A cantilever beam supports the loads shown. The cross-sectional dimensions of the shape are also shown. Assume a = 0.4 m, PA = 2.0 kN, PB = 6.0 kN, PC = 3.0 kN, d = 85 mm, bf = 105 mm, tf = 5 mm, tw = 9 mm. Determine(a) the maximum vertical shear stress.(b) the maximum compression bending stress.(c) the maximum tension bending stress.See the coordinate system for the beam in the problem figure with the origin of the x axis at the fixed support. Consider four points along the beam’s axis:Point A at x = 1.2 mPoint B at x = 0.8 mPoint C at x = 0.4 mPoint D at x = 0Break the beam into three segments: AB, BC, and CD. Enter the shear force in each segment with its correct sign based on the coordinate system in the problem figure and the sign convention for shear forces in Chapter 7.Answers:VAB = kNVBC = kNVCD = kNPart 2Enter the maximum shear force magnitude in the beam. Since this is a magnitude, enter a positive value.Answer: Vmax = kN Part 3Enter the bending moment at four critical x-locations:MD is the moment at x = 0MC is the moment at x = 0.4 mMB is the moment at x = 0.8 mMA is the moment at x = 1.2 mEnter each bending moment with its correct sign based on the coordinate system in the problem figure and the sign convention for bending moments presented in Chapter 7. Answers:MA = kN-mMB = kN-mMC = kN-mMD = kN-m Part 4Enter the maximum positive and negative bending moments for the beam. The “maximum negative bending moment” is the negative bending moment with the largest magnitude. Enter the maximum negative moment as a negative value here.Answers:Mmax, + = kN-mMmax, - = kN-mPart 5Break the cross-sectional area into two areas:(1) Top horizontal flange with rectangular cross-section 105 mm × 5 mm.(2) Vertical stem with rectangular cross-section 9 mm × 80 mm.Find the areas and the centroid locations in the y-direction for each area. Enter the centroid locations, y1 and y2, as measured with respect to a reference axis at the bottom of the cross-section. In other words, let y = 0 at the bottom edge of the vertical stem.Answers:A1 = mm2,y1 = mmA2 = mm2,y2 = mmPart 6Determine the centroid location in the y direction for the entire beam cross-section with respect to the reference axis at the bottom of the cross-section.Answer: y¯ = mmPart 7Consider again the cross-section broken into two areas, as in Step 1:(1) Top horizontal flange with rectangular cross-section 105 mm × 5 mm.(1) Top horizontal flange with rectangular cross-section 9 mm × 80 mm.Determine the moment of inertia for each area about its own centroid and the distance from the centroid of each area to the centroid of the entire cross-section. For example, Ic1 is the moment of inertia for area (1) about its own centroid, and d1 is the distance from the centroid of area (1) to the centroid of the entire cross-section calculated in Step 6. Enter the distances d1 and d2 as positive values here, regardless of whether the centroid for an area is above or below the centroid of the entire cross-section.Answers:Ic1 = (103) mm4,d1 = mmIc2 = (103) mm4,d2 = mm bfPAadaPca

Answered: wn. Assume a = 0.4 m, PA = 2.0 kN, PB =…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Part 2

Enter the maximum shear force magnitude in the beam. Since this is a magnitude, enter a positive value.

Answer: V_max = kN

Part 3 Enter the bending moment at four critical x-locations:

M_D is the moment at x = 0
M_C is the moment at x = 0.4 m
M_B is the moment at x = 0.8 m
M_A is the moment at x = 1.2 m
Enter each bending moment with its correct sign based on the coordinate system in the problem figure and the sign convention for bending moments presented in Chapter 7.

Answers:

M_A =	kN-m
M_B =	kN-m
M_C =	kN-m
M_D =	kN-m

Part 4

Enter the maximum positive and negative bending moments for the beam. The “maximum negative bending moment” is the negative bending moment with the largest magnitude. Enter the maximum negative moment as a negative value here.

Answers:

M_{max, +} =	kN-m
M_{max, -} =	kN-m

Part 5

Break the cross-sectional area into two areas:

(1) Top horizontal flange with rectangular cross-section 105 mm × 5 mm.
(2) Vertical stem with rectangular cross-section 9 mm × 80 mm.

Find the areas and the centroid locations in the y-direction for each area. Enter the centroid locations, y₁ and y₂, as measured with respect to a reference axis at the bottom of the cross-section. In other words, let y = 0 at the bottom edge of the vertical stem.

Answers:

A₁ =	mm²,	y₁ =	mm
A₂ =	mm²,	y₂ =	mm

Part 6

Determine the centroid location in the y direction for the entire beam cross-section with respect to the reference axis at the bottom of the cross-section.

Answer: y¯ = mm

Part 7

Consider again the cross-section broken into two areas, as in Step 1:

(1) Top horizontal flange with rectangular cross-section 105 mm × 5 mm.
(1) Top horizontal flange with rectangular cross-section 9 mm × 80 mm.

Determine the moment of inertia for each area about its own centroid and the distance from the centroid of each area to the centroid of the entire cross-section. For example, I_c₁ is the moment of inertia for area (1) about its own centroid, and d₁ is the distance from the centroid of area (1) to the centroid of the entire cross-section calculated in Step 6. Enter the distances d₁ and d₂ as positive values here, regardless of whether the centroid for an area is above or below the centroid of the entire cross-section.

Answers:

I_c₁ =	(10³) mm⁴,	d₁ =	mm
I_c₂ =	(10³) mm⁴,	d₂ =	mm

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V_AB =	kN
V_BC =	kN
V_CD =	kN

wn. Assume a = 0.4 m, PA = 2.0 kN, PB = 6.0 kN, PC = 3.0 kN, d = 85 mm, bf = 105 mm, tf = 5 mm, tw = 9 mm. Determine (a) the maximum vertical shear stress. (b) the maximum compression bending stress. (c) the maximum te

Part 2

Part 3

Enter the bending moment at four critical x-locations:

Part 4

Part 5

Part 6

Part 7