Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Strength of Materials Exam with Solution | ENGR 2530, Exams of Engineering

Material Type: Exam; Professor: Shephard; Class: STRENGTH OF MATERIALS; Subject: Core Engineering; University: Rensselaer Polytechnic Institute; Term: Spring 2014;

Typology: Exams

2013/2014

Uploaded on 05/12/2014

tahsinl-95
tahsinl-95 🇺🇸

6 documents

1 / 10

Toggle sidebar

Related documents


Partial preview of the text

Download Strength of Materials Exam with Solution | ENGR 2530 and more Exams Engineering in PDF only on Docsity! ENGR-2530 Strength of Materials Exam 2 RIN:------------ Section:------- Problem 1 (10 pts) Problem 2 (10 pts) Problem 3 (20 pts) Problem 4 (20 pts) Problem 5 (20 pts) Problem 6 (20 pts) Total Spring 2014 ENGR-2530 Strength of Materials Spring 2014 1- A cantilever beam is subjected to a point load P in the way shown in the figure below. Assuming that the length of AB is half of AC, considering the section of the beam at the support, determine the angle a that would create zero stresses at point A and Grsinretenesualy. Pp A B / YP aA Cc D (ope gigas Ty = te (culo? ha Te ty Sy 2 Ip 4 laa) 708 Mz on fZ Sin & M, = PA Cosa tL Sint) , PL Cos (%) _ Oo. = Te a¥ Tat i. Sinale) .. PL Cosl®) ras ATG ¥, ot nN ENGR-2530 Strength of Materials Confiauracs on Ais epreHaL } eel per each Q the. tuoe Neils < 10> b Spring 2014 Feacwo) = ¥> 2D 200 = 134.35 Ss => Se 1.44in ENGR-2530 Strength of Materials Spring 2014 4- An axial load P of magnitude 30kN is applied as shown, to a short section of a steel channel. Determine the largest distance a for which the maximum compressive stress is 60 MPa, (The load can only move on the dashed line, shown in the figure on the right) P= 30kN 150 mm} n ? X > 0x5 PERAXUSXSKATS 2 1.975 mm 1BO x54 2ZxXUGKS Mf = NTS HE cianot> AEF 25 Am My = — (30x00 (9.375 x10°%) = 297. 4 = My a p ; 4a. =-7Smm 7 Xn = X= .¢7S &, ae y, x5 x Ob 2x Ve AGS XSF LXUSH SA F2.9 a Fle ; : P 2 = 344x107 mm : Bs es t y x50 x5 +2X U2 <5 XYUS?-p 1SO = HL. 2 s : $Rxysxs x AIS- W.g7s ) = 0.955% 10% mnt My Ia _F _ Mg 1A pie Om * A ee. 44 a ENGR-2530 Strength of Materials 7 My etx a —O, a) 3 3 My. . BFF x10 ° /-A8)-RSXN-RTS AD” so sia) -F5 x 0°? 0.253 x10-© 1200.Xi0 Spring 2014 ~ & 5 2 B44 A10* (13 02-25 +66)x10 =f. 09S X10-mm =45 «197° Gu es - (-1.095 x10) | 0: 0365 =3b.5mm = 30x jo”
Docsity logo



Copyright © 2024 Ladybird Srl - Via Leonardo da Vinci 16, 10126, Torino, Italy - VAT 10816460017 - All rights reserved