E 150 mm 50 DQ = 30 kN 6 $2 80 Q= 20 kN PP. 6 50 A y 30 6 $1 B 120 N
E 150 mm 50 DQ = 30 kN 6 $2 80 Q= 20 kN PP. 6 50 A y 30 6 $1 B 120 N
Refrigeration and Air Conditioning Technology (MindTap Course List)
8th Edition
ISBN:9781305578296
Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Publisher:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Chapter1: Heat, Temperature, And Pressure
Section: Chapter Questions
Problem 17RQ: Convert 22C to Fahrenheit.
Related questions
Question
Determine the shear flow qqq for the given profile when the shear forces acting at the torsional center are Qy=30Q_y = 30Qy=30 kN and Qz=20Q_z = 20Qz=20 kN. Also, calculate qmaxq_{\max}qmax and τmax\tau_{\max}τmax.
Given:
Iy=10.5×106I_y = 10.5 \times 10^6Iy=10.5×106 mm4^44,
Iz=20.8×106I_z = 20.8 \times 10^6Iz=20.8×106 mm4^44,
Iyz=6×106I_{yz} = 6 \times 10^6Iyz=6×106 mm4^44.
Additional parameters:
αy=0.5714\alpha_y = 0.5714αy=0.5714,
αz=0.2885\alpha_z = 0.2885αz=0.2885,
γ=1.1974\gamma = 1.1974γ=1.1974.
(Check hint: τmax\tau_{\max}τmax should be approximately 30 MPa.)
Transcribed Image Text:E
150 mm
50
DQ = 30 kN
6
$2
80
Q= 20 kN
PP.
6
50
A
y
30
6
$1
B
120
N
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