Solve the Modulus of RuptureGiven:Span length of the beam(L) = 18 inchesWidth of the beam(b) = 6 inchesDepth of the beam(d) = 6 inchesMaximum load applied(P) = 9950 lb cttpCalculate modulus of rupture based on the location of fracture in the tension faceOutside MiddleOutside MiddleMiddle ThirdThird (s 5 %)Third (> 5 %)PLR =bd?3PaRbd²Discarda = average distance between lineof fracture and nearest supportR = modulus of ruptureP = maximum loadb:averagewidth (0.05")L = span lengthd = average depth (0.05")%3D4:36 / 5:00 • Remove beam >

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Answered: Solve the Modulus of Rupture Given:…

Solve the Modulus of Rupture Given: Span length of the beam(L) = 18 inches Width of the beam(b) = 6 inches Depth of the beam(d) = 6 inches Maximum load applied(P) = 9950 lb

Materials Science And Engineering Properties

1st Edition

ISBN:9781111988609

Author:Charles Gilmore

Publisher:Charles Gilmore

Chapter11: Fracture And Fatigue

Section: Chapter Questions

Problem 11.5P

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Question

Solve the Modulus of Rupture

Given:

Span length of the beam(L) = 18 inches

Width of the beam(b) = 6 inches

Depth of the beam(d) = 6 inches

Maximum load applied(P) = 9950 lb

cttp
Calculate modulus of rupture based on the location of fracture in the tension face
Outside Middle
Outside Middle
Middle Third
Third (s 5 %)
Third (> 5 %)
PL
R =
bd?
3Pa
R
bd²
Discard
a = average distance between line
of fracture and nearest support
R = modulus of rupture
P = maximum load
b:
average
width (0.05")
L = span length
d = average depth (0.05")
%3D
4:36 / 5:00 • Remove beam >

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