Quantcast Solution - dm2_080294

Share on Google+Share on FacebookShare on LinkedInShare on TwitterShare on DiggShare on Stumble Upon
Custom Search
 
  
 


Solution:
(1) Assume a plate thickness of 2-1/2 inches.
(2) Determine the elastic and plastic section moduli (per unit
width).
S = bd2/6
= (1 x 2.52/6
= 1.04 in3/in
Z = bd24
= (1 x 2.52)/4
= 1.56 in3/in
(3) Calculate the design plastic moment, Mp, of the plate.
Mp = Fdy(S + Z)/2
= 46.2(1.04 + 1.56)/2
= 60.06 K-in/in
(4) Calculate the flexural rigidity of the plate.
D = Et3/12(1 - [nu]2)
= (29 x 106 x 2.53)/12(1 - 0.32)
= 41.5 x 106 lb-in
(5) Calculate the elastic stiffness of the plate, KE.
From Figure 5-19 of NAVFAC P-397:
XD = [gamma]rH4
KE = r/X = D/[gamma]H4
For H/L = 0.58, [gamma]= 0.0089
Therefore,
KE = (41.5 x 106)/(0.0089 x 504)
= 746 psi/in
(6) Calculate the ultimate unit resistance of the plate.
From Figure 5-11 of NAVFAC P-397:
For L/H = 86/50 = 1.72, x/L = 0.355
Therefore,
x  = 0.355(86)
= 30.53 inches.
From Table 5-6 of NAVFAC P-397, ultimate unit resistance:
ru = 5(MHN + MHP)/x2
MHN = 0; MHP = Mp
ru = 5(60,060)/30.532
= 322.2 psi
2.08-277





 


Privacy Statement - Copyright Information. - Contact Us

Integrated Publishing, Inc.