood splice...Ch. 1.5 - For the joint and loading of Prob. 1.43, determine...Ch. 1.5 - Three 34-in.-diameter steel bolts are to be used...Ch. 1.5 - Three steel bolts are to be used to attach the...Ch. 1.5 - A load P is supported as shown by a steel pin that...Ch. 1.5 - A load P is supported as shown by a steel pin that...Ch. 1.5 - A steel plate 14 in. thick is embedded in a...Ch. 1.5 - Determine the factor of safety for the cable...Ch. 1.5 - Link AC is made of a steel with a 65-ksi ultimate...Ch. 1.5 - Solve Prob. 1.51, assuming that the structure has...Ch. 1.5 - Each of the two vertical links CF connecting the...Ch. 1.5 - Solve Prob. 1.53, assuming that the pins at C and...Ch. 1.5 - In the structure shown, an 8-mm-diameter pin is...Ch. 1.5 - In an alternative design for the structure of...Ch. 1.5 - Prob. 57PCh. 1.5 - The Load and Resistance Factor Design method is to...Ch. 1 - In the marine crane shown, link CD is known to...Ch. 1 - Two horizontal 5-kip forces are applied to pin B...Ch. 1 - For the assembly and loading of Prob. 1.60,...Ch. 1 - Two steel plates are to be held together by means...Ch. 1 - A couple M of magnitude 1500 N m is applied to...Ch. 1 - Knowing that link DE is 18 in. thick and 1 in....Ch. 1 - A 58-in.-diameter steel rod AB is fitted to a...Ch. 1 - In the steel structure shown, a 6-mm-diameter pin...Ch. 1 - Prob. 67RPCh. 1 - A force P is applied as shown to a steel...Ch. 1 - The two portions of member AB are glued together...Ch. 1 - The two portions of member AB are glued together...
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- Procedure:1- Cartesian system, 2D3D,type of support2- Free body diagram3 - Find the support reactions4- If you find a negativenumber then flip the force5- Find the internal force3D∑Fx=0∑Fy=0∑Fz=0∑Mx=0∑My=0\Sigma Mz=02D\Sigma Fx=0\Sigma Fy=0\Sigma Mz=05- Use method of sectionand cut the elementwhere you want to findthe internal force andkeep either side of thearrow_forwardProcedure: 1- Cartesian system, 2(D)/(3)D, type of support 2- Free body diagram 3 - Find the support reactions 4- If you find a negative number then flip the force 5- Find the internal force 3D \sum Fx=0 \sum Fy=0 \sum Fz=0 \sum Mx=0 \sum My=0 \Sigma Mz=0 2D \Sigma Fx=0 \Sigma Fy=0 \Sigma Mz=0 5- Use method of section and cut the element where you want to find the internal force and keep either side of the sectionarrow_forwardProcedure: 1- Cartesian system, 2(D)/(3)D, type of support 2- Free body diagram 3 - Find the support reactions 4- If you find a negative number then flip the force 5- Find the internal force 3D \sum Fx=0 \sum Fy=0 \sum Fz=0 \sum Mx=0 \sum My=0 \Sigma Mz=0 2D \Sigma Fx=0 \Sigma Fy=0 \Sigma Mz=0 5- Use method of section and cut the element where you want to find the internal force and keep either side of the sectionarrow_forward
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