Question

Problem 4: The frame with cross-section given supports the distributed load as shown. Answer the following: A. Support reactions at A and the force in member BC B. Find the Shear and moment equations for AD section [V(x) and M(x)] and plot the shear and moment diagrams. C. State of stress at D D. State of stress at E 4 kN/m D 20 mm 60 mm 20 mm BA E 5 m 50 mm |-1.5 m -1.5 m --- 3 m 3 m

Answer 1

Solution:-

A) Reaction at A and Force in BC:-

4 KAL RE TRCI sino. Folg 3 5 Rea Moment about C RAXIO = (12) x 6X4X7 RA = 3.4 KM & Rai Focx sino FB x ( 3/5) = (1)x 5 xu - VA FBE =6 KN

B) Shear and moment equation from A to D is given by

C) State of stress at point D ( X=3 ) :-

Shear stress is given by:-

$\tau =\frac{V(X)*A*\bar{Y}}{I_{NA}*b}$

V(3)=2.4 KN , A=50*20 mm² , Y = 40 mm and I = 4166666.66 mm⁴

$\tau =460.7996 KPa$

Bending moment at point D :-

$\sigma _{b}=\frac{Y*M(X)}{I_{NA}}$

M(3)=19.2 KN-m and Y = 30 mm

$\sigma _{b}=-138.24 MPa$ ( ( - )ve as the upper fibers are under compression )

D) State of stress at point E ( X=3 ) :-

V(1.5)=5.4 KN , A=50*20 mm² , Y = -40 mm and I = 4166666.66 mm⁴

$\tau =1036.8 KPa$

Bending moment at point E :-

M(3)=11.1 KN-m and Y = 30 mm

$\sigma _{b}=79.92 MPa$ ( Bottom fibers are under tension )

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